Complex Analysis (1)

Prévia do material em texto

�
�
“book” — 2014/6/4 — 17:28 — page i — #1
�
�
�
�
�
�
Complex Analysis
This page is intentionally left blank
�
�
“book” — 2014/6/4 — 17:28 — page iii — #3
�
�
�
�
�
�
Complex Analysis
R. Roopkumar
Department of Mathematics
Alagappa University
Karaikudi
Tamilnadu
Chennai • Delhi
�
�
“book” — 2014/11/19 — 13:37 — page iv — #4
�
�
�
�
�
�
Copyright © 2015 Pearson India Education Services Pvt. Ltd
Published by Pearson India Education Services Pvt. Ltd, CIN:
U72200TN2005PTC057128, formerly known as TutorVista Global Pvt. Ltd,
licensee of Pearson Education in South Asia.
No part of this eBook may be used or reproduced in any manner whatsoever
without the publisher’s prior written consent.
This eBook may or may not include all assets that were part of the print
version. The publisher reserves the right to remove any material in this
eBook at any time.
ISBN 978-93-325-3761-3
eISBN 978-93-325-4159-7
Head Office: A-8 (A), 7th Floor, Knowledge Boulevard, Sector 62, Noida
201 309, Uttar Pradesh, India.
Registered Office: Module G4, Ground Floor, Elnet Software City, TS-140,
Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu,
India.
Fax: 080-30461003, Phone: 080-30461060
www.pearson.co.in, Email: companysecretary.india@pearson.com
http://www.pearson.co.in
mailto:companysecretary.india@pearson.com
�
�
“book” — 2014/6/4 — 17:28 — page v — #5
�
�
�
�
�
�
Contents
Preface vii
1. Preliminaries 1
1.1 Introduction and Brief Prerequisites 1
1.2 Complex Numbers and Geometrical Representations 9
1.3 Sequences and Series of Complex Numbers 24
1.4 Some Topological Properties of the Complex Plane 36
1.5 Extended Complex Numbers and Stereographic
Projection 52
1.6 Limit and Continuity 55
2. Analytic Functions 71
2.1 Differentiability 71
2.2 Cauchy–Riemann Equations 79
2.3 Power Series and Abel’s Theorems 94
2.4 Exponential and Trigonometric Functions 102
2.5 Hyperbolic Functions 119
3. Rational Functions and Multivalued Functions 123
3.1 Polynomials and Rational Functions 123
3.2 Linear Fractional Transforms 132
3.3 Branch of a Multivalued Function 148
3.4 Conformal Mapping 156
3.5 Elementary Riemann Surface 161
4. Complex Integration 165
4.1 Line Integral 165
4.2 Winding Number and Cauchy’s Theorems 182
4.3 Cauchy’s Integral Formula 193
�
�
“book” — 2014/6/4 — 17:28 — page vi — #6
�
�
�
�
�
�
vi Contents
4.4 General Version of Cauchy’s Theorem 209
4.5 Local Correspondence Theorem and Its Consequences 212
5. Series Developments and Infinite Products 223
5.1 Taylor Series and Laurent Series 223
5.2 Zeroes, Poles, and Singularities 244
5.3 Partial Fraction of Entire Functions 257
5.4 Infinite Product 264
5.5 Gamma Function and Its Properties 278
6. Residue Calculus 287
6.1 Residue 287
6.2 Cauchy’s Residue Theorem 294
6.3 Argument Principle and Rouche’s Theorem 306
6.4 Evaluation of Real Integrals 314
6.5 Integrals of Multivalued Functions 326
7. Some Interesting Theorems 341
7.1 Mean Value Property of Harmonic Functions 341
7.2 Poisson’s Integral 354
7.3 Schwarz Reflection Principle 363
7.4 Riemann Mapping Theorem 368
7.5 Schwarz–Christoffel Formula 373
8. Elliptic Functions 391
8.1 Basic Concepts 391
8.2 Fundamental Parallelogram 398
8.3 Weierstrass ℘a,b Function 406
8.4 The Functions ζa,b and σa,b 423
8.5 Jacobi’s Elliptic Functions snk , cnk and dnk 432
Bibliography 455
Index 457
�
�
“book” — 2014/6/4 — 17:28 — page vii — #7
�
�
�
�
�
�
Preface
There are many good text books on complex analysis, some of which are
listed in the bibliography. In my opinion, most of the books are written with
the assumption that the reader can understand the intricacies of the proofs
by filling the gaps in the arguments. However, based on my personal experi-
ence of offering several courses on complex analysis to the present generation
of students, I sensed the necessity to write a book on complex analysis by
explaining each and every argument in any proof in a lucid manner so that the
book would be an ideal self study material for the students. The present book
has been written to address this need. Since many concepts in complex analy-
sis are geometrical in nature, more geometrical arguments are given, without
any compromise in rigor. While the detailed proofs presented in the book
may appear to be self-evident to the experts in complex analysis, beginner
students who try to learn the subject with rigor and without any assumptions
will find such treatment helpful. At the same time, this book may also be used
as a hand book by a young teacher who needs explanations for some tedious
theorems in complex analysis.
This text book is intended for both under graduate and post graduate
courses in complex analysis. The first chapter consists of the basic concepts
of complex numbers, operations on the complex numbers, topological prop-
erties and limiting concepts with more details. All the results given in this
chapter are used at least once in the later part of this book. For an under-
graduate course, the chapters 2,3,4,5,6 may be prescribed by omitting big
theorems. For a postgraduate course, chapters 1 to 6 may be prescribed as
a first course on complex analysis. For those who have the desired level of
exposure to complex analysis in an under graduate course, directly chapters
4,5,6,7,8 may be prescribed in the post graduate course. In the last chapter, the
properties of Jacobian elliptic functions are having very long proofs. Indeed,
these proofs have been presented to motivate the young students and urge
them persevere instead of being bewildered by the subtle nuances and com-
plications involved. However, my personal opinion is that it is not advisable
to ask the proofs of those results in any examination.
�
�
“book” — 2014/6/4 — 17:28 — page viii — #8
�
�
�
�
�
�
viii Preface
The content of the book is subdivided into 8 chapters with two objectives.
The first objective is to club similar topics under a chapter title and the second
one is to maintain the size of the chapters as uniform as possible. I have made
sure that each chapter has a sufficient number of results and problems of
varied lengths, so that it becomes suitable for setting questions of all types.
For doing this, I realized that I either had to compromise on at least one of
the above said objectives or I had to use two results from the later chapters in
earlier chapters. I have chosen the latter option. However, I have confirmed
that there is no begging of question by using a result before it is proved.
There are some concepts such as the Jordan curve Theorem, Hardamard’s
Theorem regarding the estimate of genus of an infinite product, introduction
to univalent and star-like functions that are not discussed here, since these
are to be taken up at an advanced level beyond the scope of this book. I have
taken stringent efforts to make this book error-free. Nonetheless, if any error
is found, please write to me so that they may be eliminated in future editions.
My e-mail address is roopkumar r@yahoo.co.in
I am grateful to Dr R. Vembu for his constant encouragement, fruitful dis-
cussions and valuable suggestions while preparing the content for this book.
My thanks are also due to the editors at Pearson Education who worked on
this book, for their professionalism and diligence.
R. Roopkumar
mailto:roopkumar_r@yahoo.co.in
�
�
“book” — 2014/6/4 — 17:28 — page ix — #9
�
�
�
�
�
�
Complex Analysis
This page is intentionally left blank
�
�
“book” — 2014/6/4 — 20:35 — page 1 — #1
�
�
�
�
�
�
1
Preliminaries
1.1 INTRODUCTION AND BRIEF PREREQUISITES
Every mathematical structure has some good properties as well as some
deficiencies. To overcome a particular deficiency on a mathematical structure,
it is, therefore, customary to generalize the mathematical structure. How-
ever, it should be noted that any generalized mathematical structure having a
good property that is not in the earlier structure will not satisfy another good
propertyof an earlier structure. Consider the following examples:
1. Z is an ordered abelian group, whereas N is an ordered commutative
semi-group but not a group. However, every non-empty subset of N has
a minimum in N, which is not true in Z.
2. Q is an ordered field, whereas Z is an ordered integral domain but not
a field. However, for every element of Z, we can find previous element
and next element in Z, which is not true in Q.
3. R is an ordered field with least upper bound property, whereas Q is not
having least upper bound property. However, Q is countable, whereas
R is not countable.
The set R of all real numbers is a very good setup in which we have lot of
mathematical structures such as Archimedian field with least upper bound
property, complete metric space having Heine–Borel property and Banach
space (a complete normed linear space), and so on. In fact, it is identi-
fied with the set of all points on a straight line. However, in the algebraic
point of view, not all polynomials of degree n (for some n ∈ N) over R
have n roots in R. To be specific, x2 + 1 is a polynomial of degree 2 over
R, which has no solution in R. This is the motivation for introducing the
complex number system in which every polynomial of degree n over C has
exactly n roots in C. In fact, C is the splitting field of x2 + 1, considering
1
�
�
“book” — 2014/6/4 — 20:35 — page 2 — #2
�
�
�
�
�
�
2 Introduction and Brief Prerequisites
x2 + 1 as a polynomial over R. As in the earlier generalization of number
systems (just listed above), R is an ordered field, whereas there is no order
relation on C, which makes it as an ordered field. However, complex anal-
ysis has more good and different properties that cannot be expected in real
analysis.
As R and C are metric spaces, we can say all the topological proper-
ties that are common in a metric space such as the properties of open sets,
closed sets, limit point of a set, convergent sequences and series, limit of a
function, and continuous functions etc are similar in R and C. However, it is
interesting to note some of the good differences between real and complex
analysis.
1. If x → x0 in R, then x can approach only in two directions along the
real axis. However, if z → z0 in C, then z gets closer to z0 through
uncountable number of paths. See the following diagrams for some
examples.
x0
x0
x
z0
z
z0
z
x
z
z
z0
�
�
“book” — 2014/6/4 — 20:35 — page 3 — #3
�
�
�
�
�
�
Preliminaries 3
2. We know that interior of A is connected for every connected subset A
of R. However, there is a connected subset A of C whose interior is not
connected.
3. If f = (u, v) : C → C is differentiable, then for the given u, we can
find v and for the given v, we can find u (uniquely up to a constant).
However, if f = ( f1, f2) : R → R × R is differentiable, then neither f2
can be found from f1 nor f1 can be found from f2.
4. If f : � → C is (once) differentiable (where � is an open connected
subset of C), then f is infinitely many times differentiable. However, if
f : I → R (where I is an interval) is differentiable, then f ′ need not be
even continuous.
5. If f : C → C is a bounded differentiable function, then f is a con-
stant function. However, it is not true for a bounded differentiable
real-valued function on R.
6. If f is a complex-valued differentiable function on � ⊆ C such that
f = 0 on a set having a limit point must be identically zero. However,
it is not true for a differentiable real-valued function on an interval of
R.
7. Every non-constant differentiable function f : �→ C is an open map.
However, this is not true for a differentiable real-valued function on an
interval of R.
8. Although the above points are some positive properties of complex
analysis that are not in real analysis, graph of a complex-valued
function of a complex variable cannot be visualized, whereas the graph
of a real-valued function of a real variable can be plotted. However, the
geometric properties of a complex-valued function of a complex vari-
able can be studied by seeing the image of a curve or a region under
the given function.
To discuss any branch of Mathematics, it is necessary to know the required
basic definitions and results. Therefore, to make this this book self-contained,
we carefully selected the most useful definitions and results from set theory,
real analysis, and algebra, and presented them in this section.
1. We call a collection of ‘well defined’ objects a set.
2. A set containing no elements is called an empty set. To denote this
special set, we use the notation ∅.
�
�
“book” — 2014/6/4 — 20:35 — page 4 — #4
�
�
�
�
�
�
4 Introduction and Brief Prerequisites
3. To say that x is an element of a set, we write x ∈ A.
4. We say that B is a subset of A if x ∈ A whenever x ∈ B. We write this
symbolically as B ⊆ A.
5. We say that A and B are equal if A ⊆ B and B ⊆ A.
6. Let X be a set, A, B and Aα ⊆ X , ∀α ∈ I for some index set I , then
(a) the union ∪
α∈I
Aα of {Aα}α∈I is defined by {x : x ∈ Aα , for some
α ∈ I}.
(b) the intersection ∩
α∈I
Aα of {Aα}α∈I is defined by {x : x ∈ Aα ,
∀α ∈ I}.
When I = {1, 2, 3, . . . , n}, we also use the following equivalent
notations:
∪
α∈I
Aα = n∪
α=1
Aα = A1 ∪ A2 ∪ A3 ∪ · · · ∪ An
and
∩
α∈I
Aα = n∩
α=1
Aα = A1 ∩ A2 ∩ A3 ∩ · · · ∩ An,
(c) The complement Ac = X \ A of A in X is defined by the set {x ∈
X : x � A}.
(d) A and B are said to be disjoint if A ∩ B = ∅. We also note that
A ∩ B = ∅ iff A ⊆ Bc iff B ⊆ Ac.
(e) De-Morgan’s laws
(i)
(
∪
α∈I
Aα
)c
= ∩
α∈I
Ac
α and (ii)
(
∩
α∈I
Aα
)c
= ∪
α∈I
Ac
α .
7. Let {Xi}n
i=1 be a finite collection of sets, then the Cartesian product
X1 × X2 × · · · × Xn (or)
n
�
i=1
Xi
consists of all elements of the form (x1, x2, x3, . . . , xn) called ordered
n-tuples, where xi ∈ Xi, ∀i = 1, 2, 3, . . . , n. We say that two elements
(x1, x2, x3, . . . , xn) and (y1, y2, y3, . . . , yn) are said to be equal in
n
�
i=1
Xi
if xi = yi and ∀i = 1, 2, 3, . . . , n. The members of
n
�
i=1
Xi are called
ordered pairs when n = 2 and ordered triplets when n = 3.
�
�
“book” — 2014/6/4 — 20:35 — page 5 — #5
�
�
�
�
�
�
Preliminaries 5
One can say that mathematical analysis cannot be discussed without
inequalities. Therefore, now we recall the definitions of order relation,
lower bound, upper bound, infimum, and supremum and state some useful
inequalities.
1. A relation R on a non-empty set X is a subset of X × X .
2. A relation R on X is said to be an order relation if
(a) (x, y), (y, z) ∈ R ⇒ (x, z) ∈ R;
(b) for every x, y ∈ X , one of the following is true:
(i) x = y (ii) (x, y) ∈ R (iii) (y, x) ∈ R
If R is an ordered relation on X , then it is customary to write (x, y) ∈ R
by x < y and (X ,<) is called an ordered set. In an ordered set (X ,<),
the notations x ≤ y and a ≥ b mean that (x < y or x = y) and (b < a
or a = b), respectively.
3. Let (X ,<) be an ordered set and ∅ � A ⊆ X , then
(a) an element α ∈ X is said to be an upper bound of A, if
x ≤ α, ∀x ∈ A.
(b) A is said to be bounded above in X if A has an upper bound α ∈ X .
(c) least upper bound of A is an upper bound β of A with the property
that
β ≤ γ , for every upper bound γ of A.
least upper bound of A is also called supremum of A and is denoted
by sup A in X or simply by sup A.
(d) an element α ∈ X is said to be a lower bound of A, if
α ≤ x, ∀x ∈ A.
(e) A is said to be bounded below in X if A has a lower bound α ∈ X .
(f) greatest lower bound of A is a lower bound β of A with the property
that
γ ≤ β, for every lower bound γ of A.
greatest lower bound of A is also called infimum of A and is denoted
by inf A in X or simply by inf A.
�
�
“book” — 2014/6/4 — 20:35 — page 6 — #6
�
�
�
�
�
�
6 Introduction and Brief Prerequisites
(g) • sup A is unique if it exists.
• inf A is unique if it exists.
• If α = sup A and α ∈ A, then α is called the maximum of A.
• If α = inf A and α ∈ A, then α is called the minimum of A.
(h) X is said to have least upperbound property, if for every bounded
above subset A of X , sup A exists in X .
(i) X is said to have greatest lower bound property, if for every
bounded below subset A of X , inf A exists in X .
(j) X has greatest lower bound property iff it has least upper bound
property.
(k) R is an ordered set with least upper bound property and Archime-
dian property1.
(l) For given x, y ∈ R such that x < y, there exists p ∈ Q such that
x < p < y.
(m) For a given x > 0 there exists unique y > 0 such that yn = x. This
y is called the nth root of x and is denoted by n
√
x or x
1
n .
Although the content of this book includes some topological properties of
subsets of the complex plane, mainly we concentrate on functions of a com-
plex variable. Therefore, now we recall some basic ideas from function
theory.
1. Let A and B be non-empty sets. We say that f is a function from A into
B (denoted by f : A → B) if f is a subset of A × B with the following
properties:
(a) for every a ∈ A, there exists b ∈ B such that (a, b) ∈ f .
(b) if (a, b1), (a, b2) ∈ f , then b1 = b2.
2. Let f : A → B be a function.
(a) Then A and B are called the domain and codomain of f , respec-
tively.
(b) (a, b) ∈ f is equivalently denoted by f (a) = b.
(c) If f : A → A is defined by f (x) = x, ∀x ∈ A, then f is called the
identity function on A.
1Archimedian property: Given x > 0 and y ∈ R, there exists n ∈ N such that nx > y.
�
�
“book” — 2014/6/4 — 20:35 — page 7 — #7
�
�
�
�
�
�
Preliminaries 7
(d) If A1 ⊆ A, then the image f (A1) of A1 under f is defined by f (A1) =
{f (a) : a ∈ A1}.
(e) If B1 ⊆ B, then the pre-image or inverse image f −1(B1) of B1 is
defined by f −1(B1) = {a ∈ A : f (a) ∈ B1}.
(f) If g : B → C, then we define the composition g ◦ f : A → C of f
and g by (g ◦ f )(x) = g( f (x)), ∀x ∈ A.
(g) f is invertible if there exists a function g : B → A such that g ◦ f
is the identity function on A and f ◦ g is the identity function on B.
Furthermore, g is called the inverse of f and is denoted by f −1.
3. A function f : A → B is said to be
(a) one-to-one (or injective) if f (x) � f (y) whenever x, y ∈ A with
x � y.
(b) onto (or surjective) if for every b ∈ B, there exists a ∈ A such that
f (a) = b.
(c) bijective if it is injective and surjective.
4. Let f : A → B. Then, f is invertible iff f is a bijection.
Furthermore, for a bijective function f : A → B, the inverse f −1 :
B → A of f is defined by f −1(y) = x, where x ∈ A is unique such that
f (x) = y.
5. Let A, B ⊆ R. A function f : A → B is said to be
(a) an increasing function if x, y ∈ A such that x < y, then f (x) ≤ f (y).
(b) a strictly increasing function if x, y ∈ A such that x < y, then
f (x) < f (y).
(c) a decreasing function if x, y ∈ A such that x < y, then f (x) ≥ f (y).
(d) a strictly decreasing function if x, y ∈ A such that x < y, then
f (x) > f (y).
As algebra is inevitable to discuss analysis, we briefly recall some concepts
like group, ring, field, and so on.
Let A be a non-empty set.
1. Binary operator. A function f : A×A → A is called a binary operator
on A. We also denote a binary operator f by � and the value f (a, b) by
a � b.
2. Unary operator. A function f : A → A is called a unary operator on A.
�
�
“book” — 2014/6/4 — 20:35 — page 8 — #8
�
�
�
�
�
�
8 Introduction and Brief Prerequisites
3. Let A be a non-empty set and � be a binary operator on A, then
(a) � is said to be associative if a � (b � c) = (a � b) � c, ∀a, b, c ∈ A.
(b) � is said to have an identity element in A if there exists e� ∈ A such
that a � e� = e� � a = a, ∀a ∈ A.
(c) an element a of A is said to have an inverse b ∈ A with respect to �
if a � b = b � a = e�. We know that inverse of a is unique.
(d) � is said to be commutative if a � b = b � a, ∀a, b ∈ A.
4. (A, �) is said to be an abelian group if � satisfies the above four
properties (a), (b), (c) for every element of A, and (d).
5. (A,+, ·) is said to be a field if (A,+) and (A \ {0}, ·) are abelian groups,
where 0 is the additive identity and (a + b) · c = (a · c) + (b · c), ∀
a, b, c ∈ A.
6. A four tuple (A,+, ·,<) is said to be an ordered field if (A,<) is an
ordered set and (A,+, ·) is a field with the following properties:
(a) If a, b ∈ A such that a < b, then a + c < b + c, ∀c ∈ A.
(b) If a, b ∈ A such that a > 0 and b > 0, then a · b > 0.
7. If (A,+, ·,<) is an ordered field, then
(a) a · a > 0, ∀a ∈ A \ {0}.
(b) a > 0 iff −a < 0, where −a is the additive inverse of a.
(c) a > 0 iff a−1 > 0, where a−1 is the multiplicative inverse of a.
(d) if a < b and c > 0, then ac < bc.
(e) if a < b and c < 0, then ac > bc.
(f) if 0 < a < b, then b−1 < a−1.
8. The set R of all real numbers is an ordered field with respect to usual
+, ·, and <.
(a) The modulus function on R is defined by |x| =
{
x if x ≥ 0
−x if x < 0,
∀x ∈ R.
(b) The modulus function on R satisfies the following properties:
(i) |x| ≥ 0, ∀x ∈ R, (ii) |x| = 0 iff x = 0, (iii) |xy| = |x| |y|,
(iv) |x + y| ≤ |x| + |y|, and (v) x2 ≤ y2 iff |x| < |y|.
�
�
“book” — 2014/6/4 — 20:35 — page 9 — #9
�
�
�
�
�
�
Preliminaries 9
1.2 COMPLEX NUMBERS AND GEOMETRICAL
REPRESENTATIONS
The terminology Complex numbers was due to C. F. Gauss in 1831. The rigor-
ous definition of complex number as an ordered pair of real numbers (a1, a2)
was introduced by W. R. Hamilton in 1837. The set of all complex numbers
is denoted by C. Then immediately it follows that C = R × R. In the set of
all complex numbers, there is a special element (0, 1), which is denoted by i.
Euler named this number i by the imaginary number. Note that this number i
was used as a number satisfying i2 = −1 before the rigorous construction of
complex numbers, and hence, it was called an imaginary number.
Since there is no element in R whose square is a negative number and
i2 = −1, this number is not in R and hence, it might be called an imaginary
number. As the numbers on a straight line are not imaginary, the name set of
real numbers is given to R.
We define addition and multiplication on C as follows:
(a1, a2) + (b1, b2) = (a1 + b1, a2 + b2),
(a1, a2) · (b1, b2) = (a1b1 − a2b2, a1b2 + a2b1), and
(a1, a2), (b1, b2) ∈ C.
THEOREM 1.2.1 The set C of all complex numbers is a field with respect to
addition and multiplication as defined above.
Proof: Let a = (a1, a2), b = (b1, b2), and c = (c1, c2) ∈ C. Using the fact
that R is a field, we get the following:
1. Clearly, a + b = (a1, a2) + (b1, b2)
= (a1 + b1, a2 + b2)
= (b1 + a1, b2 + a2)
= (b1, b2) + (a1, a2)
= b + a
2. a + (b + c) = (a1, a2) + [(b1, b2) + (c1, c2)]
= (a1, a2) + (b1 + c1, b2 + c2)
= (a1 + (b1 + c1), a2 + (b2 + c2))
= ((a1 + b1) + c1, (a2 + b2) + c2)
= (a1 + b1, a2 + b2) + (c1, c2)
= [(a1, a2) + (b1, b2)] + (c1, c2)
= (a + b) + c.
�
�
“book” — 2014/6/4 — 20:35 — page 10 — #10
�
�
�
�
�
�
10 Complex Numbers and Geometrical Representations
3. (0, 0) ∈ C is the identity element with respect to + because (a1, a2) +
(0, 0) = (a1 + 0, a2 + 0) = (a1, a2) and (0, 0) + (a1, a2) = (0 + a1, 0 +
a2) = (a1, a2).
4. If (a1, a2) ∈ C, then (−a1,−a2) ∈ C and (a1, a2) + (−a1,−a2) =
(a1 + (−a1), a2 + (−a2)) = (0, 0).
5. (a1, a2) · (b1, b2) = (a1b1 − a2b2, a1b2 + a2b1)
= (b1a1 − b2a2, b2a1 + b1a2)
= (b1a1 − b2a2, b1a2 + b2a1)
= (b1, b2) · (a1, a2).
6. a · (b · c)
= (a1, a2) · [(b1, b2) · (c1, c2)]
= (a1, a2) · [(b1c1 − b2c2, b1c2 + b2c1)]
= (a1(b1c1 − b2c2) − a2(b1c2 + b2c1), a1(b1c2 + b2c1)
+ a2(b1c1 − b2c2))
= (a1b1c1 − a1b2c2 − a2b1c2 − a2b2c1, a1b1c2 + a1b2c1
+ a2b1c1 − a2b2c2)
= ((a1b1 − a2b2)c1 − (a1b2 + a2b1)c2, (a1b1 − a2b2)c2
+ (a1b2 + a2b1)c1)
= ((a1b1 − a2b2), (a1b2 + a2b1)) · (c1, c2)
= ((a1, b1) · (a2, b2)) · (c1, c2)
= (a · b) · c.
7. Clearly, (1, 0) ∈ C and (a1, a2) · (1, 0) = (a1 ·1−a2 ·0, a1 ·0+a2 ·1) =
(a1, a2).
8. If (a1, a2) ∈ C � (0, 0), then at least one of a1 and a2 is non-zero.
Therefore, a2
1 + a2
2 � 0, and hence,
(
a1
a2
1 + a2
2
,
−a2
a2
1 + a2
2
)
∈ C. Now,
(a1, a2) ·
(
a1
a2
1 + a2
2
,
−a2
a2
1 + a2
2
)
=
(
a2
1 +a2
2
a2
1 + a2
2
,
−a1a2 + a2a1
a2
1 + a2
2
)
= (1, 0).
�
�
“book” — 2014/6/4 — 20:35 — page 11 — #11
�
�
�
�
�
�
Preliminaries 11
9. a · (b + c)
= (a1, a2) · ((b1, b2) + (c1, c2))
= (a1, a2) · (b1 + c1, b2 + c2)
= ((a1(b1 + c1) − a2(b2 + c2)), a1(b2 + c2) + a2(b1 + c1))
= (a1b1 + a1c1 − a2b2 − a2c2, a1b2 + a1c2 + a2b1 + a2c1)
= ((a1b1 − a2b2) + (a1c1 − a2c2), (a1b2 + a2b1)+(a1c2+a2c1))
= (a1b1 − a2b2, a1b2 + a2b1) + (a1c1 − a2c2, a1c2 + a2c1)
= ((a1, a2) · (b1, b2)) + ((a1, a2) · (c1, c2))
= (a · b) + (a · c).
Hence, (C,+, ·) is a field. �
We identify every real number x with the element (x, 0) of C. By this
identification, the set of all real numbers is regarded as a subset of C. It is
also customary to write a complex number z = (x, y) as x + iy, which is the
simplified form of the expression (x, 0)+(0, 1)·(y, 0). We call x, y the real and
imaginary parts of z and denote them by x = Re z and y = Im z, respectively.
LEMMA 1.2.2 i · i = (−1, 0).
Proof: By direct computation, we get i · i = (0, 1) · (0, 1) = (0 − 1, 0 + 0) =
(−1, 0). �
Remark 1.2.3: C is not an ordered field with respect to any order relation.
Reason: By the observation 7(a) in page 8, if C were an ordered field, then
we would get
−1 = i · i > 0 and 1 = 1 · 1 > 0,
which contradicts the observation 7(b) in page 8.
Example 1.2.4 Write the following numbers in a + ib form.
1. (3 + i5)(4 − i2).
2.
(3 + i)2
1 − i2
.
3. sin(1 + i).
�
�
“book” — 2014/6/4 — 20:35 — page 12 — #12
�
�
�
�
�
�
12 Complex Numbers and Geometrical Representations
Solution:
1. (3 + i5)(4 − i2) = (12 + 10) + i(−6 + 20) = 22 + i14.
2.
(3 + i)2
1 − i2
= (9 − 1 + i6)(1 + i2)
5
= (8 − 12) + i(16 + 6)
5
=
−4 + i22
5
.
3. sin(1 + i) = sin(1) cos(i) + cos(1) sin(i) = sin(1) cosh(1) +
i cos(1) sinh(1).
Exercise 1.2.5 Write the following numbers in a + ib form.
1.
(2 + i)(3 − i2)
1 + i4
.
2. (4 + i3)4.
3. cos(1 − i2).
4. exp(i(1 + i)).
5. cos(i) exp(i3 + 2).
Answers:
1.
4 − i33
17
.
2. −527 − i336.
3. cos(1) cosh(2) + i sin(1) sinh(2).
4. exp(−1) cos(1) + i exp(−1) sin(1).
5. cosh(1) exp(2) cos(3) + i cosh(1) exp(2) sin(3).
Definition 1.2.6 We define the complex conjugate z of z ∈ C by z = (x,−y)
whenever z = (x, y).
RESULT 1.2.7 (Properties of complex conjugate)
Let z, z1, z2 ∈ C be arbitrary.
1. z + z = 2Re z and z − z = i2Im z.
2. z = z.
�
�
“book” — 2014/6/4 — 20:35 — page 13 — #13
�
�
�
�
�
�
Preliminaries 13
3. z1 + z2 = z1 + z2.
4. z1 · z2 = z1 · z2.
5.
(
z1
z2
)
= z1
z2
, when z2 � (0, 0).
6. If z = x + iy then z · z = x2 + y2.
7. z = z if and only if z is a real number.
Proof: Let z = x + iy, z1 = x1 + iy1, and z2 = x2 + iy2.
1. z+z = (x+ iy)+(x− iy) = 2x = 2Re z and z−z = (x+ iy)−(x− iy) =
i2y = i2 Im z.
2. z = x + iy = x − iy = x + iy = z.
3. (z1 + z2) = (x1 + iy1) + (x2 + iy2)
= (x1 + x2) + i(y1 + y2)
= (x1 + x2) − i(y1 + y2)
= (x1 − iy1) + (x2 − iy2)
= z1 + z2.
4. z1 · z2 = (x1 + iy1) · (x2 + iy2)
= (x1x2 − y1y2) + i(x1y2 + x2y1)
= (x1x2 − y1y2) − i(x1y2 + x2y1)
= (x1x2 − (−y1)(−y2)) + i(x1(−y2) + x2(−y1))
= (x1 − iy1) · (x2 − iy2)
= z1 · z2.
5. First, we prove that
(
1
z
)
= 1
z
, for z � (0, 0).
(
1
z
)
=
(
1
x + iy
)
=
(
x − iy
(x − iy)(x + iy)
)
=
(
x
x2 + y2
− i
y
x2 + y2
)
�
�
“book” — 2014/6/4 — 20:35 — page 14 — #14
�
�
�
�
�
�
14 Complex Numbers and Geometrical Representations
= x
x2 + y2
+ i
y
x2 + y2
= x + iy
x2 + y2
= x + iy
(x + iy)(x − iy)
= 1
x − iy
= 1
z
.
Then, applying (4), we get(
z1
z2
)
=
(
z1 · 1
z2
)
= z1 ·
(
1
z2
)
= z1 · 1
z2
= z1
z2
.
6. By a direct computation, we get z · z = (x + iy) · (x − iy) = x2 + y2.
7. z = z iff x + iy = x − iy iff y = −y iff y = 0 iff z is a real number. �
We also have one more unary operator called the modulus or absolute value
of a complex number, which is defined by
|z| =
√
z · z or equivalently |(x, y)| =
√
x2 + y2
here, we take the positive square root. This unary operator satisfies the
following properties:
RESULT 1.2.8 Let z, z1, z2 ∈ C be arbitrary.
1. |Re z| ≤ |z| and |Im z| ≤ |z|.
2. |z| = |z|.
3. |z1 · z2| = |z1| · |z2|.
4.
∣∣∣∣ z1
z2
∣∣∣∣ = |z1|
|z2| if z2 � 0.
Proof: Let z = x + iy, z1 = x1 + iy1, and z2 = x2 + iy2.
1. x2 ≤ x2 + y2 implies that |x| ≤
√
x2 + y2 = |z|. As
√
x2 + y2 is
symmetric in x and y, by interchanging x and y, we get |y| ≤ |z|.
�
�
“book” — 2014/6/4 — 20:35 — page 15 — #15
�
�
�
�
�
�
Preliminaries 15
2. |z| =
√
x2 + y2 =
√
x2 + (−y)2 = |z|.
3. |z1 · z2| = |(x1x2 − y1y2) + i(x1y2 + x2y1)|
=
√
(x1x2 − y1y2)2 + (x1y2 + x2y1)2
=
√
x2
1x2
2 + y2
1y2
2 + x2
1y2
2 + x2
2y2
1
=
√
(x2
1 + y2
1)(x2
2 + y2
2)
= |z1| · |z2|.
.
4. As ∣∣∣∣1z
∣∣∣∣ =
∣∣∣∣ x − iy
x2 + y2
∣∣∣∣ =
√
x2 + y2
x2 + y2
= 1√
x2 + y2
= 1
|z| ,
we get ∣∣∣∣ z1
z2
∣∣∣∣ =
∣∣∣∣z1 · 1
z2
∣∣∣∣ = |z1|
∣∣∣∣ 1
z2
∣∣∣∣ = |z1|
|z2| .
�
RESULT 1.2.9 (Triangle inequality)
If z1, z2 ∈ C, then |z1 + z2| ≤ |z1| + |z2|. Furthermore, |z1 + z2| = |z1| + |z2|
iff z1 · z2 is real and is non-negative.
Proof: Using Results 1.2.7 and 1.2.8, we get
|z1 + z2|2 = (z1 + z2) · (z1 + z2)
= (z1 + z2) · (z1 + z2)
= (z1 · z1 + z1 · z2 + z2 · z1 + z2 · z2
= |z1|2 + |z2|2 + z1 · z2 + z1 · z2
= |z1|2 + |z2|2 + 2 Re (z1 · z2)
≤ |z1|2 + |z2|2 + 2|z1 · z2|
= |z1|2 + |z2|2 + 2|z1| · |z2|
= (|z1| + |z2|)2.
Hence, we get |z1 + z2| ≤ |z1| + |z2|.
Next, we find the necessary and sufficient condition to get equality in the
above inequality.
From the proof of the triangle inequality, we get
|z1 + z2| = |z1| + |z2| ⇔ |z1|2 + |z2|2 + 2Re (z1 · z2)
= |z1|2 + |z2|2 + 2|z1 · z2|
⇔ Re (z1 · z2) = |z1 · z2|
⇔ z1 · z2 is real and non-negative.
Hence, the result follows. �
�
�
“book” — 2014/6/4 — 20:35 — page 16 — #16
�
�
�
�
�
�
16 Complex Numbers and Geometrical Representations
As a corollary of the above result, we have another version of triangle
inequality, which follows.
COROLLARY 1.2.10 If z1, z2 ∈ C, then | |z1| − |z2| | ≤ |z1 − z2|.
Proof: Now, |z1| = |z1 − z2 + z2| ≤ |z1 − z2| + |z2|, which implies that
|z1| − |z2| ≤ |z1 − z2|.
Interchanging z1, z2 and using | − z| = | − 1||z| = |z|, we get
|z2| − |z1| ≤ |z2 − z1| = |z1 − z2|.
Hence, by the definition of the modulus function on R, the corollary
follows. �
RESULT 1.2.11 If zj ∈ C, 1 ≤ j ≤ n, then
∣∣∣∣∣
n∑
j=1
zj
∣∣∣∣∣ ≤
n∑
j=1
|zj|.
Proof: We prove this by induction on n. For n = 1, this result holds obviously.
Assume that this result is true for some n. Now, using triangle inequality
(Result 1.2.9) and by induction hypothesis, we get∣∣∣∣∣∣
n+1∑
j=1
zj
∣∣∣∣∣∣ =
∣∣∣∣∣∣
n∑
j=1
zj + zn+1
∣∣∣∣∣∣ ≤
∣∣∣∣∣∣
n∑
j=1
zj
∣∣∣∣∣∣+ |zn+1| ≤
n∑
j=1
|zj| + |zn+1| =
n+1∑
j=1
|zj|.
This completes the proof of this result. �
Exercise 1.2.12 Prove that
∣∣∣∣∣
n∑
j=1
zj
∣∣∣∣∣ =
n∑
j=1
|zj| iff zjzk is real and non-negative,
∀j, k ∈ {1, 2, . . . , n}.
RESULT 1.2.13 (Cauchy–Schwarz inequality)
If xi, yi ∈ C, ∀i = 1, 2, 3, . . . , n, then
∣∣∣∣∣
n∑
k=1
xkyk
∣∣∣∣∣
2
≤
n∑
k=1
|xk |2 ·
n∑
k=1
|yk |2. (1.1)
�
�
“book” — 2014/6/4 — 20:35 — page 17 — #17
�
�
�
�
�
�
Preliminaries 17
Proof: Let c =
n∑
k=1
xkyk , ‖x‖ =
(
n∑
k=1
|xk |2
) 1
2
and ‖y‖ =
(
n∑
k=1
|yk |2
) 1
2
.
Consider
0 ≤
n∑
k=1
∣∣∣(‖y‖2xk − cyk)
∣∣∣2
=
n∑
k=1
(‖y‖2xk − cyk)(‖y‖2xk − c yk)
=
n∑
k=1
(‖y‖4|xk |2 − ‖y‖2cxkyk − ‖y‖2cykxk + ccykyk)
= ‖y‖4 · ‖x‖2 − ‖y‖2 · |c|2 − ‖y‖2 · |c|2 + ‖y‖2 · |c|2
= ‖y‖2(‖x‖2‖y‖2 − |c|2)
If ‖y‖2 = 0, then equality holds in (1.1); otherwise, we get |c|2 ≤ ‖x‖2‖y‖2.
This implies (1.1). �
RESULT 1.2.14 (Schwarz’s inequaltiy)
If x = (x1, x2, . . . , xn), y = (y1, y2, . . . , yn) ∈ Cn, then ‖x + y‖ ≤ ‖x‖ + ‖y‖,
where ‖z‖ =
(
n∑
k=1
|zk |2
) 1
2
, ∀z = (z1, z2, . . . , zn) ∈ Cn.
Proof:
‖x + y‖2 =
n∑
k=1
|xk + yk |2
=
n∑
k=1
(xk + yk)(xk + yk)
=
n∑
k=1
(xkxk + xkyk + ykxk + ykyk)
= ‖x‖2 + ‖y‖2 +
n∑
k=1
xkyk +
n∑
k=1
ykxk
(note that
n∑
k=1
xkyk +
n∑
k=1
ykxk is real)
�
�
“book” — 2014/6/4 — 20:35 — page 18 —#18
�
�
�
�
�
�
18 Complex Numbers and Geometrical Representations
≤ ‖x‖2 + ‖y‖2 +
∣∣∣∣∣
n∑
k=1
xkyk +
n∑
k=1
ykxk
∣∣∣∣∣
≤ ‖x‖2 + ‖y‖2 +
∣∣∣∣∣
n∑
k=1
xkyk
∣∣∣∣∣+
∣∣∣∣∣
n∑
k=1
ykxk
∣∣∣∣∣
≤ ‖x‖2 + ‖y‖2 + ‖x‖‖y‖ + ‖x‖‖y‖
(by Cauchy–Schwarz inequality)
= (‖x‖ + ‖y‖)2.
Thus, the result follows. �
Geometrically, the set of all complex numbers can be viewed as the
set of all points in a plane. We fix a point arbitrarily and name it the
origin and denote it by (0, 0). We draw two lines on the plane that pass
through (0, 0) horizontally and vertically. We call the horizontal line and
vertical line, respectively, the real axis and the imaginary axis. Now, every
element (x, y) ∈ C is represented by the point of intersection of the
horizontal line passing through (0, y) and the vertical line passing through
(x, 0). This representation is also called the Cartesian form of a complex
number.
Every non-zero complex number z = (x, y) can also be represented in
another form r exp(iθ ) (or reiθ ) called polar form2, where r = |z| is the length
of the line segment joining (0, 0) and z (called the modulus of z), and θ is
the angle between the positive side of the x-axis and the line segment joining
(0, 0) and z, measured in the anti-clock wise sense (called an argument θ of z).
Using the Pythagoras theorem, one can get the length of the line segment as√
x2 + y2 = |(x, y)| = |z|.
Note that the modulus of a complex number is unique but the argument
is not. The reason for the second statement is that if θ is one of the values
of argument of (a, b), then for every k ∈ Z, θ + 2kπ is also an argument
of the same (a, b). Hence, we define the principal argument of a non-zero
complex number (a, b) as the argument θ of (a, b) satisfying the condition
−π < θ ≤ π .
The subsets {(a, b) : a > 0, b > 0}, {(a, b) : a < 0, b > 0}, {(a, b) :
a < 0, b < 0} and {(a, b) : a > 0, b < 0} are called first quadrant, second
quadrant, third quadrant, and fourth quadrant, respectively.
2At present, treat r exp(iθ ) as just a symbol. The reason for using this notation is explained
in Corollary 2.4.25.
�
�
“book” — 2014/6/4 — 20:35 — page 19 — #19
�
�
�
�
�
�
Preliminaries 19
Remark 1.2.15: At this juncture, the reader is advised not to use arg (a +
ib) = arctan
(
b
a
)
, for all non-zero a + ib ∈ C.
To justify this remark, if we use the above formula, then we will get arg (1 + i)
= arg (−1 − i) = π
4
, but the first point is in the first quadrant and the second
point is in the third quadrant, and hence obviously, their arguments should be
different. See the following diagram.
1 + i
−1 − i
arg(1 + i)arg(−1 − i)
Definition 1.2.16 (Principal argument)3 Let a + ib � 0.
• Case (i): If a = 0, b > 0, then θ = π
2
; If a = 0, b < 0, then θ = −π
2
;
If a < 0, b = 0, then θ = π .
• Case (ii): If a > 0, b ∈ R, then θ = arctan
(
b
a
)
.
• Case (iii): If a < 0, b > 0, then θ = π − arctan
(
b
|a|
)
.
• Case (iv): If a < 0, b < 0, then θ = −π + arctan
( |b|
|a|
)
.
3For arctan and its properties, we refer the reader to Section 2.4.
�
�
“book” — 2014/6/4 — 20:35 — page 20 — #20
�
�
�
�
�
�
20 Complex Numbers and Geometrical Representations
q
(a, b)
(a, −b)
(a, b) (a, b)
q q
−q
−p + q
(|a|, |b|)
qp q
(a, b)
Example 1.2.17 Write the given numbers in the polar form with principal
arguments: (1) 2 + i3 and (2)
1
2
(1 − √
3).
1. |3 + i4| = √
9 + 16 = √
25 = 5 and arg 3 + i4 = arctan
(
4
3
)
⇒ 3 + i4 = √
25 e
iarctan
(
4
3
)
.
2.
∣∣∣∣12(1 − √
3)
∣∣∣∣ = 1
2
√
1 + 3 = 1 and arg
1
2
(1 − √
3) = − arctan(√
3
)
= −π
3
⇒ 1
2
(1 − √
3) = e
−π
3 .
Definition 1.2.18 Given a non-zero complex number z = reiθ , we define
n
√
z = r
1
n e
i
(
θ+2πk
n
)
, where k = 0, 1, 2, . . . , n − 1.
�
�
“book” — 2014/6/4 — 20:35 — page 21 — #21
�
�
�
�
�
�
Preliminaries 21
Example 1.2.19 Find the cube roots of −8.
First, we write −8 = 8eiπ , and hence, (−8)1/3 = 2ei(π+2kπ )/3, k = 0, 1, 2.
Hence, the three values of (−8)1/3 are as follows:
1. 2eiπ/3 = 2(cos(π/3) + i sin(π/3)) = 1 + i
√
3.
2. 2eiπ = 2(cos(π ) + i sin(π )) = −2.
3. 2ei4π/3 = 2(cos(4π/3) + i sin(4π/3)) = 1 − i
√
3.
The sum of two complex numbers (a, b) and (c, d) can be viewed geomet-
rically as follows. Construct a parallelogram using the line segment joining
(0, 0), (a, b) and the line segment joining (0, 0), (c, d), as the adjacent sides.
Then the fourth vertex of the parallelogram is (a, b)+(c, d). See the following
figure.
(a, b)
(c, d )
(0, 0)
(a + c, b + d)
The product of two complex numbers z1 and z2 can be understood easily
if we represent them in polar coordinates rather than in Cartesian coordinates.
Let r1 exp(θ1), r2 exp(θ2) be the given two non-zero complex numbers. (If at
least one of them is zero, then the product is obvious.) Then the product of
these two complex numbers is the complex number r1 · r2 exp(θ1 + θ2). Geo-
metrically, r1 ·r2 exp(θ1+θ2) is obtained by rotating the line joining the origin
and r1 exp(θ1) through an angle θ2 and then multiplying it by r2. Multiplying
r1 exp(θ1 + θ2) by r2 expands (or shrinks) the length of line segment joining
(0, 0) and r1 exp(θ1 + θ2) by the scale r2, when r2 > 1 (or 0 < r2 < 1).
At this stage, take one of the end points of the resulting line segment, other
than origin, as the product of the given complex numbers. See the following
figure.
�
�
“book” — 2014/6/4 — 20:35 — page 22 — #22
�
�
�
�
�
�
22 Complex Numbers and Geometrical Representations
q1
q1
r2 exp(iq2)
r1 exp(iq1)
r1r2 exp(i(q1 + q2))
r2 exp(i(q1 + q2))
q2
The complex conjugate of a complex number (a, b) can be viewed as the
point of reflection with respect to the real axis.
z
z
Geometrically, the additive inverse of a complex number can be viewed
as the rotation of the given non-zero complex number through an angle π or
equivalently, it is the reflection of z with respect to the real axis followed by
another reflection with respect to the imaginary axis.
z
−z
Exercise 1.2.20 Geometrically explain the multiplicative inverse of a com-
plex number.
�
�
“book” — 2014/6/4 — 20:35 — page 23 — #23
�
�
�
�
�
�
Preliminaries 23
Now, we find the equation of the straight line passing through the given
two points ζ and ξ . Geometrically, the slope of the line segment joining ξ
and ζ is same as the slope of the line segment joining 0 and ζ − ξ .
z + x
z − x
z
0
Line segment joining x and z.
Line segment joining 0 and z − x.
x
−x
Therefore, the straight line passing through ξ and ζ is same as the straight
line passing through ξ with the slope as that of the line segment joining 0 and
ζ − ξ . Note that if t moves from 0 towards ∞, then t(ζ − ξ ) moves from 0
to ∞, and it passes through ζ − ξ when t = 1. If t moves from 0 to −∞,
then t(ζ − ξ ) moves from 0 to ∞ through −(ζ − ξ ) when t = −1. Hence, the
equation of the straight line passing through 0 and ζ − ξ is t(ζ − ξ ), t ∈ R.
If the straight line passing through 0 and ζ − ξ is translated by ξ , then the
resulting straight line passes through ξ and ζ . Hence, its equation becomes
ξ + t(ζ − ξ ), t ∈ R. Therefore, the equation of a straight line passing through
a and having slope as the slope of the line segment joining 0 and b, for some
a, b ∈ C with b � 0, is given by
z = a + tb, t ∈ R. (1.2)
Now, we find an implicit equation of the same straight line.
z lies on the line (1.2) iff
z − a
b
∈ R
iff
z − a
b
=
(
z − a
b
)
iff zb − zb + ab − ab = 0
iff z(ib) + zib + 2Im ab = 0.
�
�
“book” — 2014/6/4 — 20:35 — page 24 — #24
�
�
�
�
�
�
24 Sequences and Series of Complex Numbers
The circle with centre a ∈ C and radius r > 0 can be defined as the set of all
points that are at distance r units from a. More explicitly {z : |z − a| = r}.
We shall prove latter (in Corollary 2.4.25) that z − a = r(exp(iθ )) for some
θ ∈ [0, 2π]. Thus, an equation of a circle can be given by z = a + r exp(iθ ),
θ ∈ [0, 2π].
z
r
a
The implicit equation of the circle |z − a| = r is given as follows.z belongs to the circle iff |z − a|2 = r2
iff |z|2 − za − za + |a|2 − r2 = 0
iff |z|2 + z(−a) − z(−a) + |a|2 − r2 = 0.
The interesting fact is that there is a combined version of an equation for
straight lines and circles. This is simply
α|z|2 + βz + βz + γ = 0, where α, γ ∈ R and β ∈ C with not α = 0 = β.
The above equation represents a
{
straight line if α = 0.
circle if α � 0.
1.3 SEQUENCES AND SERIES OF COMPLEX
NUMBERS
In this section, first we recall some definitions and results on sequences and
series of complex numbers.
Definition 1.3.1 A sequence (an) of complex numbers is said to be
1. convergent if there exists a ∈ C such that given ε > 0, then there exists
m ∈ N such that |an − a| < ε, ∀n ≥ m. In this case, we call ‘a’ the
limit of (an) and we write an → a as n → ∞ or a = lim
n→∞ an or (an)
converges to a.
�
�
“book” — 2014/6/4 — 20:35 — page 25 — #25
�
�
�
�
�
�
Preliminaries 25
2. a Cauchy sequence if given ε > 0, then there exists m ∈ N such that
|aj − ak | < ε, ∀j, k ≥ m.
3. bounded if there exists M > 0 such that |an| ≤ M ,∀n ∈ N.
RESULT 1.3.2 If (an) is a convergent sequence, then (an) is a Cauchy
sequence.
Proof: Let (an) be a convergent sequence. Then, there exists a ∈ C such that
given ε > 0, there exists m ∈ N such that
|an − a| < ε
2
, ∀n ≥ m.
For the same m ∈ N, if j, k ≥ m, then
|aj − ak | ≤ |aj − a| + |ak − a| < ε
2
+ ε
2
= ε.
Hence, (an) is a Cauchy sequence. �
RESULT 1.3.3 If (an) is a Cauchy sequence, then (an) is a bounded sequence.
Proof: For ε = 1, there exists m ∈ N such that
|aj − ak | < 1, ∀j, k ≥ m.
Then, put
K = max{1 + |am|, |an| : n = 1, 2, 3, . . . , m − 1}.
Then, K > 0 and for every n ≤ m − 1, obviously we have |an| ≤ K and for
n ≥ m,
|an| ≤ |an − am| + |am| < 1 + |am| ≤ K,
which implies that (an) is a bounded sequence. �
From the above two results, we obtain the following corollary.
COROLLARY 1.3.4 If (an) is a convergent sequence, then (an) is a bounded
sequence.
Example 1.3.5 Every constant sequence is obviously a convergent sequence.
�
�
“book” — 2014/6/4 — 20:35 — page 26 — #26
�
�
�
�
�
�
26 Sequences and Series of Complex Numbers
Example 1.3.6
(
n
n + 1
)
converges to 1.
Given ε > 0, applying Archimedian property, we choose N >
1
ε
. If n ≥ N ,
then ∣∣∣∣ n
n + 1
− 1
∣∣∣∣ =
∣∣∣∣ 1
n + 1
∣∣∣∣ = 1
n + 1
<
1
n
< ε.
Thus,
n
n + 1
→ 1 as n → ∞.
Definition 1.3.7 Given a sequence (an), we mean (ank ) a subsequence of (an)
if (nk) is a strictly increasing sequence of natural numbers. That is, nk < nj if
k < j.
LEMMA 1.3.8 If (xn) is a Cauchy sequence and it has a subsequence (xnk )
converging to x, then (xn) also converges to the same x.
Proof: Let ε > 0 be given. Then, there exist N , j ∈ N such that
|xn − xm| < ε
2
, ∀m, n ≥ N and |xnk − x| < ε
2
, ∀k ≥ j.
If p = max{N , nj}, then for every n ≥ p, as np ≥ p ≥ N and p ≥ nj ≥ j, we
have,
|xn − x| ≤ |xn − xnp | + |xnp − x| < ε
2
+ ε
2
= ε.
Hence, the lemma follows. �
THEOREM 1.3.9 Let (an) and (bn) be complex sequences and a, b ∈ C.
(1) an → a as n → ∞ iff Re an → Re a and Im an → Im a as n → ∞.
Let an → a and bn → b as n → ∞. Then,
(2) can → ca as n → ∞.
(3) Every subsequence of (an) converges to a.
(4) an + bn → a + b as n → ∞.
(5) an · bn → a · b as n → ∞.
(6)
an
bn
→ a
b
as n → ∞ if b � 0, bn � 0, ∀n ∈ N.
�
�
“book” — 2014/6/4 — 20:35 — page 27 — #27
�
�
�
�
�
�
Preliminaries 27
Proof:
1. Assume that an → a as n → ∞. Then, for a given ε > 0, there exists
m ∈ N such that |an − a| < ε, ∀n ≥ m. For the same m ∈ N, if n ≥ m,
we have
|Re an − Re a| ≤ |an − a| < ε and |Im an − Im a| ≤ |an − a| < ε
Conversely, assume that Re an → Re a and Im an → Im a as n → ∞.
Given ε > 0, there exist m1, m2 ∈ N such that
|Re an − Re a| < ε√
2
, ∀n ≥ m1
and
|Im an − Im a| < ε√
2
, ∀n ≥ m2.
For n ≥ max{m1, m2}, we have
|an − a| =
√
|Re an − Re a|2 + |Im an − Im a|2 <
√
ε2
2
+ ε2
2
= ε.
Hence, an → a as n → ∞.
2. Using an → a as n → ∞ for a given ε > 0, we choose m ∈ N such
that
|an − a| < ε
1 + |c| , ∀n ≥ m.
For n ≥ m,
|can − ca| = |c||an − a| ≤ |c| ε
1 + |c| ≤ ε.
Therefore, can → ca as n → ∞.
3. Let (ank ) be a subsequence of the given sequence (an). Then, (nk) is
a strictly increasing sequence of natural numbers. Given ε > 0, there
exists m ∈ N such that |an − a| < ε, ∀n ≥ m. Choose j ∈ N such that
nj > m. If k ≥ j, then nk ≥ nj > m, and hence, |ank − a| < ε.
4. Given ε > 0, choose m1, m2 ∈ N such that
|an − a| < ε
2
, ∀n ≥ m1 and |bn − b| < ε
2
, ∀n ≥ m2.
�
�
“book” — 2014/6/4 — 20:35 — page 28 — #28
�
�
�
�
�
�
28 Sequences and Series of Complex Numbers
If n ≥ max{m1, m2}, then
|(an + bn) − (a + b)| ≤ |an − a| + |bn − b| < ε
2
+ ε
2
= ε.
Thus, an + bn → a + b as n → ∞.
5. Using Corollary 1.3.4, we find an M > 0 such that |an| ≤ M , ∀n ∈ N.
Given ε > 0, choose m1, m2 ∈ N, such that
|an − a| < ε
2(1 + |b|) , ∀n ≥ m1 and |bn − b| < ε
2M
, ∀n ≥ m2.
If n ≥ max{m1, m2}, then
|an · bn − a · b| = |an · bn − an · b + an · b − a · b|
≤ |an| · |bn − b| + |b| · |an − a|
< M
ε
2M
+ |b| ε
2(1 + |b|)
≤ ε
2
+ ε
2
= ε.
6. It is left as an exercise to the reader. For a hint, see Theorem 1.6.3. �
LEMMA 1.3.10 Let (an) and (bn) be real sequences such that an ≤ bn ∀n ∈ N.
If an → a and bn → b as n → ∞, then a ≤ b.
Proof: If a > b, then a−b
2 > 0. Then, for ε = a−b
2 , there exist N1, N2 ∈ N
such that
|an − a| < a − b
2
, ∀n ≥ N1 and |bn − b| < a − b
2
, ∀n ≥ N2.
If m ≥ max{N1, N2}, then
|am − a| < a − b
2
and |bm − b| < a − b
2
.
Therefore,
a − b = a − am + am − b ≤ (a − am) + (bm − b) <
a − b
2
+ a − b
2
= a − b.
which is a contradiction. Thus, a ≤ b. �
THEOREM 1.3.11 (Cauchy criterion)
Let (an) be a sequence of complex numbers. Then, (an) is a Cauchy sequence
iff (an) is a convergent sequence.
�
�
“book” — 2014/6/4 — 20:35 — page 29 — #29
�
�
�
�
�
�
Preliminaries 29
The half of the proof of the above theorem is already proved and the remain-
ing half follows from the fact that C is a complete metric space. (See Theorem
1.4.46.)
Definition 1.3.12 Let (an) be a sequence of complex numbers. We write
an → ∞ as n → ∞ if for every K > 0, there exists m ∈ N such that
|an| > K,∀n ≥ m.
Definition 1.3.13 Let (xn) be a sequence of real numbers.
1. We say that xn → +∞ as n → ∞ if for every K > 0, there exists
m ∈ N such that xn > K, ∀n ≥ m.
2. We say that xn → −∞ as n → ∞ if for every K > 0, there exists
m ∈ N such that xn < −K, ∀n ≥ m.
Definition 1.3.14 Let (xn) be a sequence of real numbers, then an extended
real number α is said to be the limit superior or upper limit, which is denoted
by lim sup
n→∞
xn, if it satisfies the following conditions:
1. There exists a subsequence (xnk ) of (xn) such that xnk → α as k → ∞.
2. Given ε > 0, there exists m ∈ N such that an < α + ε, ∀n ≥ m.
Definition 1.3.15 Let (xn) be a sequence of real numbers, then an extended
real number α is said to be the limit inferior or lower limit, which is denoted
by lim inf
n→∞ xn, if it satisfies the following conditions:
1. There exists a subsequence (xnk ) of (xn) such that xnk → α as k → ∞.
2. Given ε > 0, there exists m ∈ N such that an > α − ε, ∀n ≥ m.
LEMMA 1.3.16 For a real sequence (xn), lim inf
n→∞ xn ≤ lim sup
n→∞
xn.
Proof: Let ε > 0 be given. If α = lim inf
n→∞ xn and β = lim sup
n→∞
xn, then there
exist N1, N2 ∈ N such that
xn > α − ε, ∀n ≥ N1 and xn < β + ε, ∀n ≥ N2.
If m ≥ max{N1, N2}, then α−ε < xm and xm < β+ε imply that α−ε < β+ε.
As ε > 0 is arbitrary, we get α ≤ β. �
�
�
“book” — 2014/6/4 — 20:35 — page 30 — #30
�
�
�
�
�
�
30 Sequences and Series of Complex Numbers
RESULT 1.3.17 Let (xn) be a sequence of real numbers. (xn) converges to x
iff lim sup
n→∞
xn = lim inf
n→∞ xn = x.
The proof of the above result is straightforward from the definitions.
LEMMA 1.3.18 If (xn) and (yn) be sequences of real numbers such that xn ≤
yn,∀n ∈ N, then lim sup
n→∞
xn ≤ lim sup
n→∞
yn and lim inf
n→∞ xn ≤ lim inf
n→∞ yn.
Proof: Let x and y be the limit superiorof (xn) and (yn), respectively, then by
definition, given ε > 0, there exists N ∈ N such that yn < y + ε, ∀n ≥ N .
From the hypothesis, we get
xn ≤ yn < y + ε, ∀n ≥ N .
Let (xnk ) be a subsequence of (xn) such that
n1 ≥ N and xnk → x as k → ∞.
As the constant sequence converges to the same constant, in view of Lemma
1.3.10, from xnk < y + ε, ∀k ∈ N, we infer that lim sup
n→∞
xn = lim
k→∞
xnk x ≤
y + ε. Since ε > 0 is arbitrary, we get
lim sup
n→∞
xn = x ≤ y = lim sup
n→∞
yn.
Similarly, we can prove the other inequality. �
THEOREM 1.3.19 Let (xn) and (yn) be sequences of positive real numbers.
If (xn) converges to a positive real number, then lim sup
n→∞
(xn · yn) = lim
n→∞ xn ·
lim sup
n→∞
yn.
Proof: Let x = lim
n→∞ xn and y = lim sup
n→∞
yn. First, we assume that y ∈ R.
Given ε ∈ (0, 1), we find m1, m2 ∈ N such that
|xn − x| < ε
1 + x + y
⇔ x − ε
1 + x + y
< xn < x + ε
1 + x + y
∀n ≥ m1
and
yn < y + ε
1 + x + y
, ∀n ≥ m2.
�
�
“book” — 2014/6/4 — 20:35 — page 31 — #31
�
�
�
�
�
�
Preliminaries 31
For n ≥ max{m1, m2}, we get
xn · yn < x · y + ε(x + y)
1 + x + y
+
(
ε
1 + x + y
)2
< x · y + ε(x + y)
1 + x + y
+ ε
1 + x + y
(as
ε
1 + x + y
< 1 )
< x · y + ε(1 + x + y)
1 + x + y
= x · y + ε.
If (ynk ) is a subsequence of (yn) converges to y, then using Theorem 1.3.9, we
get (xnk · ynk ) converges to x · y. Thus, x · y = lim sup
n→∞
(xn · yn).
If y = +∞, then there exists a subsequence (ynk ) of y such that ynk → ∞
as k → ∞. Let K > 0 be given, then by assumption, there exists j ∈ N such
that ynk >
2K
x , ∀k ≥ j. For ε = x
2 > 0, we choose m such that |xn − x| < x
2 ,
∀n ≥ m. Hence,
x − xn ≤ |xn − x| < x
2
⇒ xn >
x
2
, ∀n ≥ m.
Next, choose p ∈ N such that np ≥ m. Now, for k ≥ max{j, p}, we get
xnk · ynk >
x
2
· 2K
x
= K.
Hence, xnk · ynk → ∞ as k → ∞. As we have
xn · yn <∞+ ε, ∀ε > 0, ∀n ∈ N.
we get lim sup
n→∞
(xn · yn) = ∞ = x · y. �
THEOREM 1.3.20 Let (xn) be a sequence of positive real numbers, then
lim inf
n→∞
xn+1
xn
≤ lim inf
n→∞
n
√
xn ≤ lim sup
n→∞
n
√
xn ≤ lim sup
n→∞
xn+1
xn
Proof: Let α = lim sup
n→∞
n
√
xn and β = lim sup
n→∞
xn+1
xn
. If β = +∞, then
obviously we have α ≤ β. Therefore, we assume that β < +∞. Then by
definition of limit superior, for a given ε > 0, there exists m ∈ N such that
xn+1
xn
< β + ε, ∀n ≥ m. For n ≥ m, we have
xn = xn
xn−1
× xn−1
xn−2
× · · · × xm+1
xm
× xm < xm × (β + ε)n−m.
�
�
“book” — 2014/6/4 — 20:35 — page 32 — #32
�
�
�
�
�
�
32 Sequences and Series of Complex Numbers
which implies that n
√
xn <
(
xm
(β + ε)m
)1/n
(β + ε). If c = xm
(β + ε)m
, then we
get c1/n → 1 as n → ∞. (Because if c > 1, then c1/n = 1 + δn for some
δn > 0 ⇒ c = (1 + δn)n > nδn ⇒ δn <
c
n → 0 as n → ∞ ⇒ c1/n →
1 as n → ∞. If c < 1, then c−1 > 1 ⇒ c−1/n → 1 ⇒ c1/n → 1 as n → ∞).
Therefore, using Lemma 1.3.18, we get
α = lim sup
n→∞
n
√
xn ≤ β + ε.
As ε > 0 is arbitrary, we get
lim sup
n→∞
n
√
xn = α ≤ β = lim sup
n→∞
xn+1
xn
.
Similarly, we can prove that
lim inf
n→∞
xn+1
xn
≤ lim inf
n→∞
n
√
xn.
Using Lemma 1.3.16, we get
lim inf
n→∞
n
√
xn ≤ lim sup
n→∞
n
√
xn.
Thus, the theorem follows. �
Definition 1.3.21 Let fn : A → C, ∀n ∈ N, where A ⊆ C. We say that the
sequence (fn) of functions converges
1. point-wise to f on A, if for a given z ∈ A and a given ε > 0, there exists
mz ∈ N such that |fn(z) − f (z)| < ε, ∀n ≥ mz.
2. uniformly to f on A, if given ε > 0, there exists m ∈ N such that
|fn(z) − f (z)| < ε, ∀n ≥ m and ∀z ∈ A.
Definition 1.3.22 Let x0 ∈ [a, b] and f : (a, b) × (c, d) → C. We say that
f (x, y) → � as x → x0 uniformly in y on (c, d). If given ε > 0, then
there exists δ > 0 such that 0 < |x − x0| < δ ⇒ |f (x, y) − f (x0, y)| < ε,
∀y ∈ (c, d).
�
�
“book” — 2014/6/4 — 20:35 — page 33 — #33
�
�
�
�
�
�
Preliminaries 33
It is customary to define series of complex numbers by a formal sum
∞∑
n=1
an.
Before talking about its convergence, the sum has no meaning other than in
the expression
∞∑
n=1
an, first term is a1, second term is a2, third term is a3, and
so on. In the sequence (an) of complex numbers also, we say that first term
is a1, second term is a2, third term is a3, and so on. So what is the difference
between (an) and
∞∑
n=1
an? To find the difference, we can redefine the series of
complex numbers as follows.
Definition 1.3.23 (Rigorous Definition)
Let (an) be a sequence of complex numbers. We say that the sequence (an)
is summable if (Sn) converges, where Sn =
n∑
k=1
ak , ∀n ∈ N. In this case, we
call lim
n→∞ Sn the sum of (an) and is denoted by
∞∑
n=1
an. The sum
∞∑
n=1
an is also
called a series (or) an infinite series.
However, the old definition of a series (formal sum) is in usage for a long
time and most of the readers are accustomed with using the notation
∞∑
n=1
an
before discussing its convergence, a few similar less rigorous statements on
series are also used in this book, only to facilitate the reader
Definition 1.3.24 A series
∞∑
n=0
an of complex numbers is said to be conver-
gent if sm =
m∑
k=0
ak , ∀m ∈ N and sm → s as m → ∞ for some complex
number s. In this case, we call ‘s’ the sum of the series
∞∑
n=0
an and is also
denoted by the series itself.
Definition 1.3.25 A series
∞∑
n=0
an of complex numbers is said to be absolutely
convergent if the series
∞∑
n=0
|an| converges.
RESULT 1.3.26 A series
∞∑
n=0
an converges iff lim
m→∞
∞∑
n=m
an = 0.
�
�
“book” — 2014/6/4 — 20:35 — page 34 — #34
�
�
�
�
�
�
34 Sequences and Series of Complex Numbers
Proof:
∞∑
n=0
an converges to s iff sm → s as m → ∞, where sm = a0 + a1 +
a2 + · · · + am, ∀m ∈ N and s =
∞∑
n=0
an iff sm−1 − s → 0 as m → ∞ iff
lim
m→∞
∞∑
n=m
an = 0. �
The following result gives a necessary condition for a convergent series
of complex numbers.
RESULT 1.3.27 If
∞∑
n=0
an converges, then lim
n→∞ an = 0.
Proof: Let
∞∑
n=0
an converge. If
sm = a0 + a1 + a2 + · · · + am, ∀m ∈ N,
then (sm) is a convergent sequence and by Result 1.3.2, (sm) is a Cauchy
sequence. Let ε > 0 be given. Then, there exists m ∈ N such that
|sj − sk | < ε, ∀j, k ≥ m.
If n ≥ m + 1, then n, n − 1 ≥ m, and hence, |an| = |sn − sn−1| < ε. Hence,
(an) converges to 0. �
Definition 1.3.28 (Rearrangement of a series)
Let
∞∑
n=1
an be a given series. If f : N → N is a bijection, then
∞∑
n=1
af (n) is
called a rearrangement of
∞∑
n=1
an.
RESULT 1.3.29 If
∞∑
n=1
an converges to S absolutely, then every rearrangement
of
∞∑
n=1
an converges to the same S.
Proof: Let
∞∑
n=1
af (n) be a rearrangement of
∑
n=1
an, where f : N → N is a
bijection. Let Sm =
m∑
n=1
|an| and Tm =
m∑
n=1
af (n), ∀m ∈ N. Since (Sm) con-
verges, by Cauchy criterion (Theorem 1.3.11), we have (Sm) is a Cauchy
�
�
“book” — 2014/6/4 — 20:35 — page 35 — #35
�
�
�
�
�
�
Preliminaries 35
sequence and hence for a given ε > 0, there exists N0 ∈ N such that
|Sp − Sq| =
q∑
n=p
|an| < ε
2
∀q ≥ p ≥ N0.
Since
∞∑
n=1
an convergs to s, there exists N ∈ N such that N ≥ N0 and
∣∣∣∣∣
p∑
n=1
an − S
∣∣∣∣∣ ≤ ε
2
∀p ≥ N . (1.3)
Since f (N) = N we can choose M ∈ N such that {1, 2, . . . , N} ⊆ {f (n) : 1 ≤
n ≤ M}. Now for all p ∈ N with p ≥ M , we have
∣∣Tp − S
∣∣ =
∣∣∣∣∣
p∑
n=1
af (n) −
q∑
n=1
an
∣∣∣∣∣+
∣∣∣∣∣
q∑
n=1
an − S
∣∣∣∣∣ (where q ≥ p.)
≤
∑
nεA
|an| + ε
2
(using q ≥ p ≥ M ≥ N and (1.3))
where A ⊆ N such that K ≤ n ≤ J ,∀n ε A for some K, J ε N
with N ≤ K ≤ J .
≤
J∑
K
|an| + ε
2
<
ε
2
+ ε
2
= ε,
because by the choice of M , we have N ≤ K ≤ J . Thus
∞∑
n=1
af (n) converges
to S. �
THEOREM 1.3.30 (Comparison test)
If (an) is a sequence of complex numbers and (bn) is a sequence of non-
negative real numbers such that |an| ≤ bn, ∀n ≥ m, for some m ∈ N and
∞∑
n=0
bn converges, then
∞∑
n=0
an converges absolutely.
Proof: For each n ∈ N, let
sn = |a0| + |a1| + · · · + |an| and tn = b0 + b1 + · · · + bn.
By assumption, (tn) is a convergent sequence, and hence, (tn) is a Cauchy
sequence. Then, for a given ε > 0,there exists p ∈ N such that
|tj − tk | < ε, ∀j, k ≥ p.
�
�
“book” — 2014/6/4 — 20:35 — page 36 — #36
�
�
�
�
�
�
36 Some Topological Properties of the Complex Plane
Now for the same p, if j > k ≥ p, then we have
|sj − sk | =
j∑
n=k+1
|an| ≤
j∑
n=k+1
bn = |tj − tk | < ε.
Hence, it follows that |sj − sk | < ε whenever j, k ≥ p. Therefore, (sn)
converges, by Cauchy’s criterion, and hence,
∞∑
n=0
an converges absolutely. �
Example 1.3.31 If z ∈ C with |z| < 1, then the geometric series
∞∑
n=0
zn
converges.
If z = 0, then obviously the series converges to 0. For z � 0, then we have
0 < |z| < 1. Hence, we write |z| = 1
1 + r
, where r = 1
|z| − 1 > 0. For every
n > 2, since
(1 + r)n > nC1r = nr
we get
0 < |z|n = 1
(1 + r)n
<
1
nr
→ 0 as n → ∞.
If sn = 1 + z + z2 + · · · + zn, then
sn(1 − z) = (1 + z + z2 + · · · + zn)(1 − z) = 1 − zn+1,∀n ∈ N.
Therefore, it follows that
sn = 1 − zn+1
1 − z
→ 1
1 − z
as n → ∞
and hence,
∞∑
n=0
zn = 1
1 − z
.
1.4 SOME TOPOLOGICAL PROPERTIES OF THE
COMPLEX PLANE
Although we assume that the reader is familiar with mathematical analy-
sis, we briefly recall some important definitions and results on topology of
complex plane, which will be required in the following sequel.
�
�
“book” — 2014/6/4 — 20:35 — page 37 — #37
�
�
�
�
�
�
Preliminaries 37
Definition 1.4.1 The standard metric d : C×C → R is defined by d(z, w) =
|z − w|, ∀z, w ∈ C.
From the properties of | · |, we obtain the following:
1. |z − w| ≥ 0, ∀z, w ∈ C,
2. |z − w| = |w − z|, ∀z, w ∈ C,
3. |z − w| = 0 iff z = w,
4. |z − w| ≤ |z − ζ | + |ζ − w|, ∀z, ζ , w ∈ C.
Definition 1.4.2 For a given z0 ∈ C and r > 0, by a neighbourhood of
z0 or by the open ball with center z0 and radius r > 0, we mean the set
B(z0, r) = {z ∈ C : |z − z0| < r}.
Definition 1.4.3 Let U ⊆ C and a ∈ C. We say that a is
1. a limit point of U if for every r > 0, we have (B(a, r) \ {a}) ∩ U � ∅.
2. an interior point of U if there exists ε > 0 such that B(a, ε) ⊆ U .
Definition 1.4.4 A subset U of C is said to be
1. an open set if every point of U is an interior point of U .
2. a closed set if every limit point of U is a point of U .
RESULT 1.4.5 Let U ⊆ C. U is an open set iff Uc is a closed set.
Proof: U is open iff every point of U is an interior point of U
iff for every x ∈ U , there exists rx > 0 such that
B(x, rx) ⊆ U
iff for every x � Uc, there exists rx > 0 such that
B(x, rx) ∩ Uc = ∅
iff for every x � Uc, there exists rx > 0 such that
(B(x, rx) \ {x}) ∩ Uc = ∅
iff every x � Uc is not a limit point of Uc
iff every limit point of Uc is a point of Uc
iff Uc is closed.
Hence, the result. �
�
�
“book” — 2014/6/4 — 20:35 — page 38 — #38
�
�
�
�
�
�
38 Some Topological Properties of the Complex Plane
THEOREM 1.4.6 Let Aα , Aj ⊆ C for all α ∈ � and for all j ∈ {1, 2, . . . , n},
where � is an arbitrary index set.
(i) If Aα is open ∀α ∈ �, then
⋃
α∈�
Aα is open.
(ii) If Aj is open ∀j ∈ {1, 2, . . . , n}, then
n⋂
j=1
Aj is open.
Proof: To prove
⋃
α∈�
Aα is open, let x ∈ ⋃
α∈�
Aα be arbitrary. Then x ∈ Aβ
for some β ∈ �. As Aβ is open, there exists r > 0 such that B(x, r) ⊆ Aβ ⊆⋃
α∈�
Aα . Thus,
⋃
α∈�
Aα is open.
Next, if x ∈
n⋂
j=1
Aj is arbitrary, then x ∈ Aj, ∀j ∈ {1, 2, . . . , n}. Since each
Aj is open, there exists rj > 0 such that B(x, rj) ⊆ Aj, ∀j ∈ {1, 2, . . . , n}. If
s = min{rj : 1 ≤ j ≤ n}, then s > 0, and hence, B(x, s) ⊆ B(x, rj) ⊆ Aj,
∀j ∈ {1, 2, . . . , n} ⇒ B(x, s) ⊆
n⋂
j=1
Aj. Thus,
n⋂
j=1
Aj is open. �
THEOREM 1.4.7 Let Bα , Bj ⊆ C for all α ∈ � and for all j ∈ {1, 2, . . . , n},
where � is an arbitrary index set.
(i) If Bα is closed ∀α ∈ �, then
⋂
α∈�
Bα is closed.
(ii) If Bj is closed ∀j ∈ {1, 2, . . . , n}, then
n⋃
j=1
Bj is closed.
Proof: Proof of this theorem follows from Theorem 1.4.5, De-Morgan’s laws,
and Theorem 1.4.6. �
Definition 1.4.8 Let A ⊂ C. Define closure Cl A of A by A ∪ A′, where A′ is
the set of all limit points of A.
LEMMA 1.4.9 x ∈ Cl A iff B(x, r) ∩ A � ∅ for every r > 0.
Proof of this lemma follows directly from the definition of Cl A.
RESULT 1.4.10 The closure operator satisfies the following properties:
�
�
“book” — 2014/6/4 — 20:35 — page 39 — #39
�
�
�
�
�
�
Preliminaries 39
1. A ⊆ B ⇒ Cl A ⊆ Cl B.
2. A is closed iff A = Cl A.
3. Cl A is the smallest closed set containing A.
4. Cl Cl A = Cl A.
5. Cl (A ∪ B) = (Cl A) ∪ (Cl B).
6. Cl (A ∩ B) ⊆ (Cl A) ∩ (Cl B).
Proof:
1. If A ⊂ B, then using the previous lemma, we get
x ∈ Cl A ⇒ B(x, r) ∩ A � ∅, ∀r > 0
⇒ B(x, r) ∩ B � ∅, ∀r > 0
⇒ x ∈ Cl B.
2. A = Cl A ⇔ A = A ∪ A′ ⇔ A′ ⊆ A ⇔ A is closed.
3. Let x � Cl A. Then by previous lemma, there exists r > 0 such that
B(x, r)∩A = ∅. For every y ∈ B(x, r), there exists sy = r− |x− y| > 0
such that B(y, sy) ⊆ B(x, r), and hence, B(y, sy) ∩ A = ∅. Hence, it
follows that no point of B(x, r) is a point of Cl A by the same lemma.
Therefore, B(x, r) ⊆ (Cl A)c. Thus, we have proved that (Cl A)c is an
open set, and hence, Cl A is a closed set.
Next if B is a closed subset of C such that A ⊂ B, then using the
properties (1) and (2), we have Cl A ⊆ Cl B. Therefore, Cl A ⊆ Cl
B = B. Thus, Cl A is the smallest closed set containing A.
4. Using the property (3), Cl A is a closed set, and by using the property
(2), we get Cl Cl A = Cl A.
5. Clearly, A ⊆ Cl A and B ⊆ Cl B imply that A ∪ B ⊆ (Cl A) ∪ (Cl B).
As (Cl A) ∪ (Cl B) is a closed set (being a union of two closed sets)
containing A ∪ B and Cl (A ∪ B) is the smallest closed set containing
A∪B, we get Cl (A∪B) ⊆ (Cl A)∪ (Cl B). On the other hand, using the
property (1), A ⊆ A ∪ B and B ⊆ A ∪ B imply that Cl A ⊆ Cl (A ∪ B)
and Cl B ⊆ Cl (A ∪ B). Therefore, (Cl A) ∪ (Cl B) = Cl (A ∪ B). Thus,
the property (5) holds.
�
�
“book” — 2014/6/4 — 20:35 — page 40 — #40
�
�
�
�
�
�
40 Some Topological Properties of the Complex Plane
6. Since A∩B ⊆ A and A∩B ⊆ B, using the property (1), we have Cl (A∩
B) ⊆ Cl A and Cl (A ∩ B) ⊆ Cl B. Thus, Cl (A ∩ B) ⊆ (Cl A) ∩ (Cl B).
�
Example 1.4.11 For every a ∈ C and r > 0, Cl B(a, r) = {x ∈ C :
|x − a| ≤ r}.
Suppose y � {x ∈ C : |x−a| ≤ r}. Then, |y−a| > r. If s ∈ R is such that 0 <
s < |y−a|−r, then we claim that B(y, s)∩B(a, r) = ∅. If z ∈ B(y, s)∩B(a, r),
then |z − y| < s and |z − a| < r which implies that
|y − a| ≤ |y − z| + |z − a| < s + r < |y − a|.
This is not possible. Hence, B(y, s)∩B(a, r) = ∅. Therefore, y � Cl B(a, r) by
Lemma 1.4.9. Conversely, if y � Cl B(a, r), then there exists t > 0 such that
B(y, t) ∩ B(a, r) = ∅. As y − t
2
y−a
|y−a| ∈ B(y, t), we have y − t
2
y−a
|y−a| � B(a, r),
and hence by applying |y − a| > t ⇒ 0 < t
2|y−a| < 1, we get
r ≤
∣∣∣∣y − t
2
y − a
|y − a| − a
∣∣∣∣
= |y − a|
∣∣∣∣1 − t
2|y − a|
∣∣∣∣
= |y − a|
(
1 − t
2|y − a|
)
< |y − a|.
Therefore, y � {x ∈ C : |x − a| ≤ r}, and hence, Cl B(a, r) = {x ∈ C :
|x − a| ≤ r}.
Example 1.4.12 There exists A ⊂ C and B ⊂ C such that Cl (A ∩ B) �
(Cl A) ∩ (Cl A).
We consider A = B(1, 1) and B = B(2, 1). Clearly, Cl (A ∩ B) = Cl ∅ = ∅
but (Cl A) ∩ (Cl B) = {x ∈ C : |x − 1| ≤ 1} ∩ {x ∈ C : |x − 2| ≤ 1} = {1}.
Exercise 1.4.13 Prove that x ∈ Cl A iff there exists a sequence (xn) in A such
that xn → x as n → ∞.
Exercise 1.4.14 If A is a bounded subset of R, then prove that sup A, inf A ∈
Cl A.
�
�
“book” — 2014/6/4 — 20:35 — page 41 — #41
�
�
�
�
�
�
Preliminaries 41
Definition 1.4.15 A subset A of R is dense in R if Cl A = R. Similarly, if
A ⊆ C with Cl A = C, then A is called a dense subset of C.
Example 1.4.16 The set of all rational numbers is dense in R.
To justify this example, we shall show that every point of R is a limit
point of Q. Let r > 0 be given, then using the observation 3(l) in page 6, we
choose q ∈ Q such that x < q < x + r. Therefore, q ∈ [B(x, r) \ {x}] ∩ Q.
Thus, R ⊆ Q′ ⊆ Cl Q.
Definition 1.4.17 Let A ⊂ C. The interior Int A of A is defined by the set of
all interior points of A.
Exercise 1.4.18 Let A, B ⊆ C. Then prove that
(i) A ⊆ B ⇒ Int A ⊆ Int B.
(ii) A is open ⇔ A= Int A.
(iii) Int A is the largest open set contained in A.
(iv) Int Int A = A.
(v) Int (A ∩ B) = Int A ∩ Int B.
(vi) Int (A ∪ B) ⊇ Int A ∪ Int B.
Definition 1.4.19 Let A, B ⊂ C. A and B are said to be separated if (Cl A) ∩
B = ∅ = A ∩ (Cl B).
Definition 1.4.20 Let S be a subset of C. S is said to be connected if S cannot
be a subset of union of two separated sets A and B such that A ∩ S � ∅ and
B ∩ S � ∅.
Definition 1.4.21 (Interval) A subset I of R is said to be an interval if z ∈ I
whenever x, y ∈ I and x < z < y.
There are four types of intervals, which are as follows:
1. [a, b] = {x ∈ R : a ≤ x ≤ b}.
2. [a, b) = {x ∈ R : a ≤ x < b}.
3. (a, b] = {x ∈ R : a < x ≤ b}.
4. (a, b) = {x ∈ R : a < x < b}.
�
�
“book” — 2014/6/4 — 20:35 — page 42 — #42
�
�
�
�
�
�
42 Some Topological Properties of the Complex Plane
THEOREM 1.4.22 A subset S of R is connected iff S is an interval.
Proof: Let S be a connected subset of R. Suppose S is not an interval, then
there exist x, y ∈ S and z ∈ R such that x < z < y but z � S.
Put A = (−∞, z) and B = (z,∞). Clearly, it follows that
S ⊆ A ∪ B, A ∩ S � ∅ and B ∩ S � ∅.
As
A ∩ (Cl B) = (−∞, z) ∩ [z,∞) = ∅ = (−∞, z] ∩ (z,∞) = (Cl A) ∩ B,
we get that S is disconnected, which is a contradiction. Therefore, S is an
interval.
Conversely, suppose S is an interval and if there exist subsets A, B of R such
that
S ⊆ A ∪ B, S ∩ A � ∅ � S ∩ B and (Cl A) ∩ B = ∅ = (Cl B) ∩ A.
As S ∩ A � ∅ � S ∩ B, we choose α ∈ S ∩ A and γ ∈ S ∩ B. Without loss of
generality, we assume that α < γ , since α � γ . If β = sup([α, γ ] ∩ A), then
α ≤ β ≤ γ and α ∈ Cl A. As α, γ ∈ S and S is an interval, we get β ∈ S and
hence,
β ∈ A or β ∈ B (1.4)
Using β ∈ Cl A and (Cl A) ∩ B = ∅, we obtain β � B. Then, β ∈ A, and
hence, β � Cl B. Hence, (β, γ ) � B; otherwise, β ∈ [β, γ ] = Cl (β, γ ) ⊆
Cl B, which is not possible. Therefore, there exists δ ∈ (β, γ ) such that δ � B.
From the definition of β and from the inequality β < δ, it follows that δ � A.
Thus, δ � S, as S ⊆ A∪B. This implies that S is not an interval, as α < δ < γ
and α, γ ∈ S. This is a contradiction, and hence S is connected. �
COROLLARY 1.4.23 The set R of all real numbers is a connected set.
Proof: As R itself is obviously an interval, it is connected by the previous
theorem. �
Definition 1.4.24 (Line segment) Let a, b be two points in C. We call the set
La,b = {(1 − t)a + tb : t ∈ [0, 1]} the line segment joining a and b. The point
a is called the initial point of the line segment and b is called the end point of
the line segment.
�
�
“book” — 2014/6/4 — 20:35 — page 43 — #43
�
�
�
�
�
�
Preliminaries 43
Definition 1.4.25 (Polygon) Let {ai : i = 0, 1, 2, . . . , n} be a finite set of
complex numbers. The union
n∪
i=1
Lai−1,ai of the given finite number of line
segments of the form Lai−1,ai , i = 1, 2, . . . , n is called a polygon.
THEOREM 1.4.26 If E is a subset of C such that every pair of points a, b can
be joined by a polygon Pa,b ⊆ E, then E is a connected subset of C.
Proof: Suppose E is disconnected, then there exists a pair of sets A, B in C
such that E ⊆ A∪B and A∩E � ∅ � E∩B and (Cl A)∩B = A∩ (Cl B) = ∅.
Choose a ∈ E ∩ A and b ∈ E ∩ B. By hypothesis, there exists Pa,b ⊆ E.
On examining the line segments from a to b, we can find a line segment
l = Lc,d ⊆ Pa,b such that c ∈ A and d ∈ B. Put
A1 = {t ∈ [0, 1] : (1 − t)c + td ∈ A ∩ E}
A2 = {t ∈ [0, 1] : (1 − t)c + td ∈ B ∩ E}
As the line segment l ⊆ E ⊆ A ∪ B, we get [0, 1] = A1 ∪ A2 by an easy
verification. We note that
c ∈ A ∩ E ⇒ 0 ∈ A1 ⇒ A1 � ∅
and
d ∈ B ∩ E ⇒ 1 ∈ A2 ⇒ A2 � ∅
Claim. (Cl A1) ∩ A2 = A1 ∩ (Cl A2) = ∅.
Suppose t ∈ (Cl A1) ∩ A2. Then, there exists a sequence {tn} from A1
such that tn → t as n → ∞. Therefore, (1 − tn)c + tnd ∈ A, ∀n ∈ N,
and so,
(1 − t)c + td = lim
n→∞(1 − tn)c + tnd ∈ Cl A
Clearly, (1 − t)c + td ∈ B because t ∈ A2. Similarly, if s ∈ A1 ∩ (Cl A2),
then we can show that (1 − s)c + sd ∈ A ∩ (Cl B). Therefore, (Cl A1) ∩ A2 =
A1 ∩ (Cl A2) = ∅. This gives a separation for the interval [0, 1], which is a
contradiction. �
Immediately, we get the following corollaries.
COROLLARY 1.4.27 Every line segment is connected in C.
COROLLARY 1.4.28 Every polygon is connected.
COROLLARY 1.4.29 In C, every open disc B(a, r) = {x ∈ C : |x − a| < r} is
connected.
�
�
“book” — 2014/6/4 — 20:35 — page 44 — #44
�
�
�
�
�
�
44 Some Topological Properties of the Complex Plane
Proof: Let x, y ∈ B(a, r). We shall show that Lx,y = {(1 − t)x + ty : t ∈
[0, 1]} ⊆ B(a, r).
Now, if t ∈ [0, 1], then
|(1 − t)x + ty − a| = |(1 − t)x + ty − (1 − t)a + ta|
≤ (1 − t)|x − a| + t|y − a|
< (1 − t)r + tr = r
and hence, we get (1 − t)x + ty ∈ B(a, r). Hence, the corollary follows from
the above theorem. �
Definition 1.4.30 An open connected subset of C is called a region.
THEOREM 1.4.31 If E is a region, then given a pair of points x, y of E, there
exists a polygon Qx,y joining x and y such that Qx,y ⊆ E and whose line
segments are parallel to the coordinate axes.
Proof: Throughout this proof, by Qs,t we mean a polygon joining s and t
such that Qs,t ⊆ E and whose line segments are parallel to the coordinate
axes. Assume that E is connected. Fix a ∈ E arbitrarily. Define
A = {x ∈ E : there exists a polygon Qx,a ⊆ E} and B = E \ A.
Clearly, E = A ∪ B and A � ∅. To prove this theorem, we shall show that
E = A. (Because if it is proved and if x, y ∈ E is arbitrary, then there exist
Qx,a, Qy,a ⊆ E, and hence, their union is a polygon in E joining x and y.)
Claim. (Cl A) ∩ B = ∅.
If x ∈ (Cl A)∩B, then there exists r > 0 such that B(x, r) ⊆ E (as x ∈ E
and E is open). Then, there exists y ∈ A ∩ B(x, r), using x ∈ Cl A. If
x = (x1, x2) and y = (y1, y2), then let z = (x1, y2). From
|x − z| = |(x1, x2) − (x1, y2)| = |x2 − y2| ≤ |x − y| < r
we get z ∈ B(x, r).
�
�
“book” — 2014/6/4 — 20:35 — page 45 — #45
�
�
�
�
�
�
Preliminaries 45
(x1, y2)
(y1, y2)
(x1, x2)
By applying the proof of Corollary 1.4.29 twice, we get Lx,z, Lz,y ⊆
B(x, r) ⊆ E, and hence, Px,y = Lx,z ∪ Lz,y is completely contained in E.
Since y ∈ A, there exists Py,a ⊂ E and hence Px,a = Px,y ∪ Py,a ⊆ E.
Thus, x ∈ A. This contradicts the fact that A ∩ B = ∅. Therefore,
(Cl A) ∩ B = ∅.
Claim. A ∩ (Cl B) = ∅.
If a ∈ A ∩ (Cl B), then choose s > 0 such that B(a, s) ⊆ E. As a ∈
Cl B there exists b ∈ B ∩ B(a, s). By a similar argument used in the
justification of previous claim, we can find a polygon Pa,b ⊆ E. This
implies that b ∈ A, which is a contradiction.
Hence, B = ∅, and hence, A = E. This completes the proof of the
theorem. �
Example 1.4.32 There exists a connected subset E of C such that it contains
a pair of points that cannot be joint by a polygon inside E.
If E = {z ∈ C : |z| = 1}, then E is connected because E = f ([0, 2π ]) where
f (t) = eit, t ∈ [0, 2π]; at this level, assume that the equality holds and the
fact that f is continuous. However, for any two distinct points z and w of E,
no point of Lz,w other than z and w belongs to E.
To justify the last statement, let z, w ∈ E be such that z � w. Then, there
exist θ1, θ2 ∈ [0, 2π) such that z = eiθ1 and w = eiθ2 so that 0 < |θ1 − θ2| <
2π . Hence, for any t ∈ (0, 1), (1 − t)ztw = (1 − t)tei(θ1−θ2), which is not
real and non-negative, and hence, |(1 − t)z + tw| < |(1 − t)z| + |tw| =
(1 − t)|z| + t|w| ≤ (1 − t) + t = 1. Thus, (1 − t)z + tw � E. �
�
�
“book” — 2014/6/4 — 20:35 — page 46 — #46
�
�
�
�
�
�
46 Some Topological Properties of the Complex Plane
Definition 1.4.33 Let S ⊆ C. A subset A of S is said to be a component of S
if A is a largest connected subset of C contained in S.
That is, there is no connected subset B of C such that A � B ⊆ S.
RESULT 1.4.34 The collection of all components of a set S is a partition of S.
That is, the union of all components of S is S, and components are pairwise
disjoint.
Proof: For each, x ∈ S, let CS[x] be the union of all connected subsets of S
containing x. Clearly, this union is non-empty as {x} is a connected subset ofS containing x. We claim that CS[x] is connected. Suppose, CS[x] ⊆ A ∪ B,
with CS[x] ∩ A � ∅, CS[x] ∩ B � ∅ and A ∩ Cl B = ∅ and B ∩ Cl A = ∅.
Let a ∈ A and b ∈ B. Then, there exist connected subsets Ca and Cb of S
such that a, x ∈ Ca ⊆ S and b, x ∈ Cb ⊆ S. Therefore, Ca ⊆ A ∪ B and
Cb ⊆ A ∪ B. Since Ca is a connected set either Ca ⊆ A or Ca ⊆ B, otherwise
Ca ∩ A � ∅ and Ca ∩ B � ∅, and hence, Ca becomes a disconnected set,
which is a contradiction. As a ∈ A∩Ca, we conclude that Ca ⊆ A. Similarly,
we conclude that Cb ⊆ B. Therefore, x ∈ Ca ∩ Cb ⊆ A ∩ B ⊆ A ∩ Cl B = ∅,
which is a contradiction. Hence, CS[x] is a connected set. Next we observe
that CS[x] cannot be properly contained in a connected subset of S because
if K is a connected subset of S containing CS[x], then x ∈ K, and hence by
definition of CS[x], we have K ⊆ CS[x] which implies CS[x] = K. Thus,
CS[x] is a component of S.
As for every x ∈ S, there exists a component CS[x] of S containing x;
to conclude this theorem, we show that any two component of S are either
identical or disjoint. If CS[x] and CS[y] are two components of S such that
CS[x] ∩ CS[y] � ∅, then let z ∈ CS[x] ∩ CS[y]. If CS[z] is the component of
S containing z, then CS[x] ⊆ CS[z], since CS[x] is a connected subset of S
containing z. Now, x ∈ CS[x] ⊆ CS[z] imply that CS[z] is a connected subset
of S containing x, and hence, CS[z] ⊆ CS[x] ⇒ CS[x] = CS[z]. Similarly, we
show that CS[y] = CS[z]. Thus, CS[x] = CS[z] = CS[y]. �
RESULT 1.4.35 Every non-empty connected subset A of a set S is contained
in one and only one component of S.
Proof: Let A be a non-empty connected subset S. Then, there exists x ∈ A ⊆
S. If CS[x] is the component of S containing x, then obviously CS[x] is a
component containing A, as CS[x] is the union of all connected subsets of S
containing x. Since the components of S are pairwise disjoint (by previous
result), CS[x] is the unique component of S containing A. �
�
�
“book” — 2014/6/4 — 20:35 — page 47 — #47
�
�
�
�
�
�
Preliminaries 47
Exercise 1.4.36 Let E ⊆ C. Then prove the following statements:
1. If E is written as a disjoint union of two non-empty open subsets of C,
then E is not connected.
2. If E is written as a disjoint union of two non-empty closed subsets of
C, then E is not connected.
Definition 1.4.37 A subset K of C is called a compact set if for a given
collection {Eα : α ∈ I} of open subsets of C such that K ⊂ ∪
α∈I
Eα , then
there exists α1,α2, . . . ,αn ∈ I such that K ⊂ n∪
j=1
Eαj .
THEOREM 1.4.38 Let K be a compact subset of C. Then K is closed and
bounded.
Proof: Let x � K. Then, for every y ∈ K, we have y � x. If ry = |x − y|
2
,
then ry > 0 and B(x, ry) ∩ B(y, ry) = ∅, ∀y ∈ K. As {B(y, ry) : y ∈ K}
is a collection of open sets satisfying K ⊂ ∪
y∈K
B(y, ry). Hence, there exist
y1, y2, . . . , ym ∈ K such that K ⊂ m∪
k=1
B(yk , ryk ). If
r = min{ryk : 1 ≤ k ≤ m}
then
B(x, r) ∩ K ⊆ B(x, r) ∩ m∪
k=1
B(yk , ryk ) ⊆ m∪
k=1
B(x, ryk ) ∩ B(yk , ryk ) = ∅
and hence we get B(x, r) ⊆ Kc. Thus, Kc is an open set, and hence, K is
closed.
Consider the collection of open sets {B(x, 1) : x ∈ K} with K ⊂
∪
x∈K
B(x, 1). By definition, there exist x1, x2, . . . , xn ∈ K such that K ⊂
n∪
j=1
B(xj, 1). If M = max{|xj| : j ∈ {1, 2, . . . , n}}, then we claim that
|x| ≤ M + 1, ∀x ∈ K. If x ∈ K, then x ∈ B(xj, 1) for some 1 ≤ j ≤ n.
Therefore,
|x| ≤ |x − xj| + |xj| ≤ 1 + M .
Hence, K is bounded. �
�
�
“book” — 2014/6/4 — 20:35 — page 48 — #48
�
�
�
�
�
�
48 Some Topological Properties of the Complex Plane
THEOREM 1.4.39 Every infinite subset of a compact set K has a limit point
in K.
Proof: Let S be an infinite subset of K. If S has no limit point in K, then
for each z ∈ K, there exists rz > 0 such that B(z, rz) ∩ S ⊆ {z}. Obviously,
{B(z, rz) : z ∈ K} is a collection of open sets such that ∪
z∈K
B(z, rz) = K. Since
K is compact, there exists z1, z2, . . . , zn ∈ K such that
n∪
j=1
B(zn, rzn ) = K, and
hence, S = S ∩ K = S ∩ n∪
j=1
B(zn, rzn) = n∪
j=1
S ∩ B(zn, rzn ) ⊆ {z1, z2, . . . , zn},
which is a contradiction to the assumption on S. �
THEOREM 1.4.40 Closed subset of a compact set is compact.
Proof: Let K be a compact subset of C and F ⊂ K be closed. If {Eα : α ∈ I}
is a collection of open sets such that F ⊆ ∪
α∈I
Eα , then
K ⊆ C = F ∪ Fc ⊆ ∪
α∈I
Eα ∪ Fc.
As F is closed, {Eα , Fc : α ∈ I} is a collection of open sets; therefore, using
the compactness of K, we get α1,α2, . . . ,αn ∈ I such that K ⊆ Eα1 ∪ Eα2 ∪
· · · ∪Eαn ∪Fc. Thus, F ⊆ Eα1 ∪Eα2 ∪ · · · ∪Eαn , and hence, F is compact. �
THEOREM 1.4.41 Let A be a closed subset of C and B be a compact subset
of C. If A ∩ B = ∅, then inf{|z − w| : z ∈ A, w ∈ B} > 0.
Proof: Since A ∩ B = ∅, w is not a limit point of A, ∀w ∈ B. There-
fore, there exists rw > 0 such that B(w, rw) ∩ A = ∅, ∀w ∈ B. As{
B
(
w, rw
2
)
: w ∈ B
}
is a collection of open sets such that B ⊆ ∪
w∈B
B
(
w, rw
2
)
,
there exist wk ∈ B, k = 1, 2, . . . , n such that B ⊆ n∪
k=1
B
(
wk ,
rwk
2
)
. Let
μ = min
{
rwk
2 : k = 1, 2, . . . , n
}
, then μ > 0. We claim that inf{|z − w| :
z ∈ A, w ∈ B} ≥ μ. Otherwise, there exist z ∈ A and w ∈ B such that
|z−w| < μ. Using w ∈ B, we find j ∈ {1, 2, . . . , n} such that w ∈ B
(
wj,
rwj
2
)
.
Then we have |w − wj| <
rwj
2 . Now,
|z − wj| ≤ |z − w| + |w − wj| < μ+ rwj
2
≤ rwj
�
�
“book” — 2014/6/4 — 20:35 — page 49 — #49
�
�
�
�
�
�
Preliminaries 49
which implies that z ∈ B
(
wj, rwj
)
. This is a contradiction to B
(
wj, rwj
)∩A =
∅. Hence, the theorem follows. �
THEOREM 1.4.42 (Cantor’s intersection theorem)
If (Kn) is a sequence of non-empty compact subsets of C such that Kn ⊇
Kn+1, ∀n ∈ N, and D(Kn) → 0 as n → ∞, then
∞∩
n=1
Kn is a singleton set,
where D(Kn) = sup{|z − w| : z, w ∈ Kn}, ∀n ∈ N.
Proof: First, we show that
∞∩
n=1
Kn has at most one point. If x, y ∈ ∞∩
n=1
Kn, then
x, y ∈ Kn, ∀n ∈ N. Then, 0 ≤ |x − y| ≤ D(Kn) → 0 as n → ∞, and hence,
x = y.
Suppose that
∞∩
n=1
Kn = ∅. Then K1 ∩
∞∩
n=2
Kn = ∅ ⇒ K1 ⊆
( ∞∩
n=2
Kn
)c
=
∞∪
n=2
Kc
n. As each Kn is a closed set (Theorem 1.4.38), Kc
n is open, ∀n ≥ 2.
Hence, by the compactness of K1, we have K1 ⊆
(
m∪
j=1
Kc
nj
)
=
(
m∩
j=1
Knj
)c
for
some 2 ≤ n1 < n2 < · · · < nm. Thus, Knm = K1 ∩
(
m∩
j=1
Knj
)
= ∅, which is
a contradiction. Hence, the theorem follows. �
THEOREM 1.4.43 (Heine-Borel theorem)
Every closed and bounded interval in R is compact.
Proof: Let I = [a, b] be the given interval, where a, b ∈ R with a < b.
Suppose there exists a collection C of open sets in R such that [a, b] ⊆ ∪
E∈C
E
and I is not contained in the union of any finite sub-collection of C . Let
c1 = a+b
2 be the mid-point of a and b. Then at least one of the two intervals
[a, c1] and [c1, b] is not contained in the union of any finite sub-collection
of C . Choose such an interval and denote it by I1 = [a1, b1]. If D(I) is the
diameter of I , then D(I) = b−a and D(I1) = b1 −a1 = b−a
2 . Proceeding like
this, we can find a sequence {In} of intervals with the following properties:
1. In+1 ⊆ In, ∀n ∈ N.
2. In is not contained in the union of any finite sub-collection of C .
3. D(In+1) = 1
2 D(In) = 1
2n D(I), ∀n ∈ N.
�
�
“book” — 2014/6/4 — 20:35 — page 50 — #50
�
�
�
�
�
�
50 Some Topological Properties of the Complex Plane
If In = [an, bn], ∀n ∈ N, then we have
an ≤ am+n ≤ bm+n ≤ bn,∀m, n ∈ N.
Hence, it follows that for each m ∈ N, bm is an upper bound of {an : n ∈ N}.
Using the fact that R has least upper bound property, there exists α ∈ R such
that α = sup{an : n ∈ N}. Then immediately, we have an ≤ α ≤ bm, ∀m, n ∈
N. Thus, α ∈ ∞∩
n=1
In. As α ∈ I , there exists E ∈ C such that α ∈ E. Since E
is open, there exists r > 0 such that B(α, r) ⊆ E. As
{
1
2n D(I)
}
converges to
0, there exists m ∈ N such that 1
2m D(I) < r. We now claim that Im ⊆ E. If
x ∈ Im, then |x−α| ≤ 1
2m D(I) < r, because x,α ∈ Im. Thus, x ∈ B(α, r) ⊆ E.
Hence, our claim follows. However, this is contradiction to a property (2) of
Im.Hence, [a, b] is compact. �
COROLLARY 1.4.44 Every closed and bounded subset of R is compact.
Proof: Let K be a closed and bounded subset of R. Since K is bounded,
there exists M > 0 such that |x| ≤ M , ∀x ∈ K. Therefore, K ⊆ [−M , M].
By previous theorem, [−M , M] is compact. As K is a closed subset of the
compact set [−M , M], K is compact. �
Exercise 1.4.45 Every closed and bounded subset of C is compact.
Hint: First prove that [a, b] × [c, d] is compact as in the proof of Heine–
Borel theorem. Next find positive real numbers M1 and M2 such that K ⊂
[−M1, M1]× [−M2, M2] and complete the remaining proof as in the proof of
previous corollary.
THEOREM 1.4.46 Every Cauchy sequence in C is a convergent sequence.
Proof: Let (zn) be a Cauchy sequence in C. If zn = xn + iyn, ∀n ∈ N, then
|xn − xm| ≤ |zn − zm| and |yn − ym| ≤ |zn − zm|, ∀n ∈ N.
Therefore, (xn) and (yn) are Cauchy sequences of real numbers. Then by
Result 1.3.3, (xn) and (yn) are bounded sequences. Therefore, we can find
M > 0 such that xn ∈ [−M , M], ∀n ∈ N. If (xn) has a subsequence as a
constant sequence, then obviously the subsequence is convergent (Example
1.3.5), and by Lemma 1.3.8, (xn) itself is convergent. If (xn) has no such
subsequence, then {xn : n ∈ N} is an infinite set of [−M , M], which is a
�
�
“book” — 2014/6/4 — 20:35 — page 51 — #51
�
�
�
�
�
�
Preliminaries 51
compact set by Heine–Borel theorem. Now applying Theorem 1.4.39, we get
{xn : n ∈ N} has a limit point x in [−M , M]. Hence, for each k ∈ N, there
exists xnk ∈ B(x, 1/k), with nk > nj whenever k > j. Therefore, (xn,k) is
a convergent subsequence of (xn). Again by using Theorem 1.4.39, we get
(xn) is a convergent sequence. By the same argument, we can prove that (yn)
is a convergent sequence. Thus, by Theorem 1.3.9 (2), we get that (zn) is a
convergent sequence. �
The above theorem can be rephrased as that C is a complete metric space.
THEOREM 1.4.47 (Bolzano–Weierstrass property) Every bounded infinite
subset of C has a limit point.
Proof of this theorem is a consequence of Exercise 1.4.45 and Theorem
1.4.39.
THEOREM 1.4.48 The set C of all complex numbers is second countable.
That is, there exists a countable collection B of open subsets of C such that
for every open subset G of C and z ∈ G, there exists B ∈ B such that
z ∈ B ⊆ G.
Proof: We know that the set of all rational numbers is countable and dense
in R. Therefore, the collection
B = {
B(wj, q) : wj ∈ Q × Q, q ∈ Q ∩ (0,∞)
}
is also a countable collection of open subsets of C. For z = x + iy ∈ �, there
exists a positive real number R such that B(z, R) ⊂ �. Using the fact that Q
is dense in R, we can choose a rational number r such that 0 < r < R so that
B(z, r) ⊆ B(z, R) ⊆ G. Again using the denseness of Q in R, we can choose
u, v ∈ Q such that
x < u < x + r
4
and y < v < y + r
4
.
If w = u + iv, then
|w − z| = |(u + iv) − (x + iy)| ≤ |u − x| + |v − y| < r
4
+ r
4
= r
2
.
We also have
ζ ∈ B
(
w,
r
2
)
⇒ |w − ζ | < r
2
⇒ |z − ζ | ≤ |z − w| + |w − ζ | < r
2
+ r
2
�
�
“book” — 2014/6/4 — 20:35 — page 52 — #52
�
�
�
�
�
�
52 Extended Complex Numbers and Stereographic Projection
⇒ ζ ∈ B (z, r) .
Therefore, z ∈ B
(
w, r
2
) ⊆ B (z, r) ⊆ G. We note that B
(
w, r
2
) ∈ B. Hence,
the theorem follows. �
1.5 EXTENDED COMPLEX NUMBERS AND
STEREOGRAPHIC PROJECTION
The set C of all complex numbers along with a symbol ∞, denoted by C∞,
is called the set of all extended complex numbers (or the extended complex
plane) if the symbol is assigned with the following properties:
1. a ±∞ = ∞± a = ∞, for all a ∈ C,
2. c · ∞ = ∞ · c = ∞, for all c ∈ C \ {0},
3. ∞ = ∞,
4.
a
0
= ∞, for all a ∈ C∞ \ {0},
5.
a
∞ = 0, ∀a ∈ C.
∞−∞,
∞
∞ and 0 · ∞ are not defined.
Definition 1.5.1 (Topology on C∞)
A set A ⊆ C∞ is said to be an open subset of C∞ if
1. A = C∞ \ K for some compact subset K of C or
2. A is an open subset of C.
THEOREM 1.5.2 (Stereographic projection)
There exists a bijection φ from the unit ball U in R3 onto C∞.
Proof: Given a point (x1, x2, x3) ∈ U other than the north pole (0, 0, 1), we
define the image of φ((x1, x2, x3))) by the intersection of the line passing
through (x1, x2, x3) and (0, 0, 1) with the complex plane C.
�
�
“book” — 2014/6/4 — 20:35 — page 53 — #53
�
�
�
�
�
�
Preliminaries 53
To find φ((x1, x2, x3))) explicitly, first we consider the equation of the line
passing through (x1, x2, x3) and (0, 0, 1), which is given by
(1 − t)(0, 0, 1) + t(x1, x2, x3) = (tx1, tx2, (1 − t) + tx3), t ∈ R.
This line meets the x1x2 -plane if 1 − t + tx3 = 0 ⇒ t = 1
1−x3
. Therefore,
φ((x1, x2, x3)) =
⎧⎨
⎩
(
x1
1 − x3
,
x2
1 − x3
, 0
)
if x3 � 1
∞ if x3 = 1
.
z
(0, 0, 1)
(0, 0, 0)
(x1, x2, x3)
To prove φ : U → C∞ is onto let z ∈ C. If (x1, x2, x3) ∈ U such that
z = φ(x1, x2, x3), then using x2
1 + x2
2 + x2
3 = 1, we get
|z|2 = x2
1 + x2
2
(1 − x3)2
= 1 − x2
3
(1 − x3)2
= 1 + x3
1 − x3
⇒ (1 − x3)|z|2 = (1 + x3)
⇒ x3(1 + |z|2) = |z|2 − 1
⇒ x3 = |z|2 − 1
|z|2 + 1
.
Next,
x1
1 − x3
= Re z = z + z
2
⇒ x1 = z + z
2
(
1 − |z|2 − 1
|z|2 + 1
)
= z + z
|z|2 + 1
.
Similarly, we obtain that
x2
1 − x3
= Im z = z − z
i2
⇒ x2 = z − z
i2
(
1 − |z|2 − 1
|z|2 + 1
)
= z − z
i(|z|2 + 1)
.
Next we show, that (x1, x2, x3) obtained as above belongs to U and it satisfies
φ(x1, x2, x3) = z. Now,
x2
1 + x2
2 + x2
3 = (z + z)2 + (z − z)2 + (|z|2 − 1)2
(|z|2 + 1)2
�
�
“book” — 2014/6/4 — 20:35 — page 54 — #54
�
�
�
�
�
�
54 Extended Complex Numbers and Stereographic Projection
= 2(z2 + z2) + |z|4 + 1 − 2|z|2
(|z|2 + 1)2
= |z|4 + 1 + 2|z|2
(|z|2 + 1)2
= 1.
and hence, (x1, x2, x3) ∈ U . Clearly, we have φ((x1, x2, x3)) = z. Moreover,
we have φ(∞) = (0, 0, 1). Thus, φ : U → C∞ is onto. To prove φ : U →
C∞ is one-to-one, let (x1, x2, x3), (y1, y2, y3) ∈ U such that φ((x1, x2, x3)) =
φ((y1, y2, y3)) = z. If z = ∞, then by definition, (x1, x2, x3) = (0, 0, 1) =
(y1, y2, y3). Hence, assume that z ∈ C. Then, from the previous argument,
(x1, x2, x3) =
(
z + z
|z|2 + 1
,
z − z
i(|z|2 + 1)
,
|z|2 − 1
|z|2 + 1
)
= (y1, y2, y3).
Thus, φ : U → C∞ is one-to-one. �
RESULT 1.5.3 The image of a circle on the sphere U under the map φ is
either a circle or a straight line in C∞.
Proof: Any circle on the sphere can be interpreted as the intersection of U
by a plane (say) α1x1 + α2x2 + α3x3 = α0. Using the values of x1, x2, and x3
in terms of z, we get
α1(z + z) + α2
i
(z − z) + α3(|z|2 − 1) = α0(|z|2 + 1).
Putting z = x+iy in the last equation, we have 2α1x+2α2y+α3(x2+y2−1) =
α0(x2 + y2 +1) ⇒ (α0 −α3)(x2 + y2)−2α1x−2α2y+α0 +α3 = 0, which is
the equation of a circle if (α0 − α3) � 0 and is the equation of a straight line
if (α0 − α3) = 0. �
Definition 1.5.4 Given any two points z, w ∈ C∞, define the distance D(z, w)
by the actual distance between φ−1(z) and φ−1(w) in R3.
PROBLEM 1.5.5 Find D(z, w) explicitly.
Proof:
Case 1: z, w ∈ C.
Let φ−1(z) = (x1, x2, x3) and φ−1(w) = (y1, y2, y3). Then,
(x1 − y1)2 + (x2 − y2)2 + (x3 − y3)2
�
�
“book” — 2014/6/4 — 20:35 — page 55 — #55
�
�
�
�
�
�
Preliminaries 55
= x2
1 + y2
1 − 2x1y1 + x2
2 + y2
2 − 2x2y2 + x2
3 + y2
3 − 2x3y3
= (x2
1 + x2
2 + x2
3) + (y2
1 + y2
2 + y2
3) − 2(x1y1 + x2y2 + x3y3)
= 2 − 2(x1y1 + x2y2 + x3y3)
(as x2
1 + x2
2 + x2
3 = 1 = y2
1 + y2
2 + y2
3)
= 2 − 2
(z + z)(w + w) − (z − z)(w − w) + (|z|2 − 1)(|w|2 − 1)
(|z|2 + 1)(|w|2 + 1)
= 2 − 2
(
2(zw + zw) + (|z|2|w|2 − |z|2 − |w|2 + 1)
(|z|2 + 1)(|w|2 + 1)
)
= 2 − 2
(
2(−|z|2 − |w|2 + zw + zw) + (|z|2|w|2 + |z|2 + |w|2 + 1)
(|z|2 + 1)(|w|2 + 1)
)
= 2 − 2
(−2|z − w|2 + (|z|2 + 1)(|w|2 + 1)
(|z|2 + 1)(|w|2 + 1)
)
= 4|z − w|2
(|z|2 + 1)(|w|2 + 1)
.
Case 2: z ∈ C and w = ∞.
In this case, φ−1(w) = (0, 0, 1) and let φ−1(z) = (x1, x2, x3). Therefore,
(x1 − 0)2 + (x2 − 0)2 + (x3 − 1)2 = 2 − 2x3
= 2 − 2
|z|2 − 1
|z|2 + 1
= 4
|z|2 + 1
Hence, D(z, w) =
⎧⎪⎪⎨
⎪⎪⎩
2|z − w|√
|(z|2 + 1)(|w|2 + 1)
if z, w ∈ C
2√
|z|2 + 1
if z ∈ C and w = ∞.
�
1.6 LIMIT AND CONTINUITYDefinition 1.6.1 Let � be a region, z0 ∈ �, l ∈ C and f : � → C. We say
that f (z) → l as z → z0 if given ε > 0, then there exists δ > 0 such that
|f (z) − l| < ε whenever 0 < |z − z0| < δ.
In this case, we write lim
z→z0
f (z) = l.
THEOREM 1.6.2 Let f : �→ C, z0 ∈ �, l, c ∈ C and lim
z→z0
f (z) = l. Then,
�
�
“book” — 2014/6/4 — 20:35 — page 56 — #56
�
�
�
�
�
�
56 Limit and Continuity
1. lim
z→z0
Re (f (z)) = Re l,
2. lim
z→z0
Im (f (z)) = Im l,
3. lim
z→z0
f (z) = l,
4. lim
z→z0
|f (z)| = |l|,
5. lim
z→z0
(f + c)(z) = l + c,
6. lim
z→z0
(cf )(z) = cl.
Proof: Let ε > 0. Then, there exists δ > 0 such that |f (z) − l| < ε whenever
0 < |z − z0| < δ. We choose the same δ for proving (1), (2), (3), (4), (5). If
0 < |z − z0| < δ, then
1. |Re (f (z)) − Re l| = |Re (f (z) − l)| ≤ |f (z) − l| < ε
2. |Im (f (z)) − Im l| = |Im (f (z) − l)| ≤ |f (z) − l| < ε
3. |f (z) − l| = |f (z) − l| ≤ |f (z) − l| < ε
4. | |f (z)| − |l| | ≤ |f (z) − l| < ε, by using Corollary 1.2.10
5. |(f (z) + c) − (l + c)| = |f (z) − l| < ε
6. For proving the last assertion, we first choose δ > 0 such that
|f (z) − l| < ε
|c| + 1
whenever 0 < |z − z0| < δ
If 0 < |z − z0| < δ, then |(cf (z)) − (cl)| = |c||f (z)− l| ≤ |c|ε
|c| + 1
< ε.
�
THEOREM 1.6.3 Let f and g be functions on �, z0 ∈ � and lim
z→z0
f (z) = l1,
lim
z→z0
g(z) = l2, where l1, l2 ∈ C. Then,
1. lim
z→z0
(f + g)(z) = l1 + l2,
2. lim
z→z0
(fg)(z) = l1l2,
�
�
“book” — 2014/6/4 — 20:35 — page 57 — #57
�
�
�
�
�
�
Preliminaries 57
3. lim
z→z0
(
f
g
)
(z) = l1
l2
, provided l2 � 0.
Proof: Let ε > 0 be given.
1. For a given ε > 0, there exist δ1 > 0 and δ2 > 0 such that
|f (z) − l1| < ε
2
whenever 0 < |z − z0| < δ1
and
|g(z) − l2| < ε
2
whenever 0 < |z − z0| < δ2.
Let 0 < δ < min{δ1, δ2}. If 0 < |z − z0| < δ, then 0 < |z − z0| < δ1
and 0 < |z − z0| < δ2. Thus, we get
|(f + g)(z) − (l1 + l2)| ≤ |f (z) − l1| + |g(z) − l2| < ε
2
+ ε
2
= ε
2. Consider
|(fg)(z) − (l1l2)| ≤ |f (z)g(z) − f (z)l2)| + |f (z)l2 − l1l2|
= |f (z)||g(z) − l2| + |l2||f (z) − l1|. (1.5)
Given ε > 0 choose δ1 > 0 such that
|f (z) − l1| < 1 if 0 < |z − z0| < δ1.
Hence, |f (z)| < |l1| + 1, whenever 0 < |z − z0| < δ1. For the same ε,
find δ2 > 0 such that
|g(z) − l2| < ε
2(|l1| + 1)
whenever 0 < |z − z0| < δ2.
Next choose δ3 > 0 such that
|f (z) − l1| < ε
2(|l2| + 1)
whenever 0 < |z − z0| < δ3.
If 0 < δ < min{δ1, δ2, δ3}, then from (1.5), we get
|(fg)(z) − (l1l2)| ≤ |f (z)||g(z) − l2| + |l2||f (z) − l1|
≤ (|l1| + 1)
ε
2(|l1| + 1)
+ |l2| ε
2(|l2| + 1)
<
ε
2
+ ε
2
= ε.
�
�
“book” — 2014/6/4 — 20:35 — page 58 — #58
�
�
�
�
�
�
58 Limit and Continuity
3. First we prove that lim
z→z0
1
g(z)
= 1
l2
if l2 � 0. Given ε > 0, choose δ > 0
such that
0 < |z − z0| < δ ⇒ |g(z) − l2| < min
{ |l2|
2
,
|l2|2ε
2
}
.
Thus, if 0 < |z − z0| < δ, then
| |l2| − |g(z)| | ≤ |g(z) − l2| < |l2|
2
⇒ |g(z)| > |l2|
2
⇒ 1
|g(z)| <
2
|l2| .
We also have∣∣∣∣ 1
g(z)
− 1
l2
∣∣∣∣ = |g(z) − l2|
|g(z)l2| <
|l2|2ε
2
· 2
|l2|2 = ε.
Hence, our claim follows. Next by using (2) of this theorem, we get
lim
z→z0
f (z)
g(z)
= lim
z→z0
f (z)
1
g(z)
= lim
z→z0
f (z) lim
z→z0
1
g(z)
= l1
l2
. �
Definition 1.6.4 Let f : C → C. We say that f (z) → ∞ as z → ∞ if given
M > 0, then there exists K > 0 such that |f (z)| > M whenever |z| > K. In
this case, we write lim
z→∞ f (z) = ∞.
Definition 1.6.5 Let f : C → C and a ∈ C. We say that f (z) → ∞ as z → a
if given M > 0, then there exists δ > 0 such that 0 < |z− a| < δ implies that
|f (z)| > M . We write this by lim
z→a
f (z) = ∞.
Similarly, we define lim
z→∞ f (z) = a.
THEOREM 1.6.6 Let f : � → C, g : � → C, A ∈ C and a ∈ �. If
lim
z→a
f (z) = ∞ and lim
z→a
g(z) = A, then
1. lim
z→a
(cf )(z) = ∞ for every non-zero complex number c,
2. lim
z→a
(f + g)(z) = ∞,
3. lim
z→a
(f · g)(z) = ∞ if A � 0,
4. lim
z→a
f (z)
g(z)
= ∞,
�
�
“book” — 2014/6/4 — 20:35 — page 59 — #59
�
�
�
�
�
�
Preliminaries 59
5. lim
z→a
g(z)
f (z)
= 0.
Proof: Let M > 0 be given.
1. We choose δ > 0 such that
0 < |z − a| < δ ⇒ |f (z)| > M
|c| .
If 0 < |z − a| < δ, then |cf (z)| = |c||f (z)| > |c|M
|c| = M .
2. We choose δ1 > 0 and δ2 > 0 such that
0 < |z − a| < δ2 ⇒ |f (z)| > M + 1 + |A|
and
0 < |z−a| < δ2 ⇒ |g(z)−A| < 1 ⇒ |g(z)| ≤ |g(z)−A|+|A| < 1+|A|.
If 0 < |z − a| < min{δ1, δ2}, then
|f (z) + g(z)| ≥ |f (z)| − |g(z)| > M + 1 + |A| − 1 − |A| = M .
Hence, lim
z→a
(f + g)(z) = ∞.
3. We choose δ1 > 0 and δ2 > 0 such that
0 < |z − a| < δ2 ⇒ |f (z)| > 2M
|A|
and
0 < |z − a| < δ2 ⇒ |g(z) − A| < |A|
2
⇒ |g(z)| ≥ |A|
2
.
If 0 < |z − a| < min{δ1, δ2}, then
|f (z)g(z)| > 2M
|A|
|A|
2
= M .
Hence, lim
z→a
f (z)g(z) = ∞.
�
�
“book” — 2014/6/4 — 20:35 — page 60 — #60
�
�
�
�
�
�
60 Limit and Continuity
4. We choose δ1 > 0 and δ2 > 0 such that
0 < |z − a| < δ2 ⇒ |f (z)| > M(1 + |A|)
and
0 < |z−a| < δ2 ⇒ |g(z)−A| < 1 ⇒ |g(z)| ≤ |g(z)−A|+|A| < 1+|A|.
If 0 < |z − a| < min{δ1, δ2}, then∣∣∣∣ f (z)
g(z)
∣∣∣∣ > M(1 + |A|)
1 + |A| = M .
Hence, lim
z→a
f (z)
g(z)
= ∞.
5. Given ε > 0 we choose δ1 > 0 and δ2 > 0 such that
0 < |z − a| < δ1 ⇒ |f (z)| > 1 + |A|
ε
and
0 < |z − a| < δ2 ⇒ |g(z) − A| < 1 ⇒ |g(z)| ≤ 1 + |A|.
If 0 < |z − a| < min{δ1, δ2}, then∣∣∣∣g(z)
f (z)
∣∣∣∣ < ε
1 + |A| (1 + |A|) = ε.
Hence, lim
z→a
g(z)
f (z)
= 0. �
Example 1.6.7 Let f : C → C be defined by
f (z) =
⎧⎨
⎩
xy(x + iy)
x3 + y3
z = x + iy � 0
0 z = 0
, ∀z ∈ C.
Prove that lim
z→0
f (z) does not exist.
If we allow z → 0 along the line y = cx, then we get
lim
z→0
f (z) = lim
x→0
x(cx)(x + i(cx))
x3 + (cx)3
= lim
x→0
c(1 + ic)
1 + c3
= c(1 + ic)
1 + c3
.
which is depending on c. That is, if we let z → 0 along different lines in
the family y = cx of straight lines, then f (z) approaches different values, and
hence, lim
z→0
f (z) does not exist.
�
�
“book” — 2014/6/4 — 20:35 — page 61 — #61
�
�
�
�
�
�
Preliminaries 61
LEMMA 1.6.8 Let f : � → C and let (a, b) ∈ �, where � ⊆ C is open.
If lim
(x,y)→(a,b)
f (x, y) = A for some A ∈ C, then lim
x→a
lim
y→b
f (x, y) = A =
lim
y→b
lim
x→a
f (x, y).
Proof: First we choose r > 0 such that B((a, b), r) ⊆ �. For the given ε > 0,
there exists 0 < δ < r such that
0 < |(x, y) − (a, b)| < δ ⇒ |f (x, y) − A| < ε
2
,
which implies that
|f (x, y) − f (u, v)| ≤ |f (x, y) − A| + |A − f (u, v)| < ε
2
+ ε
2
= ε. (1.6)
whenever (x, y), (u, v) ∈ B((a, b), δ) \ {(a, b)}. Now for a fixed x ∈ R such that
|x − a| < δ,
if 0 < |y − b| < δ, then |(x, y) − (x, b)| = |y − b| < δ
and hence, |f (x, y) − f (x, b)| < ε. In other words, we have proved that
lim
y→b
f (x, y) = f (x, b) for every x ∈ R with |x − a| < δ. Using the definition of
lim
(x,y)→(a,b)
f (x, b) = A , we get
0 < |x − a| < δ ⇒ |(x, b) − (a, b)| = |x − a| < δ ⇒ |f (x, b) − A| < ε.
Thus, lim
x→a
lim
y→b
f (x, y) = lim
x→a
f (x, b) = A. Similarly, we can prove that
lim
y→b
lim
x→a
f (x, y) = A. �
Exercise 1.6.9 Let f : � → C, a ∈ �, and A ∈ C. Prove that lim
z→a
f (z) = A
iff f (zn) → A as n → ∞ whenever zn → a as n → ∞ in � with zn � a,
∀n ∈ �.
Now we discuss left limit and right limit of a real-valued function on an
interval I of R.
Definition 1.6.10 Let f : I → R and x ∈ I , where I is an interval of R. We
write that
1. lim
y→x−
f (y) = l exists for some l ∈ R if given ε > 0, then there exists
δ > 0 such that (x − δ, x) ⊂ I and x − δ < y < x ⇒ |f (y) − l| < ε.
�
�
“book” — 2014/6/4 — 20:35 — page 62 — #62
�
�
�
�
�
�
62 Limit and Continuity
2. lim
y→x+
f (y) = l exists for some l ∈ R if given ε > 0, then there exists
δ > 0 such that (x, x + δ) ⊂ I and x < y < x + δ ⇒ |f (y) − l| < ε.
3. lim
y→x−
f (y) = +∞ exists if given M > 0, then there exists δ > 0 such
that (x − δ, x) ⊂ I and x − δ < y < x ⇒ f (y) > M .
4. lim
y→x+
f (y) = +∞ exists if given K > 0, then there exists δ > 0 such
that (x, x + δ) ⊂ I and x < y < x + δ ⇒ f (y) > K.
Similarly, we can define lim
y→x−
f (y) = −∞ and lim
y→x+
f (y) = −∞.
Exercise 1.6.11Let f : � → C, g : � → C, and a ∈ �. If lim
z→a
f (z) = ∞
and g is bounded in a neighbourhood of a, then prove that lim
z→a
(fg)(z) = ∞.
Remark 1.6.12: Let f : � → C, g : � → C, and a ∈ �. If lim
z→a
f (z) = ∞
and lim
z→a
g(z) = ∞, then lim
z→a
(f + g)(z) need not be ∞.
For example, for a fixed c ∈ C, if f (z) =
⎧⎨
⎩
1
z
z � 0
1 z = 0
and g(z) =
⎧⎨
⎩−
1
z
+ c z � 0
0 z = 0
, then clearly, lim
z→0
f (z) = ∞ and lim
z→0
g(z) = ∞ but
lim
z→0
(f + g)(z) = c �∞.
Definition 1.6.13 Let f : �→ C.
1. f is said to be continuous at z0 ∈ � if lim
z→z0
f (z) = f (z0).
2. f is said to be continuous on� if f is continuous at every point z0 ∈ �.
THEOREM 1.6.14 If f and g are continuous at z0 ∈ �, then f + g, fg, and
f
g
are continuous at z0.
Proof: Proof of this theorem follows from Theorem 1.6.3. �
Example 1.6.15 Every constant function on C is continuous.
Let z0 ∈ C be arbitrary. Given ε > 0, we choose δ > 0 arbitrarily. If
|z − z0| < δ, then |f (z) − f (z0)| = 0 < ε. Hence, f is continuous at z0.
�
�
“book” — 2014/6/4 — 20:35 — page 63 — #63
�
�
�
�
�
�
Preliminaries 63
Example 1.6.16 If f (z) = z, ∀z ∈ C, then f is continuous on C.
For a given ε > 0, we choose δ = ε. Then obviously, we get |z − z0| < ε
when |z − z0| < δ.
Example 1.6.17 Every polynomial is continuous on C.
Using Theorem 1.6.14 and the above two examples, we get every polynomial
is a continuous function on C.
Example 1.6.18 Re z, Im z, z, and |z| are continuous on C.
For justification, see Example 1.6.16 and Theorem 1.6.2.
LEMMA 1.6.19 f is continuous iff Re f and Im f are continuous iff f is
continuous.
Proof: Let f be continuous at z0. For a given ε > 0, there exists δ > 0 such
that |f (z) − f (z0)| < ε, whenever |z − z0| < δ. For the same δ,
if |z − z0| < δ, then |Re f (z) − Re f (z0)| ≤ |f (z) − f (z0)| < ε.
Similarly, for the same δ,
if |z − z0| < δ, then |Im f (z) − Im f (z0)| ≤ |f (z) − f (z0)| < ε.
Conversely, assume that Re f and Im f are continuous at z0. Then, for a given
ε > 0, there exist δ1 > 0 and δ2 > 0 such that |Re f (z) − Re f (z0)| < ε√
2
,
whenever |z − z0| < δ1 and |Im f (z)− Im f (z0)| < ε√
2
, whenever |z − z0| <
δ2. If δ = min{δ1, δ2}, then δ > 0, and if |z − z0| < δ, then
|f (z) − f (z0)| =
√
|Re f (z) − Re f (z0)|2 + |Im f (z) − Im f (z0)|
<
√
ε2
2
+ ε2
2
= ε.
Thus, f is continuous at z0.
Using |z| = |z|, one can prove that f is continuous iff f is continuous. �
THEOREM 1.6.20 Composition of two continuous functions is continuous.
Proof: Let f : � → C and g : f (�) → C be continuous. If h = g ◦ f , then
we show that h is continuous on �. Let z0 ∈ � be arbitrary. Then f (z0) ∈
�
�
“book” — 2014/6/4 — 20:35 — page 64 — #64
�
�
�
�
�
�
64 Limit and Continuity
f (�). Using the continuity of g at f (z0) for a given ε > 0, choose r > 0
such that
|g(w) − g(f (z0))| < ε if |w − f (z0)| < r.
Next, using the continuity of f at z0, for this r > 0, there exists δ > 0 such
that
|f (z) − f (z0)| < r if |z − z0| < δ.
Therefore, combining these two statements, we get
if |z − z0| < δ, then |h(z) − h(z0)| = |g(f (z)) − g(f (z0))| < ε.
Hence, h is continuous at z0. �
RESULT 1.6.21 Let f : A → C and x ∈ A, then f is continuous at x iff
f (xn) → f (x) as n → ∞, whenever xn → x as n → ∞.
Proof: Assume that f is continuous at x. For a given ε > 0, we can find δ > 0
such that
|y − x| < δ ⇒ |f (y) − f (x)| < ε.
If xn → x0 as n → ∞, for the δ > 0, there exists N ∈ N such that
|xn − x| < δ, ∀n ≥ N .
Therefore, if n ≥ N , then
|xn − x| < δ ⇒ |f (xn) − f (x)| < ε.
Conversely, assume that f is not continuous at x. Then, there exists ε >
0 such that for every n ∈ N, there exists xn ∈ A with |xn − x| < 1
n and
|f (x) − f (xn)| ≥ ε. Hence, there exists a sequence (xn) such that xn → x as
n → ∞ but f (xn) �→ f (x) as n → ∞. Hence, the result follows. �
Definition 1.6.22 Let m, n ∈ N, E ⊆ Rn, (x1, x2, . . . , xn) ∈ E, and f =
(f1, f2, . . . , fm) : E → Rm be continuous at (x1, x2, . . . , xn) if given ε > 0,
then there exists δ > 0 such that√√√√ m∑
k=1
|fk(t1, t2, . . . , tn) − fk(x1, x2, . . . , xn)|2 < ε
whenever
(t1, t2, . . . , tn) ∈ Rn and
√√√√ n∑
j=1
|tj − xj|2 < δ.
�
�
“book” — 2014/6/4 — 20:35 — page 65 — #65
�
�
�
�
�
�
Preliminaries 65
Limit of a function f = (f1, f2, . . . , fm) : E → Rm at a point
(x1, x2, . . . , xn) can also be defined in an analogous way. As in the earlier dis-
cussions, one can prove that f = (f1, f2, . . . , fm) : E → Rm is continuous at
(x1, x2, . . . , xn) iff fk(x1(ν), x2(ν), . . . , xn(ν)) → fk(x1, x2, . . . , xn) as ν → ∞,
∀k = 1, 2, . . . , m, whenever xj(ν) → xj as ν → ∞, ∀j = 1, 2, . . . , n.
THEOREM 1.6.23 Let φ : U → C∞ be the stereographic projection. Then
• φ is a continuous map on U \ {(0, 0, 1)} and
φ(x1, x2, x3) → ∞ as (x1, x2, x3) → (0, 0, 1),
• φ−1 is a continuous map on C and
φ−1(z) → (0, 0, 1) as z → ∞.
Proof: If φ1(x1, x2, x3) = x1
1 − x3
and φ2(x1, x2, x3) = x2
1 − x3
∀(x1, x2, x3) ∈
U \ {(0, 0, 1)}, then φ = (φ1,φ2). (Theorem 1.5.2.)
1. (a) As x3 � 1, ∀(x1, x2, x3) ∈ U \ {(0, 0, 1)}, whenever xj(ν) → xj as
ν → ∞ for all j = 1, 2, 3, we have
x1(ν)
1 − x3(ν)
→ x1
1 − x3
as ν → ∞.
and
x2(ν)
1 − x3(ν)
→ x2
1 − x3
as ν → ∞.
Therefore, φ is continuous on U \ {(0, 0, 1)}.
(b) If xj(ν) → 0 for j = 1, 2 and x3(ν) → 1 as ν → ∞, then
φk(x1(ν), x2(ν), x3(ν)) → ±∞ as ν → ∞ for k = 1, 2
in the extended real number system, and hence,
φ(x1(ν), x2(ν), x3(ν)) → ∞ in C∞ as ν → ∞.
2. As the remaining part of the proof of this theorem is similar, we leave
it as an exercise to the reader.
Therefore, φ and φ−1 are continuous functions. �
�
�
“book” — 2014/6/4 — 20:35 — page 66 — #66
�
�
�
�
�
�
66 Limit and Continuity
THEOREM 1.6.24 Let f : A → C, where A is open in C. Then, f is
continuous on A iff f −1(E) is open in A whenever E is open in C.
Proof: Assume that f is continuous on A. Let x ∈ f −1(E). Then f (x) ∈ E. As
E is open, there exists ε > 0 such that B(f (x), ε) ⊆ E. Since f is continuous
at x, for this ε > 0, there exists δ > 0 such that
|z − x| < δ ⇒ |f (z) − f (x)| < ε.
Therefore, if z ∈ B(x, δ) ⇒ f (z) ∈ B(f (x), ε). Hence,
f (B(x, δ)) ⊆ B(f (x), ε)) ⊆ E ⇒ B(x, δ) ⊆ f −1(E).
Thus, f −1(E) is open in A.
Conversely, assume that f −1(E) is open in A, whenever E is open in C.
Let x ∈ A and ε > 0 be given. Then B(f (x), ε) is an open subset of C. Then by
hypothesis, f −1(B(f (x), ε)) is open in A and x ∈ f −1(B(f (x), ε)). Then, there
exists δ > 0 such that B(x, δ) ⊆ f −1(B(f (x), ε)) ⇒ f (B(x, δ)) ⊆ B(f (x), ε).
Thus, we have proved that
|z − x| < δ ⇒ |f (z) − f (x)| < ε
and hence, f is continuous on A. �
Definition 1.6.25 Let f : E → C, where E ⊂ C. f is said to be uniformly
continuous on E if given ε > 0, there exits δ > 0 such that x, y ∈ E, |x−y| <
δ ⇒ |f (x) − f (y)| < ε.
THEOREM 1.6.26 If f : K → C is continuous and K is a compact subset of
C, then f is uniformly continuous.
Proof: Let ε > 0 be given. For every w ∈ K, there exists δw > 0 such that
|z − w| < δ ⇒ |f (z) − f (w)| < ε
3
. (1.7)
Since K is compact and
{
B
(
w, δw
2
)
: w ∈ K
}
is a collection of open sets such
that K ⊂ ∪
w∈K
B
(
w, δw
2
)
, there exist
w1, w2, w3, . . . , wn ∈ K such that K ⊂ n∪
j=1
B
(
wj,
δwj
2
)
.
�
�
“book” — 2014/6/4 — 20:35 — page 67 — #67
�
�
�
�
�
�
Preliminaries 67
If δ = min
{
δwj
2 : j = 1, 2, . . . , n
}
, then δ > 0. If x, y ∈ K with |x − y| < δ,
then x ∈ B
(
wj,
δwj
2
)
for some 1 ≤ j ≤ n. We note that y ∈ B
(
wj, δwj
)
for the
same j, because
|y − wj| ≤ |y − x| + |x − wj| ≤ δ + δwj
2
<
δwj
2
+ δwj
2
= δwj .
Hence, using (1.7), we get
|f (x) − f (y)| ≤ |f (x) − f (wj)| + |f (wj) − f (y)| < ε
3
+ ε
3
< ε.
Thus, f is uniformly continuous on K. �
THEOREM 1.6.27 Continuous image of a compact set is compact.
Proof: Let {Eα : α ∈ I} be a collection of open sets such that f (K) ⊆ ∪
α
Eα .
By Theorem 1.6.24, f −1(Eα) is open for every α ∈ I . Now, if x ∈ K, then
f (x) ∈ f (K). Then f (x) ∈ Eα for some α. This implies x ∈ f−1(Eα) for this α.
Therefore,
K ⊆ ∪
α
f −1(Eα).
As K is compact, we have K ⊆ n∪
i=1
f −1(Eαi ) for some suitable αi ∈ I , i =
1, 2, . . . , n. Then
f (K) ⊆ n∪
i=1
f (f −1(Eαi )) ⊆
n∪
i=1
Eαi .
Therefore, f (K) is compact. �
THEOREM 1.6.28 If f is a real-valued continuous function on a compact
subset K of C, then f is bounded and there exist x, y ∈ K such that
f (x) = sup f (K) and f (y) = inf f (K).
Proof: By previous theorem, f (K) is a compact subset of R, and hence, it
is closed and bounded. Thus, f is a bounded function on K. As sup f (K) ∈
Cl f (K) = f (K) (cf. Exercise 1.4.14), then there exists x ∈ K such that f (x) =
sup f (K). Similarly, we can prove that there exists y ∈ K such that f (y) =
inf f (K). �
�
�
“book” — 2014/6/4 — 20:35 — page 68 — #68
�
�
�
�
�
�
68 Limit and Continuity
THEOREM 1.6.29 Continuous image of a connected set is connected.
Proof: Let C be a connected subset of C. Suppose if f (C) is not connected in
C, then there exist non-empty separated sets A and B in C such that
f (C) ⊆ A ∪ B and f (C) ∩ A � ∅ � f (C) ∩ B.
Therefore, clearly we have
C ⊆ f −1(A) ∪ f −1(B) and C ∩ f −1(A) � ∅ � C ∩ f −1(B).
Now we claim that f −1(A) and f −1(B) are separated sets in C. Suppose
f −1(A)∩ (Clf −1(B)) � ∅, then there exists a point x ∈ f −1(A) and a sequence
(xn) in f −1(B) such that xn → x as n → ∞, indeed choose
xn ∈ B
(
x,
1
n
)
∩ f −1(B), ∀n ∈ N.
As f is continuous, using Result 1.6.21, we have
f (xn) → f (x) as n → ∞.
Note that f (x) ∈ A and f (xn) ∈ B ∀n ∈ N. Therefore, f (x) ∈ Cl B as for each
r > 0,
f (xn) ∈ B(f (x), r) ∩ B for all but finitely many n,
and hence, A∩Cl B � ∅. This contradicts the fact that A and B are separated
sets. By a similar argument, we can show that
(Cl f −1(A)) ∩ f −1(B) � ∅.
This implies that C is not connected, which is a contradiction. Therefore, f (C)
is connected. �
THEOREM 1.6.30 (Intermediate value theorem)
If f : [a, b] → R is a continuous function and λ ∈ R lies between f (a) and
f (b), then there exists c ∈ (a, b) such that f (c) = λ.
Proof: As [a, b] is a connected subset of R (as well as connected subset of
C), by previous theorem, f ([a, b]) is a connected subset of R. Hence, f ([a, b])
is an interval by Theorem 1.4.22. Thus, by definition of an interval, if λ lies
between f (a) and f (b), then λ ∈ f ([a, b]). That is, there exists c ∈ [a, b]
such that f (c) = λ. As f (a) < f (c) < f (b), we have a � c � b, and hence,
c ∈ (a, b). �
�
�
“book” — 2014/6/4 — 20:35 — page 69 — #69
�
�
�
�
�
�
Preliminaries 69
THEOREM 1.6.31 Let (fn) be a sequence of complex-valued continuous func-
tions on E ⊆ C. If fn → f as n → ∞ uniformly on E for some f : E → C,
then f is also continuous on [a, b].
Proof: Let z0 ∈ E and ε > 0 be arbitrary. By hypothesis, given ε > 0, there
exists N ∈ N such that
|fn(z) − f (z)| < ε
3
, ∀z ∈ E, ∀n ≥ N . (1.8)
As fN is continuous at z0, given ε > 0, there exists δ > 0 such that
|z − z0| < δ ⇒ |fN (z) − fN (z0)| < ε
3
. (1.9)
For the same δ > 0, if |z − z0| < δ, then using (1.8) and (1.9), we get
|f (z)−f (z0)| ≤ |f (z)−fN (z)|+|fN (z)−fN (z0)|+|fN (z0)−f (z0)| < ε
3
+ε
3
+ε
3
= ε.
Therefore, f is continuous on E. �
This page is intentionally left blank
�
�
“book” — 2014/6/4 — 20:46 — page 71 — #1
�
�
�
�
�
�
2
Analytic Functions
2.1 DIFFERENTIABILITY
Definition 2.1.1 Let� be a region, z0 ∈ � and f : �→ C. f is said to be dif-
ferentiable at z0 if lim
z→z0
f (z)−f (z0)
z−z0
exists in C or equivalently, lim
h→0
f (z0+h)−f (z0)
h
exists in C, and this limit is called the derivative of f at z0, which is denoted
by f ′(z0).
Definition 2.1.2 A function f is said to be differentiable on a region � if f is
differentiable at every point of �.
Definition 2.1.3 Let � be a region, z0 ∈ � and f : �→ C.
1. f is said to be analytic at z0 if f is differentiable on some neighbourhood
of z0.
2. f is said to be analytic on � if f is analytic at every point of �.
Remark 2.1.4: Suppose f is a function on a region �. Then, f is differen-
tiable on � iff f is analytic on �.
RESULT 2.1.5 If f is differentiable at z0, then f is continuous at z0.
Proof: Using Theorem 1.6.3, we get
lim
z→z0
( f (z) − f (z0)) = lim
z→z0
( f (z) − f (z0))
z − z0
· (z − z0)
= lim
z→z0
( f (z) − f (z0))
z − z0
· lim
z→z0
(z − z0)
= f ′(z0) · 0 = 0.
Thus, f is continuous at z0. �
71
�
�
“book” — 2014/6/4 — 20:46 — page 72 — #2
�
�
�
�
�
�
72 Differentiability
THEOREM 2.1.6 If f and g are differentiable at z0, then f + g and fg are
differentiable at z0. If g′(z0) � 0, then f
g is differentiable at z0.
Proof: By assumption, we have
lim
z→z0
f (z) − f (z0)
z − z0
= f ′(z0) and lim
z→z0
g(z) − g(z0)
z − z0
= g′(z0).
Using Theorems 1.6.2 and 1.6.3 and Result 2.1.5, we get
1. lim
z→z0
( f + g)(z) − ( f + g)(z0)
z − z0
= lim
z→z0
f (z) − f (z0)
z − z0
+ lim
z→z0
g(z) − g(z0)
z − z0
= f ′(z0) + g′(z0).
2. lim
z→z0
( fg)(z) − ( fg)(z0)
z − z0
= lim
z→z0
f (z)g(z) − f (z0)g(z) + f (z0)g(z) − f (z0)g(z0)
z − z0
= lim
z→z0
g(z)( f (z) − f (z0))
z − z0
+ lim
z→z0
f (z0)(g(z) − g(z0))
z − z0
= lim
z→z0
g(z) lim
z→z0
f (z) − f (z0)
z − z0
+ f (z0) lim
z→z0
g(z) − g(z0)
z − z0
= f ′(z0)g(z0) + f (z0)g′(z0).
3. lim
z→z0
(
f
g
)
(z) −
(
f
g
)
(z0)
z − z0
= lim
z→z0
f (z)g(z0) − f (z0)g(z)
g(z)g(z0)(z − z0)
= lim
z→z0
f (z)g(z0) − f (z0)g(z0) + f (z0)g(z0) − f (z0)g(z)
g(z)g(z0)(z − z0)
=
lim
z→z0
(
g(z0)( f (z) − f (z0))
z − z0
− f (z0)(g(z) − g(z0))
z − z0
)
lim
z→z0
g(z)g(z0)
= g(z0)f ′(z0) − f (z0)g′(z0)
(g(z0))2
.
�
�
�
“book” — 2014/6/4 — 20:46 — page 73 — #3
�
�
�
�
�
�
Analytic Functions 73
Example 2.1.7 Every constant function is differentiable on C and its deriva-
tive is 0.
Let z0 ∈ C be arbitrary. If f is a constant function, then obviously
lim
z→z0
f (z)−f (z0)
z−z0
= lim
z→z0
0
z−z0
= 0.
Example 2.1.8 If f (z) = zn, ∀z ∈ C, where n ∈ N, then f is differentiable on
C and f ′(z) = nzn−1, ∀z ∈ C.
For an arbitrary z0 ∈ C, consider
lim
z→z0
f (z) − f (z0)
z − z0
= lim
z→z0
zn − zn
0
z − z0
= lim
z→z0
(z − z0)(zn−1 + zn−2z0 + · · · + zzn−2
0 + zn−1
0 )
z − z0
= lim
z→z0
(
zn−1 + zn−2z0 + · · · + zzn−2
0 + zn−1
0
)
= nzn−1
0 .
Thus, f ′(z) = nzn−1, ∀z ∈ C.
Example 2.1.9 Every polynomial is differentiable on C.
Applying Theorem 2.1.6 repeatedly and using the above two examples, we
get every polynomial to be differentiable on C.
THEOREM 2.1.10 (Chain rule)
If f is differentiable on �, g is differentiable on f (�), and h = g ◦ f , then h is
differentiable on � and h′(z) = g′( f (z))f ′(z), ∀z ∈ �.
Proof: Let z0 ∈ � be arbitrary, then we have
lim
z→z0
f (z) − f (z0)
z − z0
= f ′(z0) and lim
w→f (z0)
g(w) − g( f (z0))
w − f (z0)
= g′( f (z0)).
Thus, if we set
R(z) = f (z) − f (z0)
z − z0
− f ′(z0) in B(z0, r) \ {z0}
and
S(w) = g(w) − g( f (z0))
w − f (z0)
− g′( f (z0)) in B( f (z0), r) \ {f (z0)}
�
�
“book” — 2014/6/4 — 20:46 — page 74 — #4
�
�
�
�
�
�
74 Differentiability
for a small r > 0, then
f (z) − f (z0) = (z − z0)( f ′(z0) + R(z)),
g(w) − g( f (z0)) = (w − f (z0))[g′( f (z0)) + S(w)],
and R(z) → 0 as z → z0, S(w) → 0 as w → f (z0). As f is continuous at z0
(Result 2.1.5), we have f (z) → f (z0) as z → z0. Therefore,
lim
z→z0
h(z) − h(z0)
z − z0
= lim
z→z0
g( f (z)) − g( f (z0))
z − z0
= lim
z→z0
[g′( f (z0)) + S( f (z))]( f (z) − f (z0))
z − z0
= lim
z→z0
[g′( f (z0)) + S( f (z))][f ′(z0) + R(z)](z − z0)
z − z0
= lim
z→z0
[g′( f (z0)) + S( f (z))]( f ′(z0) + R(z))
= g′( f (z0)) f ′(z0).
Hence, h is differentiable on � and h′(z) = g′( f (z))f ′(z), ∀z ∈ �. �
Definition 2.1.11 (Higher order derivatives)
Let f : �→ C, where � be a region.
1. The second derivative of f is defined by the derivative of f ′ and is
denoted by f (2) or f ′′.
2. The third derivative of f is defined by the derivative of f (2) and is
denoted by f (3) or f ′′′.
3. Proceeding further inductively, we define the kth derivative of f by the
derivativeof f (k−1) and is denoted by f (k), ∀k > 1.
We shall also use the notation f (0) to denote the function f . We also mean that
f is k times differentiable, by writing that f (k) exists, for k ∈ {1, 2, 3, . . .}.
Remark 2.1.12: In the above definition, the kth derivative of f could be
defined if f (k−1) is differentiable. However, we shall prove that if f is analytic,
then f (k) exists, for all k ∈ N. (See Corollary 4.3.3.)
RESULT 2.1.13 (Leibniz rule)
Let f and g be analytic functions, then ( f · g)(k) =
k∑
j=0
kCj f ( j)g(k−j), for all
�
�
“book” — 2014/6/4 — 20:46 — page 75 — #5
�
�
�
�
�
�
Analytic Functions 75
k ∈ N, where kCj = k × (k − 1) × · · · × (k − j + 1)
1 × 2 × · · · × j
Proof: We prove this theorem by induction on k. For k = 1, the result follows
from Theorem 2.1.6 immediately. Assume that this result holds for some k ∈
N. First, we note that
kCj + kCj−1
= k × (k − 1) × · · · × (k − j + 1)
1 × 2 × · · · × j
× k × (k − 1) × · · · × (k − j + 2)
1 × 2 × · · · × j − 1
= k × (k − 1) × · · · × (k − j + 2)
1 × 2 × · · · × j − 1
×
(
(k − j + 1)
j
+ 1
)
= k × (k − 1) × · · · × (k − j + 2)
1 × 2 × · · · × j − 1
× (k + 1)
j
= (k + 1)Cj.
Again applying Theorem 2.1.6 and by using nC0 = nCn = 1, ∀n ∈ N, we get
( f · g)(k+1) = (( f · g)(k))′
=
⎛
⎝ k∑
j=0
kCj f ( j) · g(k−j)
⎞
⎠
′
=
k∑
j=0
kCj
(
f ( j+1) · g(k−j) + f ( j) · g(k−j+1)
)
=
k+1∑
j=1
kCj−1 f ( j) · g(k−j+1) +
k∑
j=0
kCj f ( j) · g(k−j+1)
= kCk f (k+1) · g + kC0 f · g(k+1)
+
k∑
j=1
(
kCj + kCj−1
)
f ( j) · g(k−j+1)
�
�
“book” — 2014/6/4 — 20:46 — page 76 — #6
�
�
�
�
�
�
76 Differentiability
= (k + 1)C0 f · g(k+1) +
k∑
j=1
(k + 1)Cj f ( j) · g(k+1−j)
+ (k + 1)Ck+1 f (k+1) · g
=
(k+1)∑
j=0
(k + 1)Cj f ( j) · g(k+1−j).
Thus, the result follows. �
A function f of a complex variable z can also be viewed as a function of
two real variables x and y. Therefore, we can discuss about the partial deriva-
tives of f with respect to x and y, and the relation between the differentiability
of f and partial differentiabilities of f .
Definition 2.1.14 Let f : �→ C and (x0, y0) ∈ �, where � is a region.
1. f is said to be partially differentiable with respect to x at (x0, y0)
if lim
s→0
f (x0 + s, y0) − f (x0, y0)
s
exists and is denoted by
∂f
∂x
(x0, y0) or
fx(x0, y0).
2. f is said to be partially differentiable with respect to y at (x0, y0)
if lim
t→0
f (x0, y0 + t) − f (x0, y0)
t
exists and is denoted by
∂f
∂y
(x0, y0) or
fy(x0, y0).
In the above two limits, s and t approach 0 through reals.
THEOREM 2.1.15 If f and g are partially differentiable with respect to x (with
respect to y), then
1. f + g is partially differentiable with respect to x (with respect to y) and
( f + g)x = fx + gx (( f + g)y = fy + gy).
2. cf is partially differentiable with respect to x (with respect to y) and
(cf )x = cfx ((cf )y = cfy), where c ∈ C.
Proof: Proof of this theorem is analogous to that of Theorem 2.1.6. �
As an immediate consequence, we have the following corollary.
COROLLARY 2.1.16 Let f = u + iv, then f is partially differentiable with
respect to x (with respect to y) iff u and v are partially differentiable with
respect to x (with respect to y). In this case, fx = ux + ivx (fy = uy + ivy).
�
�
“book” — 2014/6/4 — 20:46 — page 77 — #7
�
�
�
�
�
�
Analytic Functions 77
Example 2.1.17 In general, though fxy and fyx exist at a point, they need not
be equal to each other.
Consider f (x, y) =
⎧⎨
⎩
x3y
x2 + y2
(x, y) � (0, 0)
0 (x, y) = 0
.
Now for s, t ∈ R \ {0}, we have
fx(0, 0) = lim
s→0
f (0 + s, 0) − f (0, 0)
s
= lim
s→0
0 − 0
s
= 0
fx(0, t) = lim
s→0
f (0 + s, t) − f (0, t)
s
= lim
s→0
s2t
s2 + t2
= 0
fy(0, 0) = lim
t→0
f (0, 0 + t) − f (0, 0)
t
= lim
t→0
0 − 0
t
= 0
fy(s, 0) = lim
t→0
f (s, 0 + t) − f (s, 0)
t
= lim
t→0
s3
s2 + t2
= s.
Therefore, we have
fxy(0, 0) = lim
s→0
fy(s, 0) − fy(0, 0)
s
= lim
s→0
s
s
= 1
and
fyx(0, 0) = lim
t→0
fx(0, t) − fx(0, 0)
t
= lim
t→0
0 − 0
t
= 0.
Thus, fxy(0, 0) � fyx(0, 0).
Next we recall mean-value theorem from real analysis, which will be applied
in the following sequel.
THEOREM 2.1.18 (Mean-value theorem)
Let f : [a, b] → R be a continuous function. If f is differentiable on (a, b),
then there exists x ∈ (a, b) such that f (b) − f (a) = f ′(x)(b − a).
THEOREM 2.1.19 (Complex version of mean-value theorem)
Let f : [a, b] → C be a continuous function. If f is differentiable on (a, b),
then there exists x ∈ (a, b) such that | f (b) − f (a)| ≤ | f ′(x)|(b − a).
Proof: If f = (u, v), then u and v are real-valued continuous functions on
[a, b], and they are differentiable on (a, b). Furthermore,
f ′(t) = (u′(t), v′(t)), ∀ t ∈ [a, b]
�
�
“book” — 2014/6/4 — 20:46 — page 78 — #8
�
�
�
�
�
�
78 Differentiability
Define φ : [a, b] → R by
φ(t) = xu(t) + yv(t), ∀ t ∈ [a, b].
where (x, y) = f (b) − f (a) = (u(b) − u(a), v(b) − v(a))
Therefore, φ is a real-valued continuous function on [a, b] and φ′(t) =
xu′(t) + yv′(t),∀t ∈ (a, b).
Now applying mean-value theorem (Theorem 2.1.18), we choose t ∈
(a, b) such that
φ(b) − φ(a) = φ′(t)(b − a).
Therefore,
|(x, y)|2 = x2 + y2
= x(u(b) − u(a)) + y(v(b) − v(a))
= xu(b) + yv(b) − (xu(a) + yv(a))
= φ(b) − φ(a)
= φ′(t)(b − a)
= (xu′(t) + yv′(t))(b − a)
≤
√
x2 + y2
√
(u′(t))2 + (v′(t))2 (b − a)
(by Cauchy–Schwarz inequality. )
= |(x, y)|| f ′(t)|(b − a).
If |(x, y)| � 0, then we have |(x, y)| ≤ | f ′(t)|(b − a). If |(x, y)| = 0, then
|(x, y)| = 0 ≤ | f ′(t)|(b − a). �
THEOREM 2.1.20 (Young’s theorem)
Let f : � → R and (a, b) ∈ �. If fx exists in a neighbourhood of (a, b) and
fxy is continuous at (a, b), then fyx exists at (a, b) and fyx(a, b) = fxy(a, b).
Proof: Let fx, fy, and fxy exist on B ((a, b), r). Fix (s, t) ∈ C such that (a +
s, b + t) ∈ B ((a, b), r). If F : [b, b + t] → R is defined by
F(y) = f (a + s, y) − f (a, y), ∀y ∈ [b, b + t],
then F is differentiable on [b, b + t], and hence, by applying mean-value
theorem (Theorem 2.1.18) twice, we get
�
�
“book” — 2014/6/4 — 20:46 — page 79 — #9
�
�
�
�
�
�
Analytic Functions 79
f (a + s, b + t) − f (a, b + t) − [f (a + s, b) − f (a, b)]
= F(b + t) − F(b)
= tF′(b + λt), for some λ ∈ (0, 1)
= t
[
fy(a + s, b + λt) − fy(a, b + λt)
]
= stfxy(a + μs, b + λt), for some μ ∈ (0, 1).
Using the continuity of fxy at (a, b), we obtain
fxy(a, b) = lim
(s,t)→(0,0)
fxy(a + μs, b + λt)
= lim
t→0
lim
s→0
1
t
[
f (a + s, b + t) − f (a, b + t)
s
− f (a + s, b) − f (a, b)
s
]
(using Lemma 1.6.8)
= lim
t→0
fx(a, b + t) − fx(a, b)
t
= fyx(a, b).
Thus, the theorem follows. �
2.2 CAUCHY–RIEMANN EQUATIONS
Definition 2.2.1 (Cauchy–Riemann equations)
Let u and v be real-valued functions of two real variables x and y, then we say
that u and v satisfy Cauchy–Riemann equations (or simply C–R equations) if
ux = vy and vx = −uy.
THEOREM 2.2.2 Let � be a region, z0 ∈ � and f : �→ C. If f = u + iv is
differentiable at z0, then f is partially differentiable at z0 = x0+iy0 = (x0, y0),
and u and v satisfy C–R equations.
Proof: As f is differentiable at z0, we have f ′(z0) = lim
h→0
f (z0 + h) − f (z0)
h
exists or equivalently,
f ′(x0 + iy0) = lim
s+it→0
f ((x0 + s) + i(y0 + t)) − f (x0 + iy0)
s + it
exists. (2.1)
�
�
“book” — 2014/6/4 — 20:46 — page 80 — #10
�
�
�
�
�
�
80 Cauchy–Riemann Equations
By letting t = 0 and s → 0 in equation (2.1), we get
f ′(x0 + iy0) = lim
s→0
f ((x0 + s) + iy0) − f (x0 + iy0)
s
= lim
s→0
f (x0 + s, y0) − f (x0, y0)
s
= fx(x0, y0).
If s = 0 and t → 0 in equation (2.1), then we get
f ′(x0 + iy0) = lim
t→0
f (x0 + i(y0 + t)) − f (x0 + iy0)
it
= 1
i
lim
t→0
f (x0, y0 + t) − f (x0, y0)
t
= −ify(x0, y0).
Hence, fx and fy exist at (x0, y0), and we also have fx = f ′ = −ify. As f =
u + iv, using Corollary 2.1.16, we have
ux + ivx = fx = −ify = −i(uy + ivy) = −iuy + vy.
Equating the real and imaginary parts on both sides, we get ux = vy and
vx = −uy. Thus, u and v satisfy C–R equations. �
Remark 2.2.3:From the above theorem, we conclude that a necessary con-
dition on f has to be differentiable at (x, y) such that f should satisfy the C–R
equations at (x, y) (i.e., fx = −ify).
Example 2.2.4 If f (z) = x2 − iy3 and g(z) = exp(2x + i3y), ∀z ∈ C, where
x = Re z and y = Im z, then justify that f and g are not differentiable on C.
We now check the C–R equations for f .
If u(x, y) = x2, v(x, y) = −y3, ∀(x, y) ∈ C, then f = u + iv and we have
∂u
∂x
= 2x,
∂u
∂y
= 0
∂v
∂x
= 0,
∂v
∂y
= −3y2.
Therefore,
∂u
∂x
�
∂v
∂y
. Thus, f does not satisfy C–R equations at any (x, y) ∈
C\{(0, 0)}, and hence, it is not differentiable on C.
�
�
“book” — 2014/6/4 — 20:46 — page 81 — #11
�
�
�
�
�
�
Analytic Functions 81
If u(x, y) = exp(2x) cos(3y), v(x, y) = exp(2x) sin(3y), ∀(x, y) ∈ C, then g =
u + iv and we have
∂u
∂x
= 2 exp(2x) cos(3y),
∂u
∂y
= −3 exp(2x) sin(3y)
∂v
∂x
= 2 exp(2x) sin(3y),
∂v
∂y
= 3 exp(2x) cos(3y).
Therefore, f does not satisfy C–R equations at any (x, y) ∈ C, and hence, it is
not differentiable on C.
Definition 2.2.5 Let � be a region, then a function u on � is called a
harmonic function if u satisfies the Laplace equation uxx + uyy = 0 on �.
RESULT 2.2.6 If f = u + iv and f is differentiable on �, then u and v are
harmonic functions on �.
Proof: (At present, assume the fact that partial derivatives of u and v of all
orders exist and they are continuous. This will be obtained as a consequence
of Theorem 4.3.3.) By Theorem 2.2.2, if f is differentiable on � and u and
v satisfy C–R equations, then ux = vy and uy = −vx. Now applying Young’s
theorem (Theorem 2.1.20), we get
1. uxx + uyy = ∂ux
∂x
+ ∂uy
∂y
= ∂vy
∂x
+ −∂vx
∂y
= vyx − vxy = 0. Hence, u is
harmonic.
2. vxx + vyy = ∂vx
∂x
+ ∂vy
∂y
= −∂uy
∂x
+ ∂ux
∂y
= uxy − uyx = 0. Hence, v is
harmonic. �
Definition 2.2.7 Given a harmonic function u on a region �, another har-
monic function v on � is called a harmonic conjugate of u if u + iv is a
differentiable function on �.
RESULT 2.2.8 v is a harmonic conjugate of u iff −u is a harmonic conjugate
of v.
Proof: v is a harmonic conjugate of u iff u + iv is an analytic function
iff −i(u + iv) = v − iu is an analytic function iff −u is a harmonic
conjugate of v. �
�
�
“book” — 2014/6/4 — 20:46 — page 82 — #12
�
�
�
�
�
�
82 Cauchy–Riemann Equations
Algorithm 2.2.9 (To find harmonic conjugate.)
Step 1. Let the given function be u and find ux.
Step 2. Put vy = ux and integrate vy partially with respect to y and write
v = ∫
vy dy + φ, where φ is a function of x.
Step 3. Using this v, find vx.
Step 4. From the given u, find uy.
Step 5. Using uy = vx, find the value of φ′ and then φ.
Now, we present two more algorithms (by Milne-Thompson) to find an ana-
lytic function f whose real part u (or imaginary part v) is the given harmonic
function. Replacing z by x + iy and finding the real and imaginary parts of
f (x + iy), we can also find the harmonic conjugate of u (or v).
Algorithm 2.2.10 (Finding an analytic function f = u+iv from the given u)
Step 1. Check whether the given function u is a harmonic function. If yes,
then go to next step.
Step 2. Find ux and uy.
Step 3. Put f ′(z) = ux(z, 0) − iuy(z, 0).
Step 4. Find f = ∫
f ′(z) dz + c, where c is a constant.
Step 5. If you want to find the harmonic conjugate of u, then put z = x + iy
and find real and imaginary parts of f .
Step 6. Write the harmonic conjugate of u by Im f .
Algorithm 2.2.11 (Finding an analytic function f = u+iv from the given v )
Step 1. Check whether given function v is a harmonic function.
Step 2. Find vx and vy.
Step 3. Put f ′(z) = vy(z, 0) + ivx(z, 0).
Step 4. Find f = ∫
f ′(z) dz + c, where c is a constant.
Step 5. If you want to find the harmonic conjugate of v, then put z = x + iy
and find real and imaginary parts of f .
Step 6. Write the harmonic conjugate of v by −Re f .
�
�
“book” — 2014/6/4 — 20:46 — page 83 — #13
�
�
�
�
�
�
Analytic Functions 83
Example 2.2.12 If u(x, y) = x3 − 3xy2 + 2x2 − 2y2 + x, then prove that u is
a harmonic function and find its harmonic conjugate.
As
ux = 3x2 − 3y2 + 4x + 1, uxx = 6x + 4
uy = −6xy − 4y and uyy = −6x − 4
we have
uxx + uyy = 6x + 4 − 6x − 4 = 0.
Hence, u is a harmonic function. If v is a harmonic conjugate of u, then u and
v satisfy C–R equations (i.e., ux = vy, uy = −vx). The first equation
vy = ux = 3x2 − 3y2 + 4x + 1
implies
v =
∫
vy dy =
∫
(3x2 − 3y2 + 4x + 1) dy = 3x2y − y3 + 4xy + y + φ(x)
for some function φ of x. Using uy = −vx, we get
−6xy − 4y = −6xy − 4y − φ′(x).
Therefore, φ′(x) = 0, and hence, φ is a constant. Thus,
v = 3x2y − y3 + 4xy + y + c
for some real constant c.
Example 2.2.13 If u(x, y) = sin(2x)
cosh(2y) − cos(2x)
, then find the analytic
function whose real part is u.
∂u
∂x
= (cosh(2y) − cos(2x))2 cos(2x) − sin(2x)2 sin(2x)
(cosh(2y) − cos(2x))2
= 2(cos(2x) cosh(2y) − 1)
(cosh(2y) − cos(2x))2
∂u
∂y
= − 2 sin(2x) sinh(2y)
(cosh(2y) − cos(2x))2
.
If f = u + iv, then
f ′(z) = ∂u
∂x
(z, 0) − i
∂u
∂y
(z, 0)
�
�
“book” — 2014/6/4 — 20:46 — page 84 — #14
�
�
�
�
�
�
84 Cauchy–Riemann Equations
= 2(cos(2z) − 1)
(1 − cos(2z))2
= 2
cos(2z) − 1
= 2
cos2(z) − sin2(z) − cos2(z) − sin2(z)
= 2
−2 sin2(z)
= − csc2(z)
and hence, we get f (z) = cot(z) + c for some complex constant c.
Example 2.2.14 If u(x, y) = cosh(x) sin(y), ∀(x, y) ∈ C, then prove that u
is a harmonic function and find the analytic function f whose real part is u.
Find also its harmonic conjugate.
ux = sinh(x) sin(y), uxx = cosh(x) sin(y)
uy = cosh(x) cos(y) uyy = − cosh(x) sin(y).
Therefore, uxx + uyy = cosh(x) sin(y) − cosh(x) sin(y) = 0, and hence, u is a
harmonic function.
As ux = sinh(x) sin(y) and uy = cosh(x) cos(y), we put
f ′(z) = ux(z, 0) − iuy(z, 0) = −i cosh(z).
Therefore, f (z) = ∫
f ′(z) dz = ∫ −i cosh(z) dz + c = −i sinh(z) + c.
Writing z = x + iy, we get
f (x + iy) = −i sinh(x + iy) + c
= − sin(i(x + iy)) + c
= − sin(ix − y) + c
= − sin(ix) cos(y) + sin(y) cos(ix) + c
= −i sinh(x) cos(y) + sin(y) cosh(x) + c.
Hence, the harmonic conjugate of cosh(x) sin(y) is − sinh(x) cos(y) + r for
some real constant r.
Example 2.2.15 Prove that v(x, y) = exp(x)(x sin(y) + y cos(y)), ∀(x, y) ∈ C
is a harmonic function. Find an analytic function whose imaginary part is v
and also find its harmonic conjugate.
�
�
“book” — 2014/6/4 — 20:46 — page 85 — #15
�
�
�
�
�
�
Analytic Functions 85
vx = exp(x)(x sin(y) + y cos(y)) + exp(x) sin(y)
= exp(x)((x + 1) sin(y) + y cos(y))
vxx = exp(x)((x + 1) sin(y) + y cos(y)) + exp(x) sin(y)
= exp(x)((x + 2) sin(y) + y cos(y))
vy = exp(x)(x cos(y) − y sin(y) + cos(y))
= exp(x)((x + 1) cos(y) − y sin(y))
vyy = exp(x)(−(x + 1) sin(y) − y cos(y) − sin(y))
= exp(x)(−(x + 2) sin(y) − y cos(y)).
As
vxx + vyy = exp(x)((x + 2) sin(y) + y cos(y) − (x + 2) sin(y) − y cos(y))
= 0
v is a harmonic function. We know that f ′ = fx = ux + ivx = vy + ivx;
therefore,
f ′ = exp(x)((x + 1) cos(y) − y sin(y))
+ i(exp(x)((x + 1) sin(y) + y cos(y)))
= exp(z)(z + 1) (by replacing x and y by z and 0 respectively).
Now,
f (z) =
∫
exp(z)(z + 1) dz
= (z + 1) exp(z) −
∫
exp(z) dz
= (z + 1) exp(z) − exp(z) = z exp(z).
Therefore, f (z) = z exp(z)+c, ∀z ∈ C for some constant c. Writing z = x+iy
and finding the real part of f (z), we get
u(x, y) = Re [(x + iy) exp(x + iy)]
= Re [(x + iy) exp(x)(cos(y) + i sin(y))]
= exp(x)(x cos(y) − y sin(y)).
Therefore, the harmonic conjugate of v(x, y) is
−u(x, y) = exp(x)(y sin(y) − x cos(y))
�
�
“book” — 2014/6/4 — 20:46 — page 86 — #16
�
�
�
�
�
�
86 Cauchy–Riemann Equations
Example 2.2.16 Find an analytic function f whose real and imaginary parts
are u and v, respectively, if u + v = exp(−y) (cos(x) − sin(x)).
We find
ux + vx = exp(−y) (− sin(x) − cos(x)) , (2.2)
uy + vy = − exp(−y) (cos(x) − sin(x)) ,
−vx + ux = exp(−y) (sin(x) − cos(x)) . (2.3)
Now, equations (2.2) + (2.3) and (2.2) − (2.3) imply that
ux = − exp(−y) cos(x),
vx = − exp(−y) sin(x).
Therefore,f ′(z) = ux(z, 0) + ivx(z, 0)
= −(cos(z) + i sin(z)) = − exp(iz)
and hence,
f (z) = i exp(iz) + c = exp(iz) + c
for some c ∈ C.
Exercise 2.2.17
1. Prove that the following functions are harmonic and find the harmonic
conjugates.
(a) u(x, y) = x3 − 3xy2.
(b) u(x, y) = x4 + y4 − 6x2y2 + 2xy − x.
(c) u(x, y) = exp(−x)(x sin(y) − y cos(y)).
(d) u(x, y) = cosh(x) sin(y).
(e) u(x, y) = exp(−x)(x cos(y) + y sin(y)).
2. Verify that u(x, y) = y
x2 + y2
, ∀(x, y) ∈ C \ {(0, 0)} is a harmonic
function and find an analytic function whose real part is u.
3. Verify that v(x, y) = −3 exp(2x) sin(2y), ∀(x, y) ∈ C is a harmonic
function and find an anlytic function whose imaginary part is v.
4. Verify that u(x, y) = − exp(x2 − y2) sin(2xy), ∀(x, y) ∈ C and find an
analytic function whose real part is u. Find also the harmonic conjugate
of u.
�
�
“book” — 2014/6/4 — 20:46 — page 87 — #17
�
�
�
�
�
�
Analytic Functions 87
5. Verify that v is a harmonic function and find the analytic function
whose imaginary part is v, where v(x, y) = x2 − y2 + 2y, ∀(x, y) ∈ C.
Find also its harmonic conjugate.
6. Determine the analytic function f = u + iv if
(a) (u − v)(x, y) = x3 − y3 + 3x(x − y).
(b) (u + v)(x, y) = sin(2x) + sinh(2y)
cosh(2y) + cos(2x)
.
Answers: (1) (a) v(x, y) = 3x2y − y3 + c.
(b) 4x3y − 4xy3 − (x2 − y2) − y.
(c) v(x, y) = exp(−x)( y sin(y) + x cos( y)) + c.
(d) − sinh(x) cos( y) + c.
(e) exp(−x)( y cos( y) − x sin( y)).
(2)
i
z
.
(3) −3 exp(2z).
(4) − exp(z2), exp(x2 − y2) cos(2xy).
(5) iz2 + 2z, 2xy − 2x.
(6) (a) f (z) = −iz3 + c.
(b) tan z + c.
LEMMA 2.2.18 Let φ : (a, b) → R be a differentiable function.
1. If φ′(t) = 0, ∀t ∈ (a, b), then φ is constant on (a, b).
2. If φ′(t) ≥ 0, ∀t ∈ (a, b), then φ is increasing on (a, b).
3. If φ′(t) ≤ 0, ∀t ∈ (a, b), then φ is decreasing on (a, b).
4. If φ′(t) > 0, ∀t ∈ (a, b), then φ is strictly increasing on (a, b).
5. If φ′(t) < 0, ∀t ∈ (a, b), then φ is strictly decreasing on (a, b).
Proof: Let x, y ∈ (a, b) be arbitrary such that x � y. We assume that x < y
without loss of generality. Then, φ is a differentiable function on [x, y], and
then by using mean-value theorem (Theorem 2.1.18), there exists t ∈ (x, y)
such that
φ(y) − φ(x) = φ′(t)(x − y) = 0.
1. If φ′ = 0, then φ(y) = φ(x). Thus, φ is a constant.
2. If φ′ ≥ 0, then φ(y) ≥ φ(x). Thus, φ is increasing.
The proof of the remaining assertions are similar. �
�
�
“book” — 2014/6/4 — 20:46 — page 88 — #18
�
�
�
�
�
�
88 Cauchy–Riemann Equations
THEOREM 2.2.19 If f is analytic on a region � such that f ′ = 0, then f is
constant.
Proof: Let z0 ∈ � be arbitrary. Choose r > 0 such that B(z0, r) ⊆ �. If
f = u + iv, then f ′ = ux + ivx = vy − iuy implies ux = vx = uy = vy = 0.
From ux = 0, we get that u is a constant cy on Hz ∩ B(z0, r), where Hz is the
horizontal line passing through z = (x, y) for every z ∈ B(z0, r). Similarly,
from uy = 0, we get u, which is a constant cx on Vz ∩B(z0, r), where Vz is the
vertical line passing through z = (x, y) for every z ∈ B(z0, r).
If z1 = (x1, y1) and z2 = (x2, y2) ∈ B(z0, r), then the polygon passing
through (x1, y1), (x1, y2), and (x2, y2) is completely lying inside B(z0, r). See
Corollary 1.4.29 and the following diagram.
(x1, y2)
(x2, y2)
(x1, y1)
As (x1, y1) and (x1, y2) lie on a vertical line segment in B(z0, r), we have
u(x1, y1) = u(x1, y2). As (x1, y2) and (x2, y2) lie on a horizontal line segment
in B(z0, r), we have u(x1, y2) = u(x2, y2). Thus, we get u(z1) = u(z2), and
hence, u is constant on B(z0, r).
By the same argument, from vx = vy = 0, we can prove that v is constant
on B(z0, r), and so f is a constant function on B(z0, r).
If a, b are any two points in �, then by Theorem 1.4.31, we can find a
polygon such that
1. it passes through the vertices a = a0, a1, a2, . . . an = b,
2. for each i = 1, 2, . . . , n, the line segment joining ai−1 and ai is either
vertical or horizontal,
3. for each i = 1, 2, . . . , n, the line segment joining ai−1 and ai is
contained in B(z, r) ⊆ � for some z ∈ � and r > 0.
�
�
“book” — 2014/6/4 — 20:46 — page 89 — #19
�
�
�
�
�
�
Analytic Functions 89
Therefore, f (a) = f (a0) = f (a1) = f (a2) = · · · = f (an) = f (b).
Hence, f is constant on �. �
RESULT 2.2.20 Let f = u + iv be an analytic function on a region �, then f
is a constant if any one of the following statements holds:
1. u is constant.
2. v is constant.
3. | f | is constant.
4. arg f is constant.
Proof: Assume that u is constant. Then, ux = uy = 0. As vx = −uy, we have
f ′(z) = fx = ux + ivx = 0 on �. Therefore, by using Theorem 2.2.19, we get
that f is constant.
By a similar argument, using the C–R equations, one can prove that if v is
constant, then u is constant, and hence, f is constant.
If | f | = 0, then obviously f = 0, and hence, f is constant. If | f | is con-
stant and non-zero, then we have u2 + v2 as constant and non-zero. Partially
differentiating with respect to x and y, we get
2uux + 2vvx = 0, 2uuy + 2vvy = 0.
Using C–R equations, we get
uux − vuy = 0 and vux + uuy = 0.
As determinant of the coefficient matrix
∣∣∣∣ u −v
v u
∣∣∣∣ = u2 + v2 � 0, this
system of homogeneous linear equations in the two variables ux and uy has
a unique solution and is ux = 0 = uy. Hence, u is constant. Therefore, f is
constant.
If arg f is constant, then
v
u
is constant, say c. It follows that v−cu is constant,
but v−cu = Im (1−ci)f , and hence, (1−ci)f is constant. Thus, f is constant.
�
RESULT 2.2.21 If f = u+ iv is differentiable, then we have uxvx + uyvy = 0.
Geometrically, this result is rephrased by the family u(x, y) = c of curves is
orthogonal with the family v(x, y) = d of curves.
Proof: As u and v satisfy C–R equations, we have ux = vy and vx = −uy.
Therefore, uxvx + uyvy = −uxuy + uxuy = 0. �
�
�
“book” — 2014/6/4 — 20:46 — page 90 — #20
�
�
�
�
�
�
90 Cauchy–Riemann Equations
RESULT 2.2.22 If f = u + iv is analytic, then
∂2u2
∂x2
+ ∂2u2
∂y2
= 2| f ′|2.
Proof: First we note that u and v satisfy the C–R equations,
∂u
∂x
= ∂v
∂y
and
∂u
∂y
= −∂v
∂x
.
Now,
∂2u2
∂x2
= ∂
∂x
(
2u
∂u
∂x
)
= 2
(
∂u
∂x
)2
+ 2u
∂2u
∂x2
.
Similarly, we get
∂2u2
∂y2
= 2
(
∂u
∂y
)2
+ 2u
∂2u
∂y2
. Therefore, using the fact that
u is harmonic, we get
∂2u2
∂2x2
+ ∂2u2
∂y2
= 2
[(
∂u
∂x
)2
+
(
∂u
∂y
)2
]
+ 2u
(
∂2u
∂x2
+ ∂2u
∂y2
)
= 2
[(
∂u
∂x
)2
+
(
∂u
∂y
)2
]
= 2
[(
∂u
∂x
)2
+
(
∂v
∂x
)2
]
(using C–R equations)
= 2| f ′|2
(
as f ′ = ∂f
∂x
= ∂u
∂x
+ i
∂v
∂x
)
.
�
Exercise 2.2.23 If f is an analytic function, then
1.
(
∂2
∂x2
+ ∂2
∂y2
)
| f (z)|2 = 4| f ′(z)|2
2.
[(
∂
∂x
)2
+
(
∂
∂y
)2
]
| f (z)|2 = | f ′(z)|2.
RESULT 2.2.24 (Cauchy–Riemann equations in polar form)
If u and v satisfy C–R equations, then r
∂u
∂r
= ∂v
∂θ
and r
∂v
∂r
= − ∂u
∂θ
.
�
�
“book” — 2014/6/4 — 20:46 — page 91 — #21
�
�
�
�
�
�
Analytic Functions 91
Proof: We know that C–R equations in Cartesian form are given by
∂u
∂x
= ∂v
∂y
and
∂u
∂y
= −∂v
∂x
. From the relation x = r cos(θ ) and y = r sin(θ ) between the
Cartesian coordinates and polar coordinates, we get,
r
∂u
∂r
= r
(
∂u
∂x
∂x
∂r
+ ∂u
∂y
∂y
∂r
)
= −∂v
∂x
r sin(θ ) + ∂v
∂y
r cos(θ )
= ∂v
∂x
∂x
∂θ
+ ∂v
∂y
∂y
∂θ
= ∂v
∂θ
,
r
∂v
∂r
= r
(
∂v
∂x
∂x
∂r
+ ∂v
∂y
∂y
∂r
)
= ∂u
∂x
r sin(θ ) − ∂u
∂y
r cos(θ )
= −∂u
∂x
∂x
∂θ
− ∂u
∂y
∂y
∂θ
= − ∂u
∂θ
.
Hence, C–R equations in polar form are obtained. �
RESULT 2.2.25 If f is a complex-valued function on a region � satisfying
the C–R equation
∂f
∂x
= −∂f
∂y
, then
∂f
∂z
= 0.
Proof: If z = x+ iy, then we have x = 1
2
(z+ z) and y = 1
i2
(z− z). Therefore,
∂f
∂z
= ∂f
∂x
∂x
∂z
+ ∂f
∂y
∂y
∂z
= 1
2
(
∂f
∂x
+ i
∂f
∂y
)
= 0.
Hence, the result follows. �
Example 2.2.26 There exists a function f on C such that fx and fy exist and
satisfy C–R equation at a point, but it is not differentiable at the same point.
�
�
“book” — 2014/6/4 — 20:46 — page 92 — #22
�
�
�
�
�
�
92Cauchy–Riemann Equations
Define
f (x, y) =
⎧⎨
⎩
xy(x + iy)
x2 + y2
if (x, y) � (0, 0)
0 if (x, y) = (0, 0)
, ∀(x, y) ∈ C
As f (x, 0) = 0, ∀x ∈ R, we obtain fx(0, 0) = 0, and as f (0, y) = 0, ∀y ∈ R,
we get fy(0, 0) = 0. Hence, the C–R equation fx = −ify is satisfied at (0, 0).
However,
lim
h→0
f (2h, h) − f (0, 0)
2h + ih
= lim
h→0
2h2(2h + ih)
5h2(2h + ih)
= 2
5
�
1
2
= lim
h→0
h2(h + ih)
2h2(h + ih)
= lim
h→0
f (h, h) − f (0, 0)
h + ih
Hence, f is not differentiable at (0, 0).
Example 2.2.27 There exists a function f on C such that fx and fy exist and
are continuous at a point, but it is not differentiable at the same point.
Consider the function
f (x, y) = x − iy, ∀(x, y) ∈ C
Now, fx = 1 and fy = −i, ∀(x, y) ∈ C. Clearly, fx and fy are continuous
functions. However, it is not differentiable at (0, 0) as f does not satisfy C–R
equations at (0, 0).
THEOREM 2.2.28 If f = u+iv has continuous partial derivatives with respect
to x, y on a region � and u, v satisfy C–R equations, then f is differentiable
on �.
Proof: Let z0 = (x0, y0) ∈ � be arbitrary, then by applying mean-value
theorem, we get
u(x0 + s, y0 + t) − u(x0, y0)
= u(x0 + s, y0 + t) − u(x0, y0 + t) + u(x0, y0 + t) − u(x0, y0)
= sux(x0 + ps, y0 + t) + tuy(x0, y0 + qt)
( for some p, q ∈ (0, 1))
�
�
“book” — 2014/6/4 — 20:46 — page 93 — #23
�
�
�
�
�
�
Analytic Functions 93
= s(ux(x0, y0) + r1) + t(uy(x0, y0) + r2), where r1 → 0, r2 → 0
as (s, t) → (0, 0), (by using the continuity of ux and uy)
= s(ux + r1) + t(uy + r2)
(after writing ux(x0, y0) and uy(x0, y0) simply by ux and uy).
Similarly, we get v(x0 + s, y0 + t) − v(x0, y0) = s(vx + r3) + t(vy + r4), where
r3 → 0, r4 → 0 as (s, t) → (0, 0).
Therefore,
f (x0 + s, y0 + t) − f (x0, y0)
= s(ux + r1) + t(uy + r2) + is(vx + r3) + it(vy + r4)
= s(ux + r1) + t(−vx + r2) + is(vx + r3) + it(ux + r4)
= (s + it)ux + i(s + it)vx + s(r1 + ir3) + t(r2 + ir4)
= (s + it)(ux + ivx) + s(r1 + ir3) + t(r2 + ir4)
= (s + it)fx + s(r1 + ir3) + t(r2 + ir4).
Hence,
0 ≤
∣∣∣∣ f (x0 + s, y0 + t) − f (x0, y0)
s + it
− fx
∣∣∣∣
≤
∣∣∣∣ s
s + it
(r1 + ir3) + t
s + it
(r2 + ir4)
∣∣∣∣
≤ |r1| + |r3| + |r2| + |r4| → 0 as (s, t) → (0, 0).
Thus, f ′(z0) = lim
s+it→0
f ((x0 + iy0) + (s + it)) − f (x0 + iy0)
s + it
exists and is
equal to fx(z0). �
RESULT 2.2.29 Let f : �→ C, where � is a region such that z ∈ �⇒ z ∈
�. If g(z) = f (z), ∀z ∈ �, then f is differentiable iff g is differentiable.
Proof: Let f = u + iv and g = ϕ + iψ . As
ϕ(x, y) = u(x,−y) and ψ(x, y) = −v(x,−y), ∀(x, y) ∈ �,
we get
ϕx(x, y) = ux(x,−y), ϕy(x, y) = −uy(x,−y), ∀(x, y) ∈ �
and
ψx(x, y) = −vx(x,−y), ψy(x, y) = vy(x,−y), ∀(x, y) ∈ �.
�
�
“book” — 2014/6/4 — 20:46 — page 94 — #24
�
�
�
�
�
�
94 Power Series and Abel’s Theorems
Hence, ux, uy, vx and vy exist and are continuous iff ϕx, ϕy, ψx and ψy exist
and are continuous. Furthermore, ux = vy and uy = −vx imply
ϕx(x, y) = ux(x,−y) = vy(x,−y) = ψy(x, y)
and
ϕy(x, y) = −uy(x,−y) = vx(x,−y) = −ψx(x, y).
Similarly, we can prove that ϕx = ψy and ϕy = −ψx imply ux = vy and uy =
−vx. Hence, using Theorems 2.2.2 and 2.2.28, we get that f is differentiable
iff ux, uy, vx and vy are continuous and satisfy C–R equations iff ϕx, ϕy, ψx
and ψy are continuous and satisfy C–R equations iff g is differentiable. �
2.3 POWER SERIES AND ABEL’S THEOREMS
Definition 2.3.1 The series of functions
∞∑
n=0
an(z−a)n is called a power series,
where a ∈ C and (an) is a sequence of complex numbers.
The convergence of a power series depends on the sequence (an) and the
point z.
THEOREM 2.3.2 (Abel’s theorem on convergence of a power series)
For a given power series
∞∑
n=0
an(z − a)n, we assign an extended real number
R, called the radius of convergence, satisfying the following properties:
1.
∞∑
n=0
an(z − a)n converges absolutely on B(a, R).
2. If 0 < S < R, then
∞∑
n=0
an(z − a)n converges absolutely and uniformly
on {z : |z − a| ≤ S}.
3. If |z − a| > R, then
∞∑
n=0
an(z − a)n does not converge.
Proof: Define R =
(
lim sup
n→∞
|an| 1
n
)−1
∈ [−∞,∞] with the convention that
1
0
= ∞ and
1
∞ = 0. We claim that this R has the desired properties.
�
�
“book” — 2014/6/4 — 20:46 — page 95 — #25
�
�
�
�
�
�
Analytic Functions 95
1. Let |z − a| < R. Then, choose r ∈ R such that |z − a| < r < R. As
1
R
<
1
r
, we can write
1
r
= 1
R
+ ε, where ε = 1
r
− 1
R
> 0. Then, by
definition of lim sup
n→∞
|an| 1
n , there exists m ∈ N such that
|an| 1
n <
1
R
+ ε = 1
r
, ∀n ≥ m.
Hence, it follows that |an| < 1
rn
, ∀n ≥ m. Therefore, we have
|an(z − a)n| ≤
( |z − a|
r
)n
, ∀n ≥ m
As
|z − a|
r
< 1, we get that
∞∑
n=0
( |z − a|
r
)n
converges. (See Theorem
1.3.31.) Thus, by comparison test (Theorem 1.3.30), we get that the
series
∞∑
n=0
an(z − a)n converges absolutely.
2. If |z − a| < S < R, then choose r ∈ R such that S < r < R. Arguing
as before, we get that there exists m ∈ N such that |an| < 1
rn
, ∀n ≥ m.
Hence, it follows that
|an(z − a)n| ≤
(
S
r
)n
, ∀z with |z − a| ≤ S, ∀n ≥ m.
Since
S
r
< 1, we get that the series
∞∑
n=0
(
S
r
)n
converges. Again by
comparison test, we get that
∞∑
n=0
an(z − a)n converges absolutely and
uniformly on |z − a| ≤ S as the convergence of the series
∞∑
n=0
(
S
r
)n
used in the comparison test is independent of z.
3. If |z − a| > R, then we choose t ∈ R such that |z − a| > t > R. Then,
1
t
<
1
R
. From the definition of lim sup
n→∞
|an| 1
n , there exists a subsequence
(|ank |
1
nk ) of (|an| 1
n ), which converges to
1
R
. Then for ε = 1
R
− 1
t
> 0,
there exists m ∈ N such that∣∣∣∣|ank |
1
nk − 1
R
∣∣∣∣ < 1
R
− 1
t
, ∀k ≥ m.
�
�
“book” — 2014/6/4 — 20:46 — page 96 — #26
�
�
�
�
�
�
96 Power Series and Abel’s Theorems
Therefore, we have |ank |
1
nk >
1
t
, ∀k ≥ m, which implies
|ank | >
1
tnk
, ∀k ≥ m.
Hence, for |z − a| > t and k ≥ m, we have
|ank (z − a)nk | >
( |z − a|
t
)nk
> 1, ∀k ≥ m.
Thus, the subsequence (ank (z−a)nk ) of (an(z−a)n) does not converge to
0. Hence, (an(z−a)n) itself does not converge to 0. Thus,
∞∑
n=0
an(z−a)n
diverges by using Result 1.3.27. �
THEOREM 2.3.3 Let the radius of convergence of
∞∑
n=0
an(z − a)n be R. Then
1. the radius of convergence of
∞∑
n=1
nan(z − a)n−1 is also R,
2. if f (z) =
∞∑
n=0
an(z−a)n on |z−a| < R, then f is analytic on |z−a| < R
and f ′(z) =
∞∑
n=1
nan(z − a)n−1.
Proof: Recall that
1
R
= lim sup
n→∞
|an| 1
n .
1. First, we prove that n
1
n → ∞ as n → ∞. If rn = n
1
n − 1, ∀n ∈ N, then
we have for each n ≥ 2,
n = (rn + 1)n = 1 + nC1rn + nC2r2
n + · · · + rn
n >
n(n − 1)
2
r2
n,
which implies that rn <
√
2
n − 1
→ 0 as n → ∞. Therefore, n
1
n −1 →
0 as n → ∞. Hence, our claim follows. Using Theorem 1.3.19, we get
lim sup
n→∞
n
1
n |an| 1
n = lim
n→∞ n
1
n lim sup
n→∞
|an| 1
n = R.
�
�
“book” — 2014/6/4 — 20:46 — page 97 — #27
�
�
�
�
�
�
Analytic Functions 97
2. Put g(z) =
∞∑
n=1
nan(z − a)n−1, Sm(z) =
m∑
n=0
an(z − a)n, and Rm(z) =
∞∑
n=m+1
an(z − a)n, ∀z ∈ B(a, R) and ∀m ∈ N. Obviously, we have the
following:
(a) f (z) = Sm(z) + Rm(z), ∀m ∈ N on B(a, R).
(b) S′
m(z) → g(z) as m → ∞ on B(a, R).
(c)
∞∑
n=m+1
n|an|rn−1 → 0 as m → ∞ for 0 ≤ r < R, using Result
1.3.26.
For a fixed z0 ∈ B(a, R) and any z ∈ B(a, R), choose r ∈ R such that
max{|z − a|, |z0 − a|} < r < R.
Choose m1, m2 ∈ N such that∣∣S′
m(z0) − g(z0)
∣∣ < ε
3
, ∀m ≥ m1 (2.4)
and ∞∑
n=m+1
n|an|rn−1 <
ε
3
, ∀m ≥ m2. (2.5)
For m ≥ m2, we have∣∣∣∣Rm(z) − Rm(z0)
z − z0
∣∣∣∣ ≤
∞∑
n=m+1
|an| |(z − a)n − (z0 − a)n|
|z − z0|
≤
∞∑
n=m+1
|an|
n−1∑
k=0
|(z − a)k(z0 − a)n−k−1|
(using (A − B)n = (A − B)
n−1∑
k=0
AkBn−k)
<
∞∑
n=m+1
n|an|rn−1 <
ε
3
. (2.6)
From the definition of S′
m(z0), given ε > 0, there exists δ > 0 such that∣∣∣∣Sm(z) − Sm(z0)
z − z0
− S′
m(z0)
∣∣∣∣ < ε
3
whenever 0 < |z − z0| < δ. (2.7)
�
�
“book” — 2014/6/4 — 20:46 — page 98 — #28
�
�
�
�
�
�
98 Power Series and Abel’s Theorems
For m > max{m1,m2} and for z with 0 < |z − z0| < δ, we get∣∣∣∣ f (z) − f (z0)
z − z0
− g(z0)
∣∣∣∣
=
∣∣∣∣Sm(z) − Sm(z0)
z − z0
− S′
m(z0) + Rm(z) − Rm(z0)
z − z0
+ S′
m(z0) − g(z0)
∣∣∣∣
≤
∣∣∣∣Sm(z) − Sm(z0)
z − z0
− S′
m(z0)
∣∣∣∣+
∣∣∣∣Rm(z) − Rm(z0)
z − z0
∣∣∣∣+ ∣∣S′
m(z0) − g(z0)
∣∣
<
ε
3
+ ε
3
+ ε
3
= ε
Hence, f is differentiable and f ′ = g. �
COROLLARY 2.3.4 Let the power series
∞∑
n=0
an(z−a)n converge in |z−a| < R
for some R > 0. If f (z) =
∞∑
n=0
an(z − a)n, ∀z ∈ B(a, R), then an = f (n)(a)
n! ,
∀n = 0, 1, 2, 3, . . .
Proof: Applying the previous theorem repeatedly, first we note that f (m)
exists for every m = 0, 1, 2, 3, . . . and
f (m)(z) = dm
dzm
(
m−1∑
n=0
an(z − a)n +
∞∑
n=m
an(z − a)n
)
=
∞∑
n=m
n(n − 1) · · · (n − m + 1)an(z − a)n−m
= amm! +
∞∑
n=m+1
n(n − 1) · · · (n − m + 1)an(z − a)n−m
and hence, f (m)(a) = amm!. Thus, the corollary follows. �
COROLLARY 2.3.5
∞∑
n=0
an(z − a)n =
∞∑
n=0
bn(z − a)n, ∀z ∈ B(a, R) iff an =
bn, ∀n = 0, 1, 2, 3, . . .
Proof: If an = bn, ∀n = 0, 1, 2, 3, . . ., then obviously we get
∞∑
n=0
an(z − a)n =
∞∑
n=0
bn(z − a)n.
�
�
“book” — 2014/6/4 — 20:46 — page 99 — #29
�
�
�
�
�
�
Analytic Functions 99
To prove the converse, let f (z) =
∞∑
n=0
an(z − a)n, ∀z ∈ B(a, R). If an =
bn,∀n = 0, 1, 2, 3, . . . then by using previous corollary, we get an = f (n)(a)
n!
=
bn, ∀n = 0, 1, 2, 3, . . . �
Remark 2.3.6: If f is an analytic function on B(a, R), then
∞∑
n=0
f (n)(a)
n!
(z − a)n
is called the Taylor’s series of f . The corollary 2.3.5 states that if
f (z) =
∞∑
n=0
an(z − a)n, ∀z ∈ B(a, R),
then the power series
∞∑
n=0
an(z − a)n is same as the Taylor’s series of f .
Example 2.3.7 Find the radius of convergence of the following power series:
1.
∞∑
n=0
nnzn;
2.
∞∑
n=0
n2zn;
3.
∞∑
n=0
(
1 − 1
n
)n2
zn;
4.
∞∑
n=0
(4 + i3)nzn.
Solution:
1. Let an = nn, ∀n = 0, 1, 2, . . . . As |an|1/n = n → ∞ as n → ∞, the
radius of convergence of this power series is 0.
2. Let an = n2, ∀n = 0, 1, 2, . . . . As |an|1/n = n2/n → ∞ as n → ∞, the
radius of convergence of this power series is 0.
3. As lim
n→∞
((
1 − 1
n
)n2) 1
n
= lim
n→∞
(
1 − 1
n
)n
= 1
e
, the radius of
convergence of this power series is e.
�
�
“book” — 2014/6/4 — 20:46 — page 100 — #30
�
�
�
�
�
�
100 Power Series and Abel’s Theorems
4. As lim
n→∞(|4 + i3|n)
1
n = |4 + i3| = 5, the radius of convergence of
∞∑
n=0
(4 + i3)nzn is
1
5
.
Exercise 2.3.8
1.
∞∑
n=0
2n
n!
zn;
2.
∞∑
n=0
2
√
nzn;
3.
∞∑
n=0
n + 1
(n + 2)(n + 3)
zn (Hint. Find lim
n→∞
an+1
an
which is same as lim
n→∞
n
√
an.);
4.
∞∑
n=0
1
n2
zn;
5.
∞∑
n=0
(n!)2
(2n)!
zn.
Answer: 1. ∞; 2. 1; 3. 1; 4. 1; 5. 4.
THEOREM 2.3.9 (Abel’s limit theorem)
Let
∞∑
n=0
an converge. If f (z) =
∞∑
n=0
anzn, then f (z) → f (1) as z → 1 such that
|1 − z|
1 − |z| is bounded.
Proof: First we note that, as
∞∑
n=0
an converges at 1, the radius R of conver-
gence of
∞∑
n=0
an is at least 1 by Abel’s theorem. If R > 1, then f is continuous
at 1, and hence, the theorem follows. Therefore, assume that R = 1.
Case 1: f (1) =
∞∑
n=0
an = 0.
If αn =
n∑
k=0
ak and Sn(z) =
n∑
k=0
akzk , ∀z ∈ B(0, 1) and ∀n = 0, 1, 2, . . .,
�
�
“book” — 2014/6/4 — 20:46 — page 101 — #31
�
�
�
�
�
�
Analytic Functions 101
then for z ∈ B(0, 1) with
|1 − z|
1 − |z| ≤ K, for some K > 0, we get
Sn(z) =
n∑
k=0
akzk
= α0 +
n∑
k=1
(αk − αk−1)zk
= α0 +
n∑
k=1
αkzk −
n∑
k=1
αk−1zk
= α0 +
n−1∑
k=1
αkzk + αnzn − α0z −
n−1∑
k=1
αkzk+1
= α0(1 − z) +
n−1∑
k=1
αk(zk − zk+1) + αnzn
= α0(1 − z) +
n−1∑
k=1
αkzk(1 − z) + αnzn
= (1 − z)
n−1∑
k=0
αkzk + αnzn.
Allowing n → ∞ on both sides, we get
f (z) = (1 − z)
∞∑
k=0
αkzk + lim
n→∞αnzn = (1 − z)
∞∑
k=0
αkzk ,
as αn → f (1) = 0 as n → ∞ and |z| < 1. (cf. Proof of Example
1.3.31.)
Again using αn → 0 as n → ∞, given ε > 0, there exists m ∈ N such
that |αn| < ε
2K
, ∀n ≥ m and choose δ such that
0 < δ <
ε
2
(
1 +
m−1∑
k=1
|αn|
) .
�
�
“book” — 2014/6/4 — 20:46 — page 102 — #32
�
�
�
�
�
�
102 Exponential and Trigonometric Functions
If 0 < |z − 1| < δ and
|1 − z|
1 − |z| ≤ K, then
| f (z)| ≤ |1 − z|
(
m−1∑
k=0
|αkzk | +
∞∑
k=m
|αkzk |
)
< |1 − z|
(
m−1∑
k=0
|αk | + ε
2K
∞∑
k=m
|zk |
)
< δ
m−1∑
k=0
|αk | + ε
2K
|zm||1 − z|
∞∑
k=0
|zk |
<
ε
2
+ ε
2K
|1 − z|
1 − |z|
≤ ε
2
+ ε
2
= ε.
Hence, the theorem follows in this case.
Case 2: f (1) � 0
If g(z) = f (z) − f (1) = a0 − f (1) +
∞∑
n=1
anzn, then g(1) = 0. Then, by
Case 1, we obtain
g(z) → g(1) as z → 1 with
|1 − z|
1 − |z| is bounded,
and hence,
f (z) − f (1) → 0 as z → 1 with
|1 − z|
1 − |z| is bounded.
Thus, the proof is complete. �
2.4 EXPONENTIAL AND TRIGONOMETRIC
FUNCTIONS
Definition 2.4.1 The unique solution of the initial value problem f ′ = f with
f (0) = 1 is called the exponential function and is denoted by exp.
LEMMA 2.4.2 exp(z) =
∞∑
n=0
zn
n!
for z ∈ C at which the series converges.
�
�
“book” — 2014/6/4 — 20:46 — page 103 — #33
�
�
�
�
�
�
Analytic Functions 103
Proof: Let exp(z) =
∞∑
n=0
anzn. As it is the solution of the above initial value
problem, we have
∞∑
n=0
anzn =
∞∑
n=1
an−1zn−1 =
∞∑
n=1
nanzn−1, and a0 = f (0) = 1.
Hence, an = an−1
n
, n ∈ {0, 1, 2, . . .}, and a0 = f (0) = 1. We claim that
an = 1
n!
, ∀n = 0, 1, 2, 3, . . .
We prove our claim by induction on n. For n = 0, we have a0 = 1. Assume
that the claim holds for some n ∈ N. Now,
an+1 = an
n + 1
= 1
(n + 1)n!
= 1
(n + 1)!
.
Thus, our claim follows. Therefore, exp(z) =
∞∑
n=0
zn
n!
, ∀z ∈ B(0, R), where R
is the radius of convergence of
∞∑
n=0
zn
n!
. �
LEMMA 2.4.3 The radius of convergence of exp(z) =
∞∑
n=0
zn
n!
is ∞.
Proof: First, we note that
lim
n→∞
1
(n + 1)!
1
n!
= lim
n→∞
1
n + 1
= 0.
Hence, applying Theorems 1.3.17 and 1.3.20, we get
0 = lim inf
n→∞
1
(n + 1)!
1
n!
≤ lim inf
n→∞
(
1
n!
) 1
n
≤ lim sup
n→∞
(
1
n!
) 1
n
�
�
“book” — 2014/6/4 — 20:46 — page 104 — #34
�
�
�
�
�
�
104 Exponential and Trigonometric Functions
≤ lim sup
n→∞
1
(n + 1)!
1
n!
= 0
and hence, lim sup
n→∞
(
1
n!
) 1
n
= 0.
Therefore, the radius of convergence of
∞∑
n=0
zn
n!
is ∞. �
LEMMA 2.4.4 For every z1, z2 ∈ C, exp(z1 + z2) = exp(z1) · exp(z2).
Proof: As
d
dz
(exp(z) · exp(w − z)) = exp(z) · exp(w − z) − exp(z) · exp(w − z) = 0,
we get exp(z) · exp(w − z) as a constant, say K. Putting z = 0 and using
exp(0) = 1, we get K = exp(w). Thus, we obtain
exp(z) · exp(w − z) = exp(w).
Replacing z by z1 and w by z1 + z2, we get exp(z1 + z2) = exp(z1) ·
exp(z2). �
The above property of exponential function is called the addition theorem
for exp and is the reason for denoting exp(z) by ez.
COROLLARY 2.4.5 For every z ∈ C, exp(z) � 0, and its multiplicative inverse
is exp(−z).
Proof: As for every z ∈ C, 1 = exp(0) = exp(z − z) = exp(z) · exp(−z), we
get exp(z) � 0, and its multiplicative inverse is exp(−z). �
LEMMA 2.4.6 exp : R → (0,∞) is a bijection.
Proof: First, note that exp(0) = 1, for x > 0, we have exp(x) =
∞∑
n=0
xn
n!
>
1 and by previous corollary, for x < 0, exp(x) = 1
exp(−x) ∈ (0, 1), since
−x > 0 ⇒ exp(−x) > 1. Therefore, we have exp(x) ≥ 1 if x ≥ 0 and
0 < exp(x) < 1 if x < 0. So, exp is a mapping from R into (0,∞). To prove
exp : R → (0,∞) is onto, let y > 0 be arbitrary.
�
�
“book” — 2014/6/4 — 20:46 — page 105 — #35
�
�
�
�
�
�
Analytic Functions 105
Case (1): y = 1. Then, by definition, we have 0 ∈ R such that
exp(0) = 1.
Case (2): y > 1. Then, by definition, we have exp′(x) =
∞∑
n=0
xn
n!
> x > 0, and
hence, exp(x) → ∞ as x → ∞. Using this fact, we can choose s >
0 such that exp(s) > y > 1 = exp(0). As exp is differentiable, it
is continuous. Therefore, by intermediate value theorem (Theorem
1.6.30), there exists x ∈ (0, s) such that exp(x) = y.
Case (3): y < 1. Then,
1
y
> 1. Therefore, by Case (2), there exists t > 0
such that exp(t) = 1
y
. Therefore, exp(−t) = (exp(t))−1 = y.
For every x ∈ R,we have exp′(x) = exp(x) > 0, and hence, exp is a strictly
increasing function (by Lemma 2.2.18), that is, x < y ⇒ exp(x) < exp(y).
Hence, exp is an one-to-one function. �
LEMMA 2.4.7 | exp(z)| = 1 iff z = iy for some y ∈ R.
Proof: Assume that z = iy for some y ∈ R. From the power series
representation of exp(z), it is obvious that exp(z) = exp(z), ∀z ∈ C.
Therefore,
| exp(iy)|2 = exp(iy) · exp(iy) = exp(iy − iy) = exp(0) = 1
using Lemma 2.4.4.
Conversely, assume that if z = x + iy and | exp(z)| = 1. Then, 1 =
| exp(x) · exp(iy)| = exp(x) implies that x = 0 since exp is a bijection from R
onto (0,∞). Thus, z = iy for some real y ∈ R. �
Definition 2.4.8 For every z ∈ C, define
cos(z) = exp(iz) + exp(−iz)
2
and sin(z) = exp(iz) − exp(−iz)
i2
.
By definition, both cos and sin are differentiable functions on C, and the
power series representations of these two functions are cos(z) =
∞∑
n=0
(−1)nz2n
(2n)!
and sin(z) =
∞∑
n=0
(−1)nz(2n+1)
(2n + 1)!
.
�
�
“book” — 2014/6/4 — 20:46 — page 106 — #36
�
�
�
�
�
�
106 Exponential and Trigonometric Functions
LEMMA 2.4.9 For every z ∈ C,
1. sin(−z) = − sin(z) and cos(−z) = cos(z).
2. exp(iz) = cos(z) + i sin(z). (Euler’s formula)1.
3. (cos(z) + i sin(z))n = cos nz + i sin nz, ∀n ∈ Z (de Moivre’s formula).
Proof: By direct computation, we get (1) and the Euler’s formula as given
below.
sin(−z) = exp(−iz) − exp(iz)
i2
= −exp(iz) − exp(−iz)
i2
= − sin(z)
cos(−z) = exp(−iz) + exp(iz)
2
= cos(z)
cos(z) + i sin(z) = exp(iz) + exp(−iz)
2
+ i
exp(iz) − exp(−iz)
i2
= exp(iz).
Next, by using Lemma 2.4.4 and Euler’s formula, for n ∈ N, we get
(cos(z) + i sin(z))n = (exp(iz))n = exp(niz) = cos(nz) + i sin(nz).
Hence, for n = −1, we get
(cos(z) + i sin(z))−1 = (exp(iz))−1
= exp(−iz)
= cos(−z) + i sin(−z)
= cos(z) − i sin(z).
If n is a negative integer, then −n ∈ N and n = −(−n). Therefore,
(cos(z) + i sin(z))n = (cos(z) + i sin(z))−(−n)
= (cos(−nz) + i sin(−nz))−1
= cos(nz) + i sin(nz).
Thus, the de Moivre’s formula is obtained. �
LEMMA 2.4.10 For every z ∈ C, sin2(z) + cos2(z) = 1.
1Warning! cos(z) � Re eiz and sin(z) � Im eiz for z ∈ C.
�
�
“book” — 2014/6/4 — 20:46 — page 107 — #37
�
�
�
�
�
�
Analytic Functions 107
Proof: By direct computation, we get
sin2(z) + cos2(z)
= (exp(iz) − exp(−iz))2
(i2)2
+ (exp(iz) + exp(−iz))2
4
= −(exp(iz))2 − (exp(−iz))2 + 2 + (exp(iz))2 + (exp(−iz))2 + 2
4
= 4
4
= 1. �
Example 2.4.11 Expand cos(7θ ) as a polynomial in cos(θ ).
Solution:
By de Moivre’s formula, we have
cos(7θ ) = Re (cos(7θ ) + i sin(7θ))
= Re (cos(θ ) + i sin(θ ))7
= Re
7∑
k=0
7Ck cosk(θ )(i sin(θ ))7−k
= cos7(θ ) − 21 cos5(θ ) sin2(θ ) + 35 cos3(θ ) sin4(θ )
− 7 cos(θ ) sin6(θ )
= cos7(θ ) − 21 cos5(θ )(1 − cos2(θ )) + 35 cos3(θ )(1 − cos2(θ ))2
− 7 cos(θ )(1 − cos2(θ ))3
= cos7(θ ) − 21 cos5(θ )(1 − cos2(θ ))
+ 35 cos3(θ )(1 − 2 cos2(θ ) + cos4(θ ))
−7 cos(θ)(1 − 3 cos2(θ ) + 3 cos4(θ ) − cos6(θ ))
= 64 cos7(θ ) − 112 cos5(θ ) + 56 cos3(θ ) − 7 cos(θ ).
LEMMA 2.4.12 For every z, w ∈ C, (1) sin(z + w) = sin(z) cos(w) +
cos(z) sin(w) and (2) cos(z + w) = cos(z) cos(w) − sin(z) sin(w).
Proof: As
i2 sin(z + w) = exp(i(z + w)) − exp(−i(z + w))
= exp(iz) exp(iw) − exp(−iz) exp(−iw)
= (cos(z) + i sin(z))(cos(w) + i sin(w))
− (cos(z) − i sin(z))(cos(w) − i sin(w))
= (cos(z) cos(w) − sin(z) sin(w)) + i(cos(z) sin(w)
�
�
“book” — 2014/6/4 — 20:46 — page 108 — #38
�
�
�
�
�
�
108 Exponential and Trigonometric Functions
+ sin(z) cos(w)) − (cos(z) cos(w) − sin(z) sin(w))
+ i(cos(z) sin(w) + sin(z) cos(w))
= i2(cos(z) sin(w) + sin(z) cos(w)),
we get sin(z + w) = sin(z) cos(w) + cos(z) sin(w).
Similarly,
2 cos(z + w) = exp(i(z + w)) + exp(−i(z + w))
= exp(iz) exp(iw) + exp(−iz) exp(−iw)
= (cos(z) + i sin(z))(cos(w) + i sin(w))
+ (cos(z) − i sin(z))(cos(w) − i sin(w))
= (cos(z) cos(w) − sin(z) sin(w)) + i(cos(z) sin(w)
+ sin(z) cos(w)) + (cos(z) cos(w) − sin(z) sin(w))
− i(cos(z) sin(w) + sin(z) cos(w))
= 2(cos(z) cos(w) − sin(z) sin(w)).
�
COROLLARY 2.4.13 Prove that for z, w ∈ C, sin(z − w) = sin(z) cos(w) −
cos(z) sin(w) and cos(z − w) = cos(z) cos(w) + sin(z) sin(w).
Proof: In the previous lemma, replacing w by −w and using
sin(−w) = − sin(w), cos(−w) = cos(w),
we obviously get this corollary. �
LEMMA 2.4.14 (1)
d
dz
sin(z) = cos(z) and (2)
d
dz
cos(z) = − sin(z).
Proof: By direct calculation, we obtain
1.
d
dz
sin(z) = d
dz
(
exp(iz) − exp(−iz)
i2
)
= i exp(iz) + i exp(−iz)
i2
= exp(iz) + exp(−iz)
2
= cos z.
2.
d
dz
cos z = d
dz
(
exp(iz) + exp(−iz)
2
)
= i exp(iz) − i exp(−iz)
2
= −
(
exp(iz) − exp(−iz)
i2
)
= − sin z.
�
�
�
“book” — 2014/6/4 — 20:46 — page 109 — #39
�
�
�
�
�
�
Analytic Functions 109
RESULT 2.4.15 There exists a least r0 > 0 such that cos(r0) = 0.
Proof: For every r > 0, by mean-value theorem, we have sin(r) − sin(0) =
cos(t), for some t ∈ (0, r), which implies that sin(r) ≤ r, ∀r > 0 as | cos(t)| ≤
| exp(it)| = 1 and sin(0) = exp(0) − exp(0)
i2
= 0. Applying
r∫
0
on both sides
of sin(r) < r, we get
[− cos(r)]r
0 ≤
[
r2
2
]r
0
⇒ − cos(r) + 1 ≤ r2
2
⇒ cos(r) ≥ 1 − r2
2
.
Applying
r∫
0
on both sides of the resulting inequalities repeatedly, we get
sin(r) ≥ r − r3
6
⇒ − cos(r) + 1 ≥ r2
2
− r4
24
⇒ cos(r) ≤ 1 − r2
2
+ r4
24
.
Therefore, cos(
√
3) ≤ 1 − 3
2
+ 9
24
< 0, and from the definition, we obtain
cos(0) = 1. Therefore, by intermediate value theorem (Theorem 1.6.30),
there exists r0 ∈ (0,
√
3) such that cos(r0) = 0.
Next, we claim that if 0 < r < r0 <
√
3, then cos(r) � 0. From one of the
above inequalities, we get
sin(r) ≥ r − r3
6
= r
(
1 − r2
6
)
> r
(
1 − (
√
3)2
6
)
= r
2
> 0.
Therefore,
d
dr
cos(r) = − sin(r) < 0, and hence, cos is strictly decreasing on
(0, r0), applying Lemma 2.2.18. Since sin(r) > 0, ∀r ∈ (0, r0) and sin2(r) +
cos2(r) = 1, we get sin, and it is strictly increasing on (0, r0). Therefore,
0 < sin(r) < sin(r0) =
√
1 − cos2(r0) = 1.
Therefore, cos2(r) = 1 − sin2(r) > 0, and hence, cos(r) � 0. �
Definition 2.4.16 Define π = 2r0, where r0 is the least positive real number
such that cos(r0) = 0.
Example 2.4.17 Prove that cos
(π
2
)
= 0, sin
(π
2
)
= 1, cos(π ) = −1,
sin(π ) = 0, cos(2π ) = 1, and sin(2π) = 0.
�
�
“book” — 2014/6/4 — 20:46 — page 110 — #40
�
�
�
�
�
�
110 Exponential and Trigonometric Functions
By definition, π = 2r0, where r0 is the least positive real number such that
cos(r0) = 0. Therefore,
cos
(π
2
)
= cos(r0) = 0, sin
(π
2
)
= ±
√
1 − cos2
(π
2
)
= ±1,
but sin
(π
2
)
= sin(r0) > 0, we have sin
π
2
= 1. Therefore, exp
(
i
π
2
)
= i,
which implies that
exp(iπ ) = i2 = −1 and exp(i2π ) = (−1)2 = 1.
Equating real and imaginary parts in the above two equations, we can get the
required identities.
Exercise 2.4.18 Prove that for every z ∈ C in the following:
1. sin
(
z + π
2
) = cos(z).
2. cos
(
z + π
2
) = − sin(z).
3. sin
(
z − π
2
) = − cos(z).
4. cos
(
z − π
2
) = sin(z).
5. sin (z + π) = − sin(z).
6. cos (z + π) = − cos(z).
7. sin (z − π) = − sin(z).
8. cos (z − π) = − cos(z).
Hint. Apply Result 2.4.12 and Corollary 2.4.13.
LEMMA 2.4.19 If 0 < t < 2π, then exp(it) � 1.
Proof: If 0 < t < 2π and exp(it/4) = u + iv, then using the following:
1. 0 <
t
4
<
π
2
,
2. exp(it/4) = cos(t/4) + i sin(t/4),
3.
π
2
is the least positive real number such that cos
(π
2
)
= 0,
4. sin > 0 on
(
0,
π
2
)
,
�
�
“book” — 2014/6/4 — 20:46 — page 111 — #41
�
�
�
�
�
�
Analytic Functions 111
we get 0 < u < 1 and 0 < v < 1. As u + iv is not equal to any of 1,−1, i,
and −i, which are the solutions of z4 = 1, we obtain that exp(it/4) = u + iv
is not a solution of z4 = 1. Therefore, exp(it) � 1. �
RESULT 2.4.20 exp(z) = 1 iff z = 2kiπ for some k ∈ Z.
Proof: Let z = 2kiπ for some k ∈ Z. If k = 0, then by definition,
exp(2kiπ ) = exp(0) = 1. If k is a positive integer, then
exp(2kiπ ) = [exp(i2π )]k = 1.
If k is a negative integer, then k = −(−k) and −k is a positiveinteger.
Therefore, exp(2kiπ ) = [exp(2(−k)iπ )]−1 = 1.
To prove the converse, let z be such that exp(z) = 1. Then, 1 = | exp(z)|.
Therefore, using Lemma 2.4.7, we have z = iy for some y ∈ R. We claim that
y = 2kπ for some k ∈ Z. Suppose if it is not so, then there exists a unique
integer n such that 2nπ < y < 2(n + 1)π . Therefore,
1 = exp(iy)
= exp(i(y − 2nπ + 2nπ))
= exp(i(y − 2nπ)) exp(i2nπ)
= exp(i(y − 2nπ)),
which is a contradiction to Lemma 2.4.19 as 0 < y − 2nπ < 2π . Thus,
z = 2kiπ for some k ∈ Z. �
Now, we define periodic function, which will be studied in detail in
Chapter 8.
Definition 2.4.21 Let f : 
 → C and let a ∈ 
. We say that ‘a’ is a period
of f if f (z + a) = f (z), ∀z ∈ 
, where 
 = C or R. A function is said to be
a periodic function if it has a non-zero period.
Remark 2.4.22: From addition theorem of exp function and Result 2.4.20,
we conclude that i2kπ are periods of exp.
Now, we prove a simple but useful lemma on the integrals of a periodic
function.
LEMMA 2.4.23 If f : R → R is a periodic function of period p > 0 and
a ∈ R, then
a+p∫
a
f (x) dx =
p∫
0
f (x) dx.
�
�
“book” — 2014/6/4 — 20:46 — page 112 — #42
�
�
�
�
�
�
112 Exponential and Trigonometric Functions
Proof: By Archimedian property, we can find a unique m ∈ Z such that
m ≤ a
p < (m + 1). Then, we have (m + 1)p < a + p and 0 ≤ a − mp < p.
Hence,
a+p∫
a
f (x) dx =
(m+1)p∫
a
f (x) dx +
a+p∫
(m+1)p
f (x) dx
=
p∫
a−mp
f (y − mp) dy +
a−mp∫
0
f (z − (m + 1)p) dz
(by putting y = x − mp and z = x − (m + 1)p)
=
p∫
a−mp
f (y) dy +
a−mp∫
0
f (z) dz (since p is a period of f )
=
p∫
0
f (y) dy.
Hence, the lemma follows. �
RESULT 2.4.24 |z| = 1 iff z = exp(iy) for some y ∈ R.
Proof: If z = exp(iy) for some y ∈ R, then using Lemma 2.4.7, we get
|z| = 1. Conversely, assume that |z| = 1. As cos(0) = 1 and cos(π2 ) = 0 (see
Example 2.4.17), by intermediate value theorem (Theorem 1.6.30), for every
t ∈ [0, 1], there exists θ ∈ [0, π2 ] such that cos θ = t. Since |Re z| ≤ |z|, we
have |Re z| ≤ 1; similarly, we have |Im z| ≤ 1.
Case 1: If z = 1, then we choose θ = 0; obviously, we have exp(0) = 1.
If z = i, then we choose θ = π
2 so that exp
(
iπ2
) = cos
(
π
2
)+i sin
(
π
2
) =
i as cos
(
π
2
) = 0 and sin
(
π
2
) = 1.
If z = −1, then we choose θ = π so that exp(iπ ) = cos(π )+i sin(π ) =
−1 as cos(π ) = −1 and sin(π ) = 0.
If z = −i, then we choose θ = −π
2 so that exp
(−iπ2
) = cos
(
π
2
) −
i sin
(
π
2
) = −i as cos
(
π
2
) = 0 and sin
(
π
2
) = 1.
Case 2: 0 < Re z < 1, 0 < Im z < 1. Then, there exists θ ∈ (
0, π2
)
such
that Re z = cos(θ ). Using sin(θ ) > 0 in
(
0, π2
)
, we get
Im z =
√
1 − (Re z)2 =
√
1 − cos2(θ ) = sin(θ ),
�
�
“book” — 2014/6/4 — 20:46 — page 113 — #43
�
�
�
�
�
�
Analytic Functions 113
and hence,
exp(iθ ) = cos(θ ) + i sin(θ ) = Re z + iIm z = z
Case 3: −1 < Re z < 0, 0 < Im z < 1. Then, −Re z = cos(θ ) for some
θ ∈ (0, π2
)
. As sin(θ ) > 0 in
(
0, π2
)
, we get
Im z =
√
1 − (Re z)2 =
√
1 − cos2(θ ) = sin(θ ),
and hence,
exp(i(π − θ )) = exp(iπ ) exp(−iθ )
= (cos(π ) + i sin(π ))(cos(θ ) − i sin(θ ))
= − cos(θ ) + i sin(θ )
= Re z + iIm z
= z.
Case 4: −1 < Re z < 0, −1 < Im z < 0. Then, −Re z = cos(θ ) for some
θ ∈ (0, π2
)
. As sin(θ ) > 0 in
(
0, π2
)
, we get
−Im z =
√
1 − (Re z)2 =
√
1 − cos2(θ ) = sin(θ ),
and hence,
exp(i(−π + θ )) = exp(−iπ ) exp(iθ )
= (−1)(cos(θ ) + i sin(θ ))
= − cos(θ ) − i sin(θ )
= Re z + iIm z
= z.
Case 5: 0 < Re z < 1, −1 < Im z < 0. Then, Re z = cos(θ ) for some
θ ∈ (0, π2 ). As sin(θ ) > 0 in (0, π2 ), we get
−Im z =
√
1 − (Re z)2 =
√
1 − cos2(θ ) = sin(θ ),
and hence,
exp(−iθ ) = cos(θ ) − i sin(θ ) = Re z + iIm z = z.
Hence, the result follows. �
�
�
“book” — 2014/6/4 — 20:46 — page 114 — #44
�
�
�
�
�
�
114 Exponential and Trigonometric Functions
COROLLARY 2.4.25 Every non-zero complex number, z can be written as
z = |z| exp(iθ ) for some θ ∈ (−π ,π ].
Proof: As z � 0, we write z = |z| z
|z| , where
∣∣∣∣ z
|z|
∣∣∣∣ = 1. Therefore, from the
proof of previous result, we get
z
|z| = exp(iy) for some y ∈ (−π ,π ]. Hence,
the corollary follows. �
RESULT 2.4.26 If θ ∈ (0, π2 ), then, in a right-angled triangle with an angle θ ,
1. sin θ is equal to opposite side divided by hypotenuse.
2. cos θ is equal to adjacent side divided by hypotenuse.
Proof: Let the given triangle be �ABC with ∠A = θ , ∠B = π
2 , and
hypotenuse | �AC| = r. Fix the origin as A and the real axis as the line passing
through A and B.
A = (0,0) r cos (q )
r sin (q )
B
r
C
q
Then by Euler’s formula, we have C = r exp(iθ ) = (cos(θ ) + i sin(θ )).
(Note that the polar form r exp(iθ ) was used just as a symbol in Chapter 1
and by the previous corollary it is justified that it actually means in terms
of the exponential function. Therefore, we can use the Euler’s formula.)
Hence, AB = r cos(θ ) and | �BC| = r sin(θ ). Therefore, sin(θ ) = | �BC|
| �AC| =
opposite side
hypotenuse
and cos(θ ) = | �AB|
| �AC| =
adjacent side
hypotenuse
. �
�
�
“book” — 2014/6/4 — 20:46 — page 115 — #45
�
�
�
�
�
�
Analytic Functions 115
Definition 2.4.27 Define the following:
1. tan(z) = sin(z)
cos(z)
, ∀z ∈ C \ {z : cos(z) = 0}.
2. csc(z) = 1
sin(z)
, ∀z ∈ C \ {z : sin(z) = 0}.
3. sec(z) = 1
cos(z)
, ∀z ∈ C \ {z : cos(z) = 0}.
4. cot(z) = 1
tan(z)
= cos(z)
sin(z)
, ∀z ∈ C \ {z : sin(z) = 0}.
Example 2.4.28 Find the domains of tan, csc, sec, and cot.
We have already proved that
π
2
is the least positive real number at which cos
vanishes. See Result 2.4.15. Therefore, for any k ∈ Z,
cos
(π
2
+ kπ
)
=
exp
(
i
(π
2
+ kπ
))
+ exp
(
−i
(π
2
+ kπ
))
2
= i exp (ikπ)− i exp (−ikπ)
2
= i(−1)k − i(−1)−k
2
= 0 (since ± 1 = (±1)−1).
Conversely, if z ∈ C such that cos(z) = 0, then sin(z) = ±1 from sin2(z) +
cos2(z) = 1. Hence, using Result 2.4.20, it follows that
cos(z) + i sin(z) = ±i ⇒ exp(iz) = exp
(
i
π
2
)
or exp
(
i
3π
2
)
⇒ exp
(
i
(
z − π
2
))
= 1 or exp
(
i
(
z − 3
π
2
))
= 1
⇒ z − π
2
= 2nπ or z − 3π
2
= 2nπ , n ∈ Z
⇒ z = (4n + 1)
π
2
or z = (4n + 3)
π
2
, n ∈ Z
⇒ z = (2(2n) + 1)
π
2
or
z = (2(2n + 1) + 1)
π
2
, n ∈ Z
�
�
“book” — 2014/6/4 — 20:46 — page 116 — #46
�
�
�
�
�
�
116 Exponential and Trigonometric Functions
⇒ z = (2k + 1)
π
2
, k ∈ Z
⇒ z = kπ + π
2
, k ∈ Z.
Therefore,
{z ∈ C : cos(z) = 0} =
{
kπ + π
2
: k ∈ Z
}
.
Similarly, we get
sin(z) = 0 ⇔ cos(z) = ±1
⇔ exp(iz) = ±1 = exp(0) or exp(iπ )
⇔ exp(iz) = 1 or exp(i(z − π )) = 1
⇔ z = 2nπ or z = 2nπ + π , n ∈ Z
⇔ z = kπ , k ∈ Z.
Therefore,
{z ∈ C : sin(z) = 0} = {kπ : k ∈ Z} .
Therefore, the domain of tan and sec is same as C \
{
kπ + π
2
: k ∈ Z
}
and
that of cot and csc is C \ {kπ : k ∈ Z}.
THEOREM 2.4.29 lim
x→ π
2
−
tan(x) = +∞, lim
x→− π
2
+
tan(x) = −∞, and tan :(
−π
2
,
π
2
)
→ R is a bijection.
Proof: We know that
cos(0) = 1, cos
(π
2
)
= 0, and cos(x) � 0, ∀x ∈
(
0,
π
2
)
.
Hence, cos(x) > 0, ∀x ∈
(
0,
π
2
)
because if cos(y) < 0 for some 0 < y <
π
2
,
then by intermediate value theorem, we get cos, which vanishes at a point
between 0 and y, which is a contradiction to the fact that cos � 0 on
(
0,
π
2
)
.
As sin′ = cos > 0 on
(
0,
π
2
)
, by Lemma 1.6.8, we have sin, which is strictly
increasing on
(
0,
π
2
)
. As sin(0) = 0, we have sin > 0 on
(
0,
π
2
)
. Now
using the continuity of these functions at
π
2
, we get lim
x→ π
2
sin(x) = 1 and
lim
x→ π
2
cos(x) = 0. Therefore, for a given M > 0, we choose δ1 > 0 and δ2 > 0
such that
0 <
∣∣∣x − π
2
∣∣∣ < δ1 ⇒ | sin(x) − 1| < 1
2
⇒ | sin(x)| > 1 − 1
2
= 1
2
�
�
“book” — 2014/6/4 — 20:46 — page 117 — #47
�
�
�
�
�
�
Analytic Functions 117
and
0 <
∣∣∣x − π
2
∣∣∣ < δ2 ⇒ | cos(x)| < 1
2M
.
If δ = min{δ1, δ2, π2 }, then δ > 0, and
π
2
− δ < x <
π
2
⇒ | sin(x)|
| cos(x)| =
sin(x)
cos(x)
>
1
2 cos(x)
> M .
Therefore, lim
z→ π
2
−
tan(x) = +∞.
Similarly, using the continuity of cos and sin at
π
2
, we obtainthat for a given
K < 0, there exists δ3 > 0 such that
0 <
∣∣∣x − π
2
∣∣∣ < δ3 ⇒ | cos(x)| < 1
−2K
⇒ 2 cos x <
1
−k
⇒ 1
2 cos x
> −k ⇒ 1
−2 cos x
< k.
If r = min{δ1, δ3, π2 }, then r > 0, and −π
2
< x < −π
2
+ r implies
| sin(x)|
| cos(x)| =
− sin(x)
cos(x)
>
1
2 cos(x)
⇒ sin(x)
cos(x)
<
1
−2 cos(x)
< K
since sin < 0 and cos > 0 on
(
−π
2
, 0
)
. Therefore, lim
z→− π
2
+
tan(x) = −∞.
We know that tan(0) = sin(0)
cos(0)
= 0
1
= 0 and is continuous on
(
−π
2
,
π
2
)
.
Given any y > 0, using lim
x→ π
2
−
tan(x) = +∞, we find t ∈
(
0,
π
2
)
such that
tan(t) > y. Then by using intermediate value theorem, there exists x ∈ (0, t)
such that tan(x) = y. Similarly, for a given y < 0, using lim
x→− π
2
+
tan(x), we can
find x ∈
(
−π
2
, 0
)
such that tan(x) = y. Thus, tan :
(
−π
2
,
π
2
)
→ R is onto.
Since, tan′(x) = cos(x) cos(x) + sin(x) sin(x)
cos2(x)
= 1
cos2(x)
and cos(x) � 0, ∀x ∈(
−π
2
,
π
2
)
, we get that tan is strictly increasing on
(
−π
2
,
π
2
)
. Therefore, tan :(
−π
2
,
π
2
)
→ R is one-to-one. Thus, the theorem follows. �
THEOREM 2.4.30 The inverse arctan of tan is continuous from R onto(
−π
2
,
π
2
)
.
�
�
“book” — 2014/6/4 — 20:46 — page 118 — #48
�
�
�
�
�
�
118 Exponential and Trigonometric Functions
Proof: Let x ∈ R be arbitrarily fixed. Since tan :
(
−π
2
,
π
2
)
→ R is a
bijection, there exists unique t ∈
(
−π
2
,
π
2
)
such that tan(t) = x. For a given
ε > 0, we choose ε1 > 0 such that 0 < ε1 < min
{
ε,
π
2
− t, t + π
2
}
. As
tan :
(
−π
2
,
π
2
)
→ R is continuous and strictly increasing, we obtain the
image of (t− ε1, t+ ε1) under tan is exactly equal to (tan(t − ε1), tan(t + ε1))
and it contains x = tan(t). Hence, we find δ > 0 such that
(x − δ, x + δ) ⊆ (tan(t − ε1), tan(t + ε1)) = tan((t − ε1, t + ε1))
which implies that
arctan ((x − δ, x + δ)) ⊆ (t − ε1, t + ε1) = (arctan(x) − ε1, arctan(x) + ε1)
In other words, we have
|y − x| < δ ⇒ | arctan(y) − arctan(x)| < ε1 < ε
As x ∈ R is arbitrary, the theorem follows. �
Exercise 2.4.31 Prove that 1 + tan2(z) = sec2(z), 1 + cot2(z) = csc2(z), and
tan(z + w) = tan(z) + tan(w)
1 − tan(z) tan(w)
, ∀z, w ∈ C.
Exercise 2.4.32 Prove that arctan x → π
2
as x → +∞ and arctan x → −π
2
as x → −∞.
Exercise 2.4.33
1. Prove that tan(−z) = − tan(z), cot(z) = − cot(z), sec(−z) = sec(z),
and csc(−z) = − csc(z).
2. Prove the following identities:
(a) tan
(
z − π
2
) = − cot(z)
(b) cot
(
z − π
2
) = − tan(z)
(c) tan
(
z + π
2
) = − cot(z)
(d) cot
(
z + π
2
) = − tan(z)
�
�
“book” — 2014/6/4 — 20:46 — page 119 — #49
�
�
�
�
�
�
Analytic Functions 119
(e) sec
(
z − π
2
) = csc(z)
(f) csc
(
z − π
2
) = − sec(z)
(g) sec
(
z + π
2
) = − csc(z)
(h) csc
(
z + π
2
) = sec(z)
(i) tan (z − π) = tan(z)
(j) cot (z − π) = cot(z)
(k) tan (z + π) = tan(z)
(l) cot (z + π) = cot(z)
(m) sec (z − π) = − sec(z)
(n) csc (z − π) = − csc(z)
(o) sec (z + π) = − sec(z)
(p) csc (z + π) = − csc(z).
Hint. Use Exercise 2.4.18.
2.5 HYPERBOLIC FUNCTIONS
Definition 2.5.1 Define
cosh(z) = exp(z) + exp(−z)
2
and sinh(z) = exp(z) − exp(−z)
2
, ∀z ∈ C.
LEMMA 2.5.2 For every z ∈ C, we have cosh(z) + sinh(z) = exp(z).
Proof: For each z ∈ C, we get
cosh(z) + sinh(z) = exp(z) + exp(−z)
2
+ exp(z) − exp(−z)
2
= 2 exp(z)
2
= exp(z).
Hence, the lemma follows. �
LEMMA 2.5.3 cosh2(z) − sinh2(z) = 1, ∀z ∈ C.
�
�
“book” — 2014/6/4 — 20:46 — page 120 — #50
�
�
�
�
�
�
120 Hyperbolic Functions
Proof: For z ∈ C, we have
cosh2(z) − sinh2(z)
= (exp(z) + exp(−z))2
4
− (exp(z) − exp(−z))2
4
= (exp(z))2 + (exp(−z))2 + 2 − (exp(z))2 − (exp(−z))2 + 2
4
= 4
4
= 1. �
LEMMA 2.5.4 sinh and cosh are differentiable and sinh′ = cosh and cosh′ =
sinh.
Proof: As exp is a differentiable function on C, cosh and sinh are differen-
tiable functions on C and
cosh′(z) = d
dz
(
exp(z) + exp(−z)
2
)
= exp(z) − exp(−z)
2
= sinh(z)
sinh′(z) = d
dz
(
exp(z) − exp(−z)
2
)
= exp(z) + exp(−z)
2
= cosh(z).
Hence, the lemma follows. �
Exercise 2.5.5 Prove that the power series representations of cosh(z) and
sinh(z) are
∞∑
n=0
z2n
(2n)!
and
∞∑
n=0
z2n+1
(2n + 1)!
.
Exercise 2.5.6 For every z, w ∈ C, prove the following
1. sinh(−z) = − sinh(z).
2. cosh(−z) = cosh(z).
3. sinh(z + w) = sinh(z) cosh(w) + sinh(z) cosh(w).
4. cosh(z + w) = cosh(z) cosh(w) + sinh(z) sinh(w).
5. sinh(z − w) = sinh(z) cosh(w) − sinh(z) cosh(w).
6. cosh(z − w) = cosh(z) cosh(w) − sinh(z) sinh(w).
�
�
“book” — 2014/6/4 — 20:46 — page 121 — #51
�
�
�
�
�
�
Analytic Functions 121
Definition 2.5.7 Define the following:
1. tanh(z) = sinh(z)
cosh(z)
, ∀z ∈ {z ∈ C : cosh(z) � 0}.
2. csch (z) = 1
sinh(z)
, ∀z ∈ {z ∈ C : sinh(z) � 0}.
3. sech(z) = 1
cosh(z)
, ∀z ∈ {z ∈ C : cosh(z) � 0}.
4. coth(z) = 1
tanh(z)
, ∀z ∈ {z ∈ C : sinh(z) � 0}.
LEMMA 2.5.8 For z, w ∈ {z ∈ C : cosh(z) � 0},
tanh(z + w) = tanh(z) + tanh(w)
1 + tanh(z) tanh(w)
and tanh(z − w) = tanh(z) − tanh(w)
1 − tanh(z) tanh(w)
.
Proof: As
tanh(z + w) = sinh(z + w)
cosh(z + w)
= sinh(z) cosh(w) + sinh(z) cosh(w)
cosh(z) cosh(w) + sinh(z) sinh(w)
,
dividing the numerator and denominator by cosh(z) cosh(w), we get
tanh(z + w) =
sinh(z)
cosh(z)
+ sinh(w)
cosh(w)
1 + sinh(z)
cosh(z)
sinh(w)
cosh(w)
= tanh(z) + tanh(w)
1 + tanh(z) tanh(w)
. (2.8)
Since
tanh(−w) = sinh(−w)
cosh(−w)
= − sinh(w)
cosh(w)
= − tanh(w),
by replacing w by −w in equation (2.8), we get
tanh(z − w) = tanh(z) − tanh(w)
1 − tanh(z) tanh(w)
�
Exercise 2.5.9 Prove that for every z ∈ C, 1 − tanh2(z) = sech2 (z) and
coth2(z) − 1 = csch2 (z).
�
�
“book” — 2014/6/4 — 20:46 — page 122 — #52
�
�
�
�
�
�
122 Hyperbolic Functions
Finally, we present the relation between the trigonometric functions and
hyperbolic functions.
RESULT 2.5.10 For every z ∈ C, sin(iz) = i sinh(z) and cos(iz) = cosh(z).
The proof of the above result is easy, and hence, it is left to the reader as
a simple exercise.
�
�
“book” — 2014/6/4 — 20:56 — page 123 — #1
�
�
�
�
�
�
3
Rational Functions
and Multivalued
Functions
3.1 POLYNOMIALS AND RATIONAL FUNCTIONS
Definition 3.1.1 A polynomial is a function P : C → C, which is of the form
P(z) = a0 + a1z + a2z2 + · · · + anzn, ∀z ∈ C, where ai ∈ C, ∀i = 1, 2, . . . , n
and n ∈ N.
1. If n is the largest positive integer such that an � 0, then we call n
the degree of the polynomial P. We define the degree of a constant
polynomial by 0. We denote the degree of P by deg P.
2. If z0 ∈ C such that P(z0) = 0, then we call z0 a zero of P.
Now we state (without proof) a particular version of the Euclidean algorithm,
which is an useful result from algebra.
RESULT 3.1.2 (Division algorithm for polynomials)
Let P and Q be polynomials with complex coefficients. Then there exist
unique polynomials A and B such that P = AQ + B with either B = 0 or
deg B < deg Q.
LEMMA 3.1.3 If z0 is a zero of P, then we can write P(z) = (z − z0)Q(z),
∀z ∈ C for some polynomial Q.
123
�
�
“book” — 2014/6/4 — 20:56 — page 124 — #2
�
�
�
�
�
�
124 Polynomials and Rational Functions
Proof: By division algorithm, we can find polynomials Q(z) and R(z) such
that
P(z) = (z − z0)Q(z) + R(z), ∀z ∈ C
with 0 < deg R(z) < deg (z − z0) or R(z) = 0, ∀z ∈ C. As deg (z − z0) is 1,
R(z) must be a constant and is equal to
R(z0) = P(z0) − (z0 − z0)Q(z0) = 0.
Therefore, the lemma follows. �
Definition 3.1.4 If m ∈ N is such that
P(z) = (z − z0)mQ(z), ∀z ∈ C
for some polynomial Q such that Q(z0) � 0, then m is called the order of the
zero z0 of P.
RESULT 3.1.5 If m is the order of a zero of a non-constant polynomial P,
then P(k)(z0) = 0, ∀k ∈ {1, 2, 3, . . . , m − 1} and P(m)(z0) � 0.
Proof: Let z0 be a zero of order m, then by definition, we have
P(z) = (z − z0)mQ(z), ∀z ∈ C
for some polynomial Q. Using the Leibniz rule(Result 2.1.13), we have
dk
dzk
(
(z − z0)mQ(z)
) = k∑
j=0
kCj
dj
dzj
(z − z0)m · dk−j
dzk−j
Q(z), ∀k ∈ N.
For 1 ≤ k ≤ m − 1 and 0 ≤ j ≤ k, we havedj
dzj
(z − z0)m = m(m − 1) · · · (m − j + 1)(z − z0)m−j and m − j > 0
and hence, we get P(k)(z0) = 0, 1 ≤ k ≤ m− 1. Furthermore, using the same
observation, we obtain
P(m)(z0) =
[
dm
dzm
(z − z0)m
]
Q(z) = m!Q(z0) � 0.
Thus, the result follows. �
Converse of the above result is also true and is given by the following result.
�
�
“book” — 2014/6/4 — 20:56 — page 125 — #3
�
�
�
�
�
�
Rational Functions and Multivalued Functions 125
RESULT 3.1.6 Let P be a non-constant polynomial and z0 ∈ C. If m ∈ N is
such that P(k)(z0) = 0, ∀k ∈ {1, 2, 3 . . . , m − 1} and P(m)(z0) � 0, then the
order of zero of P at z0 is m.
Proof: If m = 1, then using Lemma 3.1.3, we have
P(z) = (z − z0)P1(z), ∀z ∈ C
for some polynomial P1. Using
P′(z) = (z − z0)P′
1(z) + P1(z), ∀z ∈ C,
we get 0 � P′(z0) = P1(z0).
If m = 2, using P(z0) = 0, then we write
P(z) = (z − z0)P1(z), ∀z ∈ C
and we get 0 = P′(z0) = P1(z0). Therefore,
P1(z) = (z − z0)P2(z), ∀z ∈ C
for some polynomial P2, and hence,
P(z) = (z − z0)2P2(z), ∀z ∈ C.
Using
P′′(z) = (z − z0)2P′′
2(z) + 4(z − z0)P′
2(z) + 2P2(z), ∀z ∈ C,
we get 0 � P′′(z0) = P2(z0).
Proceeding further at the (m − 1)st stage using P(m−1)(z0) = 0, we get
P(z) = (z − z0)mPm(z), ∀z ∈ C
for some polynomial Pm. Therefore, for each z ∈ C, we have
P(m)(z) =
m∑
j=0
mCj
dj
dzj [(z − z0)m] P(m−j)
m (z)
=
m∑
j=0
mCj
[
j−1∏
ν=0
(m − ν)
]
(z − z0)m−jP(m−j)
m (z).
Thus, 0 � P(m)(z0) = m!Pm(z0) ⇒ Pm(z0) � 0. Therefore, m is the order of
zero at z0 for P. �
�
�
“book” — 2014/6/4 — 20:56 — page 126 — #4
�
�
�
�
�
�
126 Polynomials and Rational Functions
Remark 3.1.7: Another simple proof of the previous result is given in
Corollary 6.1.4.
Next we discuss a theorem that states a relation between the zeroes of a poly-
nomial P with the zeroes of its derivative. For that purpose, now we discuss
how to represent a half plane described by a straight line. We know that the
equation of a straight line, which has the slope as the argument of a non-zero
complex number b and it passes through the complex number a, is given by
z = a + tb, t ∈ R. Hence, we have
z lies on the straight line iff
z − a
b
is a real number
iff Im
z − a
b
= 0.
Therefore, z is not on the straight line iff Im
z − a
b
� 0. Hence,
{
z ∈ C : Im
z − a
b
> 0
}
and
{
z ∈ C : Im
z − a
b
< 0
}
are the two half planes determined by the straight line z = a + tb, t ∈ R.
THEOREM 3.1.8 (Lucas’ theorem)
If every zero of a polynomial P is enclosed by a closed polygon K in C, then
every zero of P′ is also enclosed by K.
Proof: Let the closed polygon K is formed by the straight lines z = aj +
tbj, t ∈ R, j = 1, 2, . . . , m. That is, the region enclosed by the given polygon
is the intersection of m half planes each of which is determined by one of
these m straight lines z = aj + tbj, t ∈ R.
Closed polygon
Therefore, so to conclude this theorem, we shall show that every zero of P
lies in a half plane described by z = aj + tbj, t ∈ R and every zero of P′ also
lies in the same half plane for all j = 1, 2, . . . , m. If α1,α2,α3, . . . ,αn ∈ C
�
�
“book” — 2014/6/4 — 20:56 — page 127 — #5
�
�
�
�
�
�
Rational Functions and Multivalued Functions 127
are the zeroes of P including multiplicities. Then, using Lemma 3.1.3 and
fundamental theorem of algebra (Theorem 4.3.12), we get
P(z) = c(z − α1)(z − α2) · · · (z − αn), ∀z ∈ C and for some c ∈ C.
Therefore, either by direct calculation or by finding
d
dz
log P(z) (with less
rigor), we get
P′(z)
P(z)
= 1
z − α1
+ 1
z − α2
+ · · · + 1
z − αn
, ∀z ∈ C. (3.1)
Let 1 ≤ j ≤ m and let Hj =
{
z ∈ C : Im
z − aj
bj
> 0
}
be a half plane
described by the straight line z = aj + tbj, t ∈ R. Now assume that αk ∈ Hj,
∀k ∈ {1, 2, . . . , n}. Therefore, we have
Im
αk − aj
bj
> 0, ∀1 ≤ k ≤ n.
If z0 � Hj, then Im
z0 − aj
bj
≤ 0, and hence,
Im
z0 − αk
bj
= Im
z0 − aj
bj
− Im
αk − aj
bj
< 0.
As (Im w) × (Im w−1) < 0, for any complex number w with Im w � 0,
we have Im
bj
z0 − αk
> 0. Therefore,
Im
bjP′(z)
P(z)
=
n∑
k=1
Im
bj
z0 − bk
> 0 ⇒ bjP
′(z0) � 0 ⇒ P′(z0) � 0.
Thus, if P′(z0) = 0, then z0 ∈ Hj. Therefore, every zero of P′ lies in Hj. As j
is arbitrary, every zero of P′ lies in the closed polygon K. �
Definition 3.1.9 A function R : C∞ → C∞ is called a rational function if it
is a quotient of two polynomials.
Usually, we write a rational function in the reduced form in the sense that
if R = P
Q
, then P and Q are polynomials without common zeroes.
�
�
“book” — 2014/6/4 — 20:56 — page 128 — #6
�
�
�
�
�
�
128 Polynomials and Rational Functions
LEMMA 3.1.10 If R is any rational function, then lim
z→∞ R(z) exists in C∞.
Proof: Let R = P
Q
, where
P(z) =
m∑
k=0
akzk and Q(z) =
n∑
j=0
bjz
j, ∀z ∈ C
with am � 0 and bn � 0. We also assume that P and Q have no common
zeroes. Then, we have
lim
z→∞ R(z) = lim
z→∞
m∑
k=0
akzk
n∑
j=0
bjzj
= lim
z→∞ zm−n
m∑
k=0
akzk−m
n∑
j=0
bjzj−n
= lim
z→∞ zm−n am
bn
=
⎧⎨
⎩
an
bm
if n = m
0 if n > m
∞ if n < m.
Definition 3.1.11 The value of a rational function R at ∞ is defined by
lim
z→∞ R(z).
Definition 3.1.12 Let R be a rational function and z0 ∈ C∞, then
1. z0 is said to be a zero of R if R(z0) = 0.
2. z0 is said to be a pole of R if R(z0) = ∞.
Note that if R = P
Q is written in the reduced form, then every zero of P is a
finite zero of R and every zero of Q is a finite pole of R. Furthermore, ∞ is a
zero of R if deg P < deg Q and is a pole if deg P > deg Q.
Definition 3.1.13 Let R = P
Q
be a rational function, z0 ∈ C∞ be a zero of R,
and k ∈ N.
�
�
“book” — 2014/6/4 — 20:56 — page 129 — #7
�
�
�
�
�
�
Rational Functions and Multivalued Functions 129
1. If z0 ∈ C, then we say that R has a zero of order k at z0 if z0 is a zero
of order k for the polynomial P; and we say that R has a pole of order
k at z0 if z0 is a zero of order k for the polynomial Q.
2. We say that R has a zero of order k at ∞ if 0 is a zero of order k for the
rational function R1; and we say that R has a pole of order k at ∞ if 0
is a zero of order k for the rational function R1, where R1(z) = R
(
1
z
)
.
Note that if R = P
Q
, then R has a zero of order k at ∞ if deg P = k + deg Q
and has a pole of order k at ∞ if deg Q = k + deg P.
Remark 3.1.14: Let zj, j = 1, 2, . . . , n be the distinct zeroes (poles) of a
rational function R. If kj is the order of the zero (pole) at zj, ∀j = 1, 2, . . . , n,
then we mean the number of zeroes (poles) of R including multiplicities by
n∑
j=1
kj.
RESULT 3.1.15 Let R = P
Q be in the reduced form, where P and Q are poly-
nomials of degree m and n, respectively. If μ is the number of zeroes of R
including multiplicities in C∞ and ν is the number of poles of R including
multiplicities in C∞, then μ = ν = max{m, n}.
Proof: Obviously, there are m finite zeroes of R including multiplicities and
n finite poles of R including multiplicities.
Case 1: m = n. In this case, ∞ is neither a pole nor a zero for R. Therefore,
μ = m = n = ν.
Case 2: m > n. In this case, it is easy to verify that ∞ is a pole for R of order
m − n. Therefore, μ = m and ν = n + m − n = m.
Case 3: m < n. In this case, it is easy to verify that ∞ is a zero for R of order
n − m. Therefore, μ = m + n − m = n and ν = n.
From the above discussions, we get μ = ν = max{m, n}. �
LEMMA 3.1.16 Let R be a rational function. R is a polynomial iff the only
possible pole of R is ∞.
Proof: Let R be a polynomial. If R is a constant polynomial, then it has no
poles. If R is a non-constant polynomial, then obviously R(∞) = ∞, and
�
�
“book” — 2014/6/4 — 20:56 — page 130 — #8
�
�
�
�
�
�
130 Polynomials and Rational Functions
hence, ∞ is a pole of R. For every c ∈ C, as lim
z→c
R(z) = R(c) � ∞, it has
no other poles. Conversely, assume that R has only one pole at ∞. If R = P
Q
,
where P and Q are polynomials without common zeroes. If Q has a zero z0
in C, then z0 becomes a pole for R, which is a contradiction. Therefore, Q
must not have a zero inC, and hence, by the fundamental theorem of algebra
(Theorem 4.3.12), we get that Q is a constant. Thus, R is a polynomial. �
Definition 3.1.17 Let z1, z2, . . . , zn be the distinct finite poles of a rational
function R, then the expression
R(z) = G(z) +
n∑
j=1
Fj
(
1
z − zj
)
, ∀z ∈ C
is called the partial fraction expansion of R if G is a polynomial and Fj are
polynomials with Fj(0) = 0.
THEOREM 3.1.18 Every rational function has partial fraction expansion.
Proof: Let R = P
Q
be in reduced form. Hence, by division algorithm for
polynomials (Result 3.1.2), we can write P = AQ + B, where A and B are
polynomials such that either B = 0 or deg B < deg Q. Therefore, R = A+ B
Q
.
Let C = A − A(0) and D = A(0) + B
Q
. Hence, we can write R = C + D,
where C is a polynomial with C(0) = 0 and D is a rational function such that
D(∞) �∞, as D = A(0)Q + B
Q
and degree of (A(0)Q + B) ≤ degree of Q.
We note that if R(∞) is finite, then C = 0, and if R(∞) = ∞, then degree of
P > degree of Q ⇒ C is a non-constant polynomial. . . . . . . . . . . . . (I)
For every j = 1, 2, . . . , n, let Rj(w) = R
(
zj + 1
w
)
, ∀w ∈ C∞, where
z1, z2, . . . , zn are distinct finite poles of R. Note that Rj(∞) = ∞; then by
previous argument, there exist non-constant polynomial Cj with Cj(0) = 0
and rational function Dj with Dj(∞) �∞ such that
Rj(w) = Cj(w) + Dj(w), ∀w ∈ C∞, ∀j = 1, 2, . . . , n.
�
�
“book” — 2014/6/4 — 20:56 — page 131 — #9
�
�
�
�
�
�
Rational Functions and Multivalued Functions 131
Applying the change of variable w = 1
z − zj
, we get
R(z) = R
(
zj + 1
w
)
= Rj(w)
= Cj(w) + Dj(w)
= Cj
(
1
z − zj
)
+ Dj
(
1
z − zj
)
= Ej(z) + Fj(z), (say) ∀z ∈ C∞
and Fj(zj) �∞, j = 1, 2, . . . , n. Now the rational function
R − C −
n∑
j=1
Ej (3.2)
has possible poles from {z1, z2, . . . , zn,∞}.
Note that
1. R(zj) = ∞,
2. from Rj(∞) = ∞, by the observation (I), we get Cj is a non-constant
polynomial, and hence, Cj(∞) = ∞ ⇒ Ej(zj) = Cj(∞) = ∞.
Although R(zj) = ∞ and Ej(zj) = ∞, as R(zj) − Ej(zj) = Fj(zj) � ∞, it
follows that zj is not a pole for the rational function in equation(3.2). Next
from the following statements:
1. If R(∞) = ∞, then using (I), we get C(∞) = ∞, but as
(R − C)(∞) = D(∞) �∞,
we get that ∞ is not a pole for R.
2. If R(∞) � ∞, then using (I), we get C = 0, and hence,
(R − C)(∞) �∞.
we conclude that ∞ is not a pole for equation(3.2).
Hence, R − C −
n∑
j=1
Ej has no poles in C∞. Therefore, by Result 3.1.15,
we get R − C −
n∑
j=1
Ej is a constant say α. Hence,
R = (C + α) +
n∑
j=1
Ej ⇒ R(z) = G(z) +
n∑
j=1
Fj
(
1
z − zj
)
, ∀z ∈ C
where G = C + α. �
�
�
“book” — 2014/6/4 — 20:56 — page 132 — #10
�
�
�
�
�
�
132 Linear Fractional Transforms
3.2 LINEAR FRACTIONAL TRANSFORMS
Definition 3.2.1 A linear fractional transform (or Möbius transfrom) T is of
the form T(z) = az + b
cz + d
, where a, b, c, and d ∈ C such that ad − bc � 0.
The condition ad−bc � 0 states that az+b and cz+d do not have a common
zero. For if ad − bc = 0, then
−b
a
= −d
c
. However, as
−b
a
is the zero of
az + b and
−d
c
is the zero of cz + d, it would follow that az + b and cz + d
have a common zero.
Definition 3.2.2 (Elementary linear fractional transforms)
1. Translation: T(z) = z + a, ∀z ∈ C∞ for some a ∈ C.
2. Rotation: T(z) = eiθ z, ∀z ∈ C∞ for some θ ∈ [0, 2π).
3. Scaling: T(z) = rz, ∀z ∈ C∞ for some r > 0.
(a) If r > 1, then this transform is called magnification.
(b) If r < 1, then this transform is called contraction.
4. Inversion: T(z) = 1
z
, ∀z ∈ C∞.
LEMMA 3.2.3 Every linear fractional transform can be written as a suitable
composition of these elementary linear fractional transforms.
Proof: Let T(z) = az + b
cz + d
be given.
Case 1: If c = 0, then T(z) = a
d
z + b
d
. Here, we note that a � 0 and d � 0
from ad − bc � 0. Let
a
d
= reiθ . Therefore, T = T1 ◦ T2 ◦ T3, where
T3(z) = rz, T2(z) = eiθ z, and T1(z) = z + b
d
.
Case 2: If c � 0, then
T(z) =
a
c
z + b
c
z + c
d
�
�
“book” — 2014/6/4 — 20:56 — page 133 — #11
�
�
�
�
�
�
Rational Functions and Multivalued Functions 133
=
a
c
(
z + d
c
)
+ b
c
− ad
c2
z + d
c
= a
c
+
(
bc − ad
c2
)
1
z + d
c
.
As
bc − ad
c2
� 0, we can write
bc − ad
c2
= ρeiφ . Therefore, T =
T1 ◦T2 ◦T3 ◦T4 ◦T5, where T5(z) = z+ d
c
, T4(z) = 1
z
, T3(z) = eiφz,
T2(z) = ρz, and T1(z) = z + a
c
. �
THEOREM 3.2.4 Every linear fractional transform maps circles and lines to
circles and lines.
Proof: To prove this theorem, we shall first show that the elementary linear
fractional transforms map circles and lines to circles and lines.
1. The translation operator w = z + c maps the line z = a + tb into the
line w = (a + c) + tb, and it maps the circle z = a + reit into the circle
w = (a + c) + reit. Geometrically,
(a) the line passing through a with slope m is mapped into the line
passing through a + c with slope m.
(b) the circle with centre a and radius r is mapped to the circle with
centre a + c and radius r.
2. The rotation operator w = eiθ z maps the line z = a + tb into the line
w = (eiθa) + t(eiθb), and it maps the circle z = a + reit into the circle
z = (eiθa) + reit. Geometrically,
(a) the line passing through a with slope m is mapped to the line
passing through eiθa with slope m + θ .
(b) the circle with centre a and radius r is mapped to the circle with
centre eiθa and radius r.
3. The scaling operator ρz maps the line z = a + tb into the line
z = (ρa) + tb, and it maps the circle z = a + reit into the circle
z = (ρa) + rρeit. Geometrically,
�
�
“book” — 2014/6/4 — 20:56 — page 134 — #12
�
�
�
�
�
�
134 Linear Fractional Transforms
(a) the line passing through a with slope m is mapped to the line
passing through ρa with slope m.
(b) the circle with centre a and radius r is mapped to the circle with
centre ρa and radius rρ.
4. The combined equation of a line and circle (with non-zero centre) is
azz + bz + bz + c = 0, whose image under the inversion operator
w = 1
z
is cww+bw+bw+a = 0, which is again a combined equation
of a line and circle (with non-zero centre). As azz + bz + bz + c = 0
represents a circle (with non-zero centre) if a � 0 and represents a line
if a = 0 and it passes through c, geometrically we have the following
statements:
(a) a line passing through 0 is mapped onto a line passing through 0.
(b) a line not passing through 0 is mapped onto a circle (with non-zero
centre) passing through 0.
(c) a circle (with non-zero centre) passing through 0 is mapped onto a
line not passing through 0.
(d) a circle (with non-zero centre) not passing through 0 is mapped
onto a circle (with non-zero centre) not passing through 0.
If we consider a circle with centre zero radius r > 0, then it is mapped
to a circle with centre zero and radius
1
r
, as
z = r exp(iθ ), θ ∈ [−π ,π ] ⇒ w = 1
r
exp(−iθ ), θ ∈ [−π ,π ].
As every linear fractional transform is a composition of these elementary
linear fractional transforms, the theorem follows. �
RESULT 3.2.5 Let T and S be linear fractional transforms, then
1. T ◦ S is a linear fractional transform.
2. T−1 exists, and T−1 is also a linear fractional transform.
�
�
“book” — 2014/6/4 — 20:56 — page 135 — #13
�
�
�
�
�
�
Rational Functions and Multivalued Functions 135
Proof: Let T(z) = a1z + b1
c1z + d1
and S(z) = a2z + b2
c2z + d2
, where ai, bi, ci, and
di ∈ C with aidi − bici � 0, for i = 1, 2. As
(T ◦ S)(z) =
a1
(
a2z + b2
c2z + d2
)
+ b1
c1
(
a2z + b2
c2z + d2
)
+ d1
= (a1a2 + b1c2)z + (a1b2 + b1d2)
(c1a2 + c2d1)z + (c1b2 + d1d2)
,
which is of the form
Az + B
Cz + D
, and
AD − BC = (a1a2 + b1c2)(c1b2 + d1d2) − (a1b2 + b1d2)(c1a2 + c2d1)
= a1a2c1b2 + a1a2d1d2 + b1c2c1b2 + b1c2d1d2
− a1b2c1a2 − a1b2c2d1 − b1d2c1a2 − b1d2c2d1
= a1a2d1d2 + b1c2c1b2 − a1b2c2d1 − b1d2c1a2
= a1d1(a2d2 − b2c2) + b1c1(b2c2 − a2d2)
= (a1d1 − b1c1)(a2d2 − b2c2) � 0.
Therefore, T ◦ S is a linear fractional transform.
The equation w = az + b
cz + d
implies that
w(cz+ d) = az + b ⇒ z(cw − a) = −dw + b ⇒ z = dw − b
−cw + a
which is denoted by T−1(w). Clearly, T−1 is a linear fractional transform as
da − (−b)(−c) � 0. �
From the previous result, we get the collection of linear fractional trans-
forms, which becomes a group with respect to composition ‘◦’ of functions,
as the other axioms of group are obvious. It is not an abelian group because
if T(z) = 2z and S(z) = 1
z
, ∀z ∈ C∞, then T ◦ S � S ◦ T .
THEOREM 3.2.6 If T is a linear fractional transform such that T is not
identity, then T has at most two fixed points in C.
Proof: Let T(z) = az + b
cz + d
, ∀z ∈ C∞ with ad − bc � 0. The fixed points of
T are the solutions of z = az + b
cz + d
⇔ cz2 + (d − a)z − b = 0.
If c � 0, then the equation has at most two solutions.
�
�
“book” — 2014/6/4 — 20:56 — page 136 — #14
�
�
�
�
�
�
136 Linear Fractional Transforms
If c = 0 and d − a � 0, then from ad − bc � 0, we get a � 0 and d � 0.
Hence,
T(z) − z =
(a
d
− 1
)
z + b
d
with 0 �
a
d
� 1.
Therefore, z = (
1 − a
d
)−1 b
d is the only fixed point of T in C.
If c = 0, d − a = 0, and b � 0, then d � 0, and hence, T(z) = z + b
d with
b
d � 0, which has no fixed point.
The case c = 0, d − a = 0, and b = 0 is not possible as T is not the identity
transformation. Thus, the theorem follows. �
Example 3.2.7 If T : C∞ → C∞ is a bijection such that T maps lines and
circles to lines and circles, then T need not be a linear fractional transform.
If T(z) = z, ∀z ∈ C∞, then T is a bijection from C∞ onto itself and T maps
the line (circle) a|z|2+bz+bz+c = 0 onto the line (circle) a|z|2+bz+bz+c =
0. By previous theorem, T is not a linear fractional transform, as it has every
real number as a fixed point.
THEOREM 3.2.8 Every linear fractional transform w = T(z), which has only
two distinct fixed points α,β ∈ C, satisfies the equation
w − α
w − β = γ
z − α
z − β
for some γ ∈ C.
Proof: Let w = T(z) = az + b
cz + d
, ∀z ∈ C∞. As α and β are the fixed points of
T , we have α = aα + b
cα + d
and β = aβ + b
cβ + d
. Therefore,
w − α = az + b
cz + d
− aα + b
cα + d
= (ad − bc)(z − α)
(cz + d)(cα + d)
w − β = az + b
cz + d
− aβ + b
cβ + d
= (ad − bc)(z − β)
(cz + d)(cβ + d)
Therefore,
w − α
w − β = γ
z − α
z − β
where γ = cβ + d
cα + d
. �
�
�
“book” — 2014/6/4 — 20:56 — page 137 — #15
�
�
�
�
�
�
Rational Functions and Multivalued Functions 137
THEOREM 3.2.9 Every linear fractional transform w = T(z), which has only
one fixed point α ∈ C, satisfies the equation
1
w − α = 1
z − α + γ for some
γ ∈ C.
Proof: As in the proof of previous theorem, if w = T(z) = az + b
cz + d
, ∀z ∈ C∞,
then we have
α = aα + b
cα + d
⇒ α(cα − a) = b − dα.
Since,
w − α = az + b
cz + d
− α
= az + b − αcz − αd
cz + d
= (a − cα)z + α(cα − a)
cz + d
= (a − cα)(z − α)
cz + d
we have
1
w − α − 1
z − α = 1
z − α
(
cz + d
a − cα
− 1
)
= 1
z − α
cz + d − a + cα
a − cα
.
As, α is the double root of the quadratic equation, cz2 + (d − a)z− b = 0, we
have 2α = a − d
c
⇒ a − d = 2αc. Therefore,
1
w − α − 1
z − α = 1
z − α
cz − 2αc + cα
a − cα
= 1
z − α
c(z − α)
a − cα
= c
a − cα
= γ (say)
Hence,
1
w − α = 1
z − α + γ . �
PROBLEM 3.2.10 Find the image of |z − 2| < 2 under the linear fractional
transform T(z) = z
2z−8 .
�
�
“book” — 2014/6/4 — 20:56 — page 138 — #16
�
�
�
�
�
�
138 Linear Fractional Transforms
Solution: As the point 4 lies on the circle |z − 2| = 2 and T(4) = ∞, the
image of the circle is a straight line. We see that 0 and 2 + 2i lie on the circle
and f (0) = 0, f (2+2i) = 2+2i
4i−4 = −i
2 . Hence, the image of the circle |z−2| = 2
is the imaginary axis. As the centre 2 of this circle is mapped to T(2) = −1
2 ,
which lies in the left half plane, the image of |z − 2| < 2 under the linear
fractional transform T(z) = z
2z−8 is the left half plane {z ∈ C : Re z < 0}.
�
PROBLEM 3.2.11 Determine the image of {z : Re z < 0 and Im z > 0} under
the linear fractional transform S(z) = z+i
z−i .
Solution: We first find the images of the real and imaginary axes under z+i
z−i .
As |T(x)| =
∣∣∣ x+i
x−i
∣∣∣ = 1, for all x ∈ R, the image of the real axis is the
unit circle. As T(i) = ∞, the image of the imaginary axis under the linear
fractional transform is a straight line. Seeing T(2i) = 3 and T(3i) = 2, we
conclude that the imaginary axis is mapped onto the real axis. As −1 + i lies
on the second quadrant {z : Re z < 0 and Im z > 0} and T(−1 + i) = 1 − 2i
(which lies outside the circle and below the real axis), we conclude that the
image of the second quadrant under z+i
z−i is {z ∈ C : |z| > 1 and Im z < 0}.
�
�
�
“book” — 2014/6/4 — 20:56 — page 139 — #17
�
�
�
�
�
�
Rational Functions and Multivalued Functions 139
Definition 3.2.12 For given four distinct extended complex numbers z1, z2,
z3, and z4, we define the cross ratio (z1, z2, z3, z4) by T(z1), where T is a linear
fractional transform that maps z2, z3, and z4 to 1, 0, and ∞, respectively. More
explicitly,
(z1, z2, z3, z4) =
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
z1 − z3
z1 − z4
× z2 − z4
z2 − z3
if z1, z2, z3, z4 ∈ C
z2 − z4
z2 − z3
if z1 = ∞, z2, z3, z4 ∈ C
z1 − z3
z1 − z4
if z2 = ∞, z1, z3, z4 ∈ C
z2 − z4
z1 − z4
if z3 = ∞, z1, z2, z4 ∈ C
z1 − z3
z2 − z3
if z4 = ∞, z1, z2, z3 ∈ C.
Example 3.2.13 Evaluate the following cross ratios:
1. (1 + i, 2 − i, 3,−i).
2. (2 + i, i, 5 − i2,∞).
3. (1 + i2, 2 + i3,∞, i).
4. (2,∞, 1 − i, 3 + i).
5. (∞, 1 − i, 1 + i, i).
Solution:
1. (1 + i, 2 − i, 3,−i) = 1 + i − 3
1 + i + i
× 2 − i + i
2 − i − 3
= −4 + i2
1 − i3
= −1 − i.
�
�
“book” — 2014/6/4 — 20:56 — page 140 — #18
�
�
�
�
�
�
140 Linear Fractional Transforms
2. (2 + i, i, 5 − i2,∞) = 2 + i − 5 + i2
i − 5 + i2
= −3 − i3
−5 + i3
= 3
17
(4 − i).
3. (1 + i2, 2 + i3,∞, 1) = 2 + i3 − 1
1 + i2 − 1
= 1 + i3
i2
= 1
2
(3 − i).
4. (2,∞, 1 − i, 3 + i) = 2 − 1 + i
2 − 3 − i
= 1 + i
−1 − i
= −1.
5. (∞, 1 − i, 1 + i, i) = 1 − i − i
1 − i − 1 − i
= 1 − i2
−i2
= 1
2
(2 + i).
Exercise 3.2.14 Find the following cross ratios:
1. (2 + i3, 1 + i, 3 − i, 1 − i4).
2. (1 + i3, 5 + i, 2 − i3,∞).
3. (3 + i2, 1 − i2,∞, i5).
4. (1 − i2,∞, 1 + i4, 2 + i).
5. (∞, 2 − i, 3 + i4, i3).
Answers: 1. 19
20 − i 1
5 , 2. 1
25 (21 + i22), 3. 4
3 − i, 4. 1
5 (9 + i3), 5. 1
13 (9 + i7).
RESULT 3.2.15 Let z1, z2, z3, and z4 be extended complex numbers, then
1. (z1, z2, z3, z4) = (z1, z2, z3, z4).
2. (S(z1), S(z2), S(z3), S(z4)) = (z1, z2, z3, z4) for every linear fractional
transform S.
�
�
“book” — 2014/6/4 — 20:56 — page 141 — #19
�
�
�
�
�
�
Rational Functions and Multivalued Functions 141
Proof: Let T be the linear fractional transform that maps z2, z3, and z4 to 1, 0,
and ∞, respectively.
1. If T(z) = az + b
cz + d
and if T1(z) = az + b
cz + d
, then T1(z) = T(z) for every
extended complex number z. Hence,
T1(z2) = T(z2) = 1 = 1,
T1(z3) = T(z3) = 0 = 0,
T1(z4) = T(z4) = ∞ = ∞.
Therefore, T1 is the linear fractional transform that maps z2, z3, and
z4 to 1, 0, and ∞, respectively. Therefore, (z1, z2, z3, z4) = T1(z1) =
T(z1) = (z1, z2, z3, z4).
2. Using Result 3.2.5, we get T ◦ S−1, which is a linear fractional trans-
form. Clearly, we have T ◦ S−1 maps Sz2, Sz3, and Sz4 to 1, 0, and ∞,
respectively. Therefore, (S(z1), S(z2), S(z3), S(z4)) = (T ◦S−1)(S(z1)) =
T(z1) = (z1, z2, z3, z4). �
THEOREM 3.2.16 Let z1, z2, z3, and z4 be distinct extended complex num-
bers, then z1, z2, z3, and z4 lie on a circle or on a straight line iff (z1, z2, z3, z4)
is real.
Proof: Let T be the linear fractional transform that maps z2, z3, and z4 to
1, 0, and ∞, respectively, then T(z1) = (z1, z2, z3, and z4). Assume that
(z1, z2, z3, z4) is real. Then, T(z1), T(z2) = 1, T(z3) = 0, T(z4) = ∞ lie on
the real line. As, T−1 is also a linear fractional transform, by Theorem 3.2.4,
it maps the real line to a circle or to a straight line. In particular,
T−1(T(z1)) = z1, T−1(T(z2)) = z2, T−1(T(z3)) = z3, and T−1(T(z4)) = z4
lie on a circle or on a straightline.
Conversely, assume that z1, z2, z3, and z4 lie on a circle or on a straight
line. Since T is a linear fractional transform, T(z1), T(z2) = 1, T(z3) =
0, and T(z4) = ∞ lie on a circle or on a straight line, say C. As ∞ lies
on C, it follows that C is a straight line. Using the fact that it passes through
1 and 0, we conclude that C must be the real line. Hence, T(z1) lies on the
real line itself. Thus, z1, z2, z3, and z4 are real. �
LEMMA 3.2.17 Let C be a circle and z and w be any two points not on C.
If (w, z1, z2, z3) = (z, z1, z2, z3) for some three distinct points z1, z2, and z3
�
�
“book” — 2014/6/4 — 20:56 — page 142 — #20
�
�
�
�
�
�
142 Linear Fractional Transforms
on C, then (w, a1, a2, a3) = (z, a1, a2, a3) for any other three distinct points
a1, a2, and a3 on C.
Proof: Let T be the linear fractional transform that maps z1, z2, and z3 to
1, 0, and ∞, respectively. Therefore, (w, z1, z2, z3) = (z, z1, z2, z3) can be
rewritten as T(w) = T(z). As, a1, z1, z2, and z3 lie on C, by Theorem 3.2.16,
T(a1) is real. Similarly, we can show that T(aj) is real for j = 2 and 3. Hence,
using Result 3.2.15, we get
(z, a1, a2, a3) = (T(z), T(a1), T(a2), T(a3))
= (T(z), T(a1), T(a2), T(a3))
= (T(z), T(a1), T(a2), T(a3))
= (T(w), T(a1), T(a2), T(a3))
= (w, a1, a2, a3).
Hence, the lemma follows. �
Definition 3.2.18 Let C be a circle or a straight line. If z, w � C, then z and
w are said to be symmetric with respect to C if (w, z1, z2, z3) = (z, z1, z2, z3)
for any three distinct points z1, z2, and z3 on C.
By previous lemma, the symmetry of the points does not depend on the choice
of the three points on C.
Example 3.2.19 Explain symmetry with respect to a straight line.
If C is a straight line and z and w are symmetric with respect to C, then by
definition, we have
(w, z1, z2,∞) = (z, z1, z2,∞)
for any pair of distinct points z1 and z2 in C. Hence, we have
z − z2
z1 − z2
=
(
w − z2
z1 − z2
)
.
This implies that |z − z2| = |w − z2| for every z2 on C and
Im
(
z − z2
z1 − z2
)
= −Im
(
w − z2
z1 − z2
)
which imply geometrically that z and w lie on different half planes determined
by the line C, and they are at equal distance from an arbitrary point on C.
�
�
“book” — 2014/6/4 — 20:56 — page 143 — #21
�
�
�
�
�
�
Rational Functions and Multivalued Functions 143
C
z
w
Example 3.2.20 Explain symmetry with respect to a circle.
If C is the circle |z − a| = r and z and w are symmetric with respect to C,
then we have (w, z1, z2, z3) = (z, z1, z2, z3) for any three distinct points on C.
Using Result 3.2.15, we get
(w, z1, z2, z3) =
(
r2
w − a
,
r2
z1 − a
,
r2
z2 − a
,
r2
z3 − a
)
=
(
r2
w − a
, z1 − a, z2 − a, z3 − a
)
=
(
r2
w − a
+ a, z1, z2, z3
)
=
((
r2
w − a
+ a
)
, z1, z2, z3
)
.
Therefore, z =
(
r2
w − a
+ a
)
, which implies that (z − a)(w − a) = r2 and
w−a = r2 z − a
|z − a|2 . Hence, |z−a| · |w−a| = r2 and arg (w−a) = arg (z−a),
provided z, w ∈ C. If z = a, then w = ∞.
Geometrically, one of the two points z and w lies inside the circle and the
other point lies outside the circle satisfying |z− a||w− a| = r2. Furthermore,
a, w, and z are collinear, such that both z and w lie on a same ray starting
from a.
C
z
w
a
RESULT 3.2.21 (Principle of symmetry)
If z and w are symmetric with respect to a circle or to a straight line C, then
T(z) and T(w) are symmetric with respect to T(C).
Proof: Let z1, z2, and z3 be three distinct points on C. As z and w are sym-
metric with respect to C, we have (w, z1, z2, z3) = (z, z1, z2, z3). By Result
3.2.15, we get
�
�
“book” — 2014/6/4 — 20:56 — page 144 — #22
�
�
�
�
�
�
144 Linear Fractional Transforms
(T(w), T(z1), T(z2), T(z3)) = (T(z), T(z1), T(z2), T(z3)).
Using Result 3.2.5, we get that T is one-to-one, and hence,
T(z1), T(z2), and T(z3) are three distinct points on T(C). Hence, T(z)
and T(w) are symmetric with respect to T(C). �
Algorithm 3.2.22 (To find the LFT that maps a, b, and c to p, q, and r ,
respectively.)
Step 1: Let T be the linear fractional transform that maps a, b, and c into
1, 0, and ∞, respectively. That is, T(z) = (z − b)
(z − c)
(a − c)
(a − b)
.
Step 2: Let S be the linear fractional transform that maps p, q, and r into
1, 0, and ∞, respectively. That is, S(w) = (w − q)
(w − r)
(p − r)
(p − q)
.
Step 3: Then S−1◦T is the required linear fractional transform, which can be
obtained by solving for w from
(w − q)
(w − r)
(p − r)
(p − q)
= (z − b)
(z − c)
(a − c)
(a − b)
in terms of z.
Example 3.2.23 Find the linear fractional transform that maps 1, 2, and 3 to
0, i, and − i, respectively.
Solution:
The linear fractional transform that maps 1, 2, and 3 into 1, 0, and ∞,
respectively, is given by
(z − 2)
(z − 3)
(1 − 3)
(1 − 2)
, and the linear fractional trans-
form that maps 0, i, and − i into 1, 0, and ∞, respectively, is given by
(w − i)
(w + i)
(0 + i)
(0 − i)
.
(w − i)
(w + i)
(0 + i)
(0 − i)
= (z − 2)
(z − 3)
(1 − 3)
(1 − 2)
⇒ − (w − i)
(w + i)
= 2
(z − 2)
(z − 3)
⇒ −(w − i)(z − 3) = 2(z − 2)(w + i)
⇒ −w(z − 3) − 2w(z − 2) = i2(z − 2) − i(z − 3)
⇒ w(−3z + 7) = i(z − 1)
⇒ w = i(z − 1)
−3z + 7
.
�
�
“book” — 2014/6/4 — 20:56 — page 145 — #23
�
�
�
�
�
�
Rational Functions and Multivalued Functions 145
Hence, R(z) = i(z − 1)
−3z + 7
, ∀z ∈ C∞, is the required linear fractional
transform.
Exercise 3.2.24 Find the linear fractional transform that maps a, b, and c to
p, q, and r, respectively, in the following:
1. 1, 0, and − 1 to 1, 1 − i2, and − 1
2. 1,−1, and i to 1, i, and − 1
3. 0, 1, and ∞ to −1,−i, and 1
4. ∞,−1, and − 2 to 1,∞, and 3
5. 1 − i, i, and 0 to −i, 1 + i, and 1
6. 1,∞, and − 2 to −1, 2, and
1
2
7. −i, 0 and ∞ to ∞,−i2, and 1
8. −1, 0, and 2 to −1, 4, and 2
9. −1,∞, and i to 3 − i3, 2 + i3, and − 4 + i2
10. 1 + i, 1, and i to
3 + i4
5
,
−1 + i
2
, and
2 + i3
2
.
Answers:
1.
(1 + i2)z + 1
z + (1 + i2)
, 2.
(i − 1)z + (3 + i)
(1 + i3)z + (1 − i)
,
3.
−iz − 1
−iz + 1
, 4.
z − 1
z + 1
,
5.
1
−z + 1
, 6.
2z + 1
z − 4
,
7.
z + 2
z + i
, 8.
3z + 4
2z + 1
,
9.
(2 + i3)z − 1
z + (1 − i)
, 10.
2z − 3
z + i
.
�
�
“book” — 2014/6/4 — 20:56 — page 146 — #24
�
�
�
�
�
�
146 Linear Fractional Transforms
Example 3.2.25 Check the symmetry of given z, and w with respect to the
given C in the following:
1. z = −4 + i7, w = 8 + i and C is the straight line y = 2x.
2. z = 2 + i5, w = 3 and C is the straight line 3x − 4x + 1 = 0.
3. z = 8 + i7, w = 17 + i19 and C is the circle |z − 5 − i3| = 10.
Solution:
1. We choose the three points ∞, 0, and 1 + 2i on y = 2x. As
(−4+ i7,∞, 0, 1+ 2i) = 11 − i3
10
and (8+ i,∞, 0, 1+ 2i) = 11 + i3
10
,
−4 + i7 and 8 + i are symmetric with respect to y = 2x.
2. We choose the three points ∞, i, and 5 + 4i on 3x − 4y + 1 = 0. As
(2 + i5,∞, i, 5 + i4) = 1 + 7i
5
and (3,∞, i, 5 + 4i) = 1 + 7i
10
,
2 + i5 and 3 are not symmetric with respect to 3x − 4y + 1 = 0.
3. We choose the three points 15 + 3i,−5 + 3i, and 5 + 13i on |z −
5 − i3| = 10. As (8 + 7i, 15 + i3,−5 + i3, 5 + i13) = 7 + 5i
6
and
(17+ i19, 15+ i3,−5+ i3, 5+ i13) = 7 − 5i
6
, 8+ 7i and 17+ 19i are
symmetric with respect to |z − 5 − i3| = 10.
Example 3.2.26 Prove that if w = T(z) = r(z − a)
r2 − az
, then T maps |z − a| = r
onto |w| = 1 and T(B(0, r)) = B(0, 1).
Proof: Let |z| = r. If w = T(z), then we can write z = reiθ . Therefore,
|w| =
∣∣∣∣ r(z − a)
r2 − az
∣∣∣∣ =
∣∣∣∣ r(reiθ − a)
r2 − areiθ
∣∣∣∣ =
∣∣∣∣ r − ae−iθ
r − aeiθ
∣∣∣∣ = 1.
Therefore, T maps |z| = r into |w| = 1. As every linear fractional transform
maps circles to circles, we conclude that T maps |z − a| = r onto |w| = 1.
Since T(B(0, r)) is a connected subset of {w ∈ C : |w| � 1} and the latter
is having exactly two components namely, B(0, 1) and {w ∈ C : |w| > 1}.
�
�
“book” — 2014/6/4 — 20:56 — page 147 — #25
�
�
�
�
�
�
Rational Functions and Multivalued Functions 147
Using the observation T(a) = 0, it follows that T(B(0, r)) ⊆ B(0, 1). Since,
T−1is also a linear fractional transform, such that T−1 maps |w| = 1 onto
|z− a| = r and T−1(0) = a, by a similar argument, we obtain T−1(B(0, 1)) ⊆
B(0, r). Thus, we get T(B(0, r)) = B(0, 1). �
PROBLEM 3.2.27 Find all linear fractional transform that maps |z| = r onto
|w| = ρ.
Solution: Let w = T(z) = az+b
cz+d be a required transformation, where
a, b, c, d ∈ C with ad − bc � 0. Clearly, 0 and ∞ are symmetric points
with respect to the circle |w| = ρ; therefore, by symmetric principle, if
T−1(0) = α and T−1(∞) = β, then α and β are symmetric points with
respect to |z| = r.
Case 1: 0 � α �∞.
By one of the properties of the symmetric points of a circle, we
have β = r2
α
. Putting w = 0 and w = ∞ in z = dw−b
−cw+a , we get
α = − b
a and r2
α
= − d
c . From these observations, we note that c � 0,
otherwise α = 0, which contradicts the fact that α � 0. Therefore,
w = a
c
z+ b
a
z+ d
c
= a
c
z−α
z− r2
α
= aα
c
z−α
αz−r2 . As the image T(r) of r must lie on
the circle |w| = ρ, we have ρ = |T(r)| =
∣∣∣ aα
c
r−α
αr−r2
∣∣∣ = ∣∣ aα
cr
∣∣, which
implies
∣∣∣ aα
crρ
∣∣∣ = 1. Hence, by Result 2.4.24, we can find θ ∈ R such
that aα
crρ = exp(iθ ). Thus, T(z) = aα
c
z−α
αz−r2 = rρ exp(iθ ) z−α
αz−r2 .
Case 2: α = 0
In this case, immediately, we have β = ∞ using the symmetry of α
and β with respect to |z| = r. Now, T(0) = 0 and T(∞) = ∞ imply
b = 0 and c = 0, and hence, a � 0 � d. As T maps |z| = r onto
|w| = ρ, we have ρ = |w| = |T(z)| = ∣∣ a
d
∣∣ r. Therefore, there exists
ϕ ∈ R such that a
d = ρ
r exp(iϕ). Thus, T(z) = ρ
r exp(iϕ)z.
Case 3: α = ∞.
Then, we have β = 0. Now, T(0) = ∞ and T(∞) = 0 imply that
d = 0 and a = 0. As T maps |z| = r onto |w| = ρ, we have
ρ = |w| = |T(z)| =
∣∣∣ b
c
∣∣∣ 1
r . Therefore, there exists ψ ∈ R such that
b
c = rρ exp(iψ). Thus, T(z) = rρ exp(iψ) 1
z .
Remark 3.2.28: The linear fractional transforms obtained in Case 1 and
Case 2 of the previous problem map |z| < r onto |w| < ρ. The
�
�
“book” — 2014/6/4 — 20:56 — page 148 — #26
�
�
�
�
�
�
148 Branch of a Multivalued Function
transforms obtained in Case 3 of the previous problem map |z| < r onto
|w| > ρ.
PROBLEM 3.2.29 Find all linear fractional transform that maps the upper
half-plane onto the unit circle.
Solution: Let w = T(z) = az+b
cz+d be a linear fractional transform, which maps
upper half-plane onto the unit circle. Clearly, 0 and ∞ are symmetric points
with respect to the circle |w| = 1. If T−1(0) = α and T−1(∞) = β, then α
and β are symmetric points with respect to the real axis. Therefore, obviously
we have α,β � {0,∞} and β = α. Using w = 0 and w = ∞ in z = dw−b
−cw+a ,
we get α = − b
a and α = − d
c . Therefore, w = a
c
z+ b
a
z+ d
c
= a
c
z−α
z−α . As T(0) lies on
|w| = 1, we have 1 = |T(0)| = ∣∣ a
c
∣∣, and hence, a
c = exp(iθ ) for some θ ∈ R.
Thus, T(z) = exp(iθ ) z−α
z−α .
3.3 BRANCH OF A MULTIVALUED FUNCTION
We shall start this section with the definition logarithm and then we introduce
the concept of branch of a multivalued function.
Definition 3.3.1 The inverse function of exp : R → (0,∞) is called the
natural logarithm and is denoted by ln.
Note that the above definition is well defined as exp : R → (0,∞) is a
bijection. See Lemma 2.4.6.
LEMMA 3.3.2 If a, b ∈ (0,∞), then ln(ab) = ln a + ln b and ln
(a
b
)
=
ln a − ln b.
Proof: Let x = ln a and y = ln b, then exp(x) = a and exp(y) = b. By
Theorem 2.4.4, we have
exp(x + y) = exp(x) exp(y) and
exp(x − y) = exp(x) exp(−y) = exp(x)
exp(y)
.
Therefore,
ln a + ln b = x + y = ln(exp(x + y)) = ln(exp(x) exp(y)) = ln(ab)
�
�
“book” — 2014/6/4 — 20:56 — page 149 — #27
�
�
�
�
�
�
Rational Functions and Multivalued Functions 149
and
ln a − ln b = x − y = ln(exp(x − y)) = ln
(
exp(x)
exp(y)
)
= ln
(a
b
)
.
Hence, the lemma follows. �
RESULT 3.3.3 ln : (0,∞) → R is differentiable and ln′ a = 1
a
.
Proof: First, we prove that ln : (0,∞) → R is continuous on (0,∞).
For arbitrary a, b ∈ [1,∞), we choose x, y ∈ [0,∞) such that exp(x) = a and
exp(y) = b and then we get x = ln a, y = ln b. Now,
|a − b| = | exp(x) − exp(y)|
=
∣∣∣∣∣
∞∑
n=1
xn − yn
n!
∣∣∣∣∣
= |x − y|
∣∣∣∣∣∣∣∣∣
∞∑
n=1
n−1∑
k=0
xkyn−1−k
n!
∣∣∣∣∣∣∣∣∣
= |x − y|
∞∑
n=1
n−1∑
k=0
xkyn−1−k
n!
≥ |x − y|
= | ln a − ln b|.
Therefore, ln b → ln a as b → a, whenever a, b ∈ [1,∞).
Suppose, a, b ∈ (0, 1], then
1
a
,
1
b
∈ [1,∞), and hence,
b → a ⇒ 1
b
→ 1
a
⇒ ln
1
b
→ ln
1
a
⇒ − ln b → − ln a ⇒ ln b → ln a.
Hence, ln is continuous on (0,∞).
To prove ln is differentiable on (0,∞), fix a, b ∈ (0,∞) arbitrarily. Then let
x = ln a and ln b = y. Now, using b → a ⇒ y → x, we get
lim
b→a
ln b − ln a
b − a
= lim
y→x
y − x
exp(y) − exp(x)
�
�
“book” — 2014/6/4 — 20:56 — page 150 — #28
�
�
�
�
�
�
150 Branch of a Multivalued Function
= lim
y→x
1
exp(y) − exp(x)
y − x
= 1
lim
y→x
exp(y) − exp(x)
y − x
= 1
exp′(x)
= 1
exp(x)
= 1
a
.
Thus, ln is differentiable on (0,∞) and ln′ a = 1
a
. �
We would like to define complex logarithm as the inverse function of
exp on C. However, unfortunately, exp : C → C is neither one-to-one nor
onto. However, we can define the complex logarithm of a non-zero complex
number as a multivalued function as follows.
Definition 3.3.4 For every w ∈ C \ {0}, we define log w = ln |w| + iarg w.
Note that in the above definition, the imaginary part of log is not single
valued.
Definition 3.3.5 Let f be a multivalued function on a subset of �. We mean
a branch (an analytic branch) of f by a single-valued continuous (or analytic)
function g on a sub-region of �, if value of g at a point is one of the multi-
values of f at the same point.
To define an analytic branch of log, we have to choose a unique value of
its imaginary part so that log becomes a single-valued analytic function.
LEMMA 3.3.6 The principal argument given in Definition 1.2.16 is continu-
ous on C \ {(a, 0) : a ≤ 0}.
Proof: Using the continuity of arctan, we get that arg is continuous on each
of the following regions.
1. {(a, b) : a > 0, b ∈ R}
2. {(a, b) : a < 0, b > 0}
3. {(a, b) : a < 0, b < 0}
Hence to conclude the proof of this lemma, we shall show that arg is
continuous on the positive imaginary axis and on the negative imaginary axis.
�
�
“book” — 2014/6/4 — 20:56 — page 151 — #29
�
�
�
�
�
�
Rational Functions and Multivalued Functions 151
Let (0, b) be fixed, where b > 0 is arbitrary. If (an, bn) → (0, b) as n → ∞,
with (an, bn) � (0, b), ∀n ∈ N, then, an → 0 and bn → b as n → ∞, and
hence, there exists M ∈ N such that bn > 0 for all n ≥ M .
Case 1: an > 0, ∀n ∈ N
As
bn
an
→ +∞ as n → ∞, we have arctan
bn
an
→ π
2
as n → ∞.
Therefore,
arg ((0, b)) = π
2
= lim
n→∞ arctan
(
bn
an
)
= lim
n→∞ arg ((an, bn)).
Case 2: an < 0, ∀n ∈ N
Therefore, using the fact that arctan
(
bn
|an|
)
→ π
2
as n → ∞, we
obtain
arg ((0, b)) = π
2
= π − π
2
= π − lim
n→∞ arctan
(
bn
|an|
)
= lim
n→∞ arg ((an, bn)).
Case 3: an ∈ R, ∀n ∈ N.
Given ε > 0 using Case 1 and Case 2, we find N1, N2 ∈ N such that
N1 > M , N2 > M , and
|arg ((an, bn)) − arg ((0, b))| =
∣∣∣arg ((an, bn)) − π
2
∣∣∣ < ε,
∀n ≥ N1 if an > 0
|arg ((an, bn)) − arg ((0, b))| =
∣∣∣arg ((an, bn)) − π
2
∣∣∣ < ε
∀n ≥ N2 if an < 0.
As
|arg ((an, bn)) − arg ((0, b))| =
∣∣∣π
2
− π
2
∣∣∣ = 0 < ε,
∀n ∈ N if an = 0,
it follows that
∣∣∣arg ((an, bn)) − π
2
∣∣∣ < ε, ∀n ≥ max{N1, N2} if
an ∈ R. �
�
�
“book” — 2014/6/4 — 20:56 — page 152 — #30
�
�
�
�
�
�
152 Branch of a Multivalued Function
Remark 3.3.7: The principal branch of arg is not continuous at any point of
negative real axis.
Indeed, if (x, 0) ∈ C with x < 0, we have(
x + 1
n
, 0
)
→ (x, 0) and
(
x − 1
n
, 0
)
→ (x, 0) as n → ∞
but
arg
(
x + 1
n
, 0
)
→ π
2
and arg
(
x − 1
n
, 0
)
→ −π
2
as n → ∞.
LEMMA 3.3.8 If 0 � w ∈ C, then there exists a region containing w in which
some suitable branch of arg is continuous.
Proof: If 0 � w ∈ C, then we can define argw on�w = C\{r exp(iθ ) : r > 0},
by argw(z) = arg(z)+π+θw,∀z∈ �w, where arg(z) is the principal argument
of z as in Definition 1.2.16, and θw ∈ R \ {arg(w)+ 2kπ : k ∈ Z}. Being argw
is a translate of principal argument, it is also continouous on �w. �
RESULT 3.3.9 (Principal branch for log)
If � = C \ {(x, 0) : x ≤ 0} and if log(w) = ln |w| + iarg (w), with Im log(w)
∈ (−π ,π ), then log : �→ C is analytic.
Proof: First, we note that for every w ∈ �, principal argument of w belongs
to (−π ,π ), and hence, log is well defined on�. As ln is continuous on (0,∞)
and | · | is continuous on C (Example 1.6.18), we have Re log, which is con-
tinuous on �. By Lemma 3.3.6, we have Im log, which is continuous on �.
Therefore, log is continuous on � by Lemma 1.6.19. Next, we prove that log
is differentiable on�. For w0, w ∈ �, using the continuity of log on�, we get
lim
w→w0
log(w) − log(w0)
w − w0
= lim
z→z0
z − z0
exp(z) − exp(z0)
,
where z = log(w) and z0 = log(w0)
= lim
z→z0
1
exp(z) − exp(z0)
z − z0
= 1
lim
z→z0
exp(z) − exp(z0)
z − z0
�
�
“book” — 2014/6/4 — 20:56 — page 153 — #31
�
�
�
�
�
�
Rational Functions and Multivalued Functions 153
= 1
exp′(z0)
= 1
exp(z0)
= 1
z0
.
Thus, log is differentiable on � and log′(a) = 1
a
, ∀a ∈ �. �
Definition 3.3.10 For a non-zero complex number z, we define w = √
z by
the solution of the equation w2 = z. More explicitly,
√
z = ±√
r exp
(
i
θ
2
)
,
where z = r exp(iθ ).
RESULT 3.3.11 (Analytic branch for
√·)
If � = C \ {(x, 0) : x ≤ 0} and if
√
z is defined by the unique value of
√
z
whose real part is positive, then
√· : �→ C is analytic.
Proof: First, we note that every z ∈ � is never real and negative, and hence,√
z is never purely imaginary. Therefore, Re
√
z � 0. If
√
z is defined by the
unique value of
√
z whose real part is positive, and then
√
z is single valued.
First, we claim that
√· is continuous on �. Let z0, z ∈ � and w = √
z, w0 =√
z0. Hence, w2 = z and w2
0 = z0. Using the definition of
√
z, we get
|z − z0| = |w2 − w2
0| = |w − w0| · |w + w0| ≥ |w − w0| · Re (w + w0)
> |w − w0| · Re w0
which implies that
|√z − √
z0| = |w − w0| < 1
Re w0
|z − z0|.
Hence,
√
z → √
z0 as z → z0. Thus,
√· is continuous on �.
As
lim
z→z0
√
z − √
z0
z − z0
= lim
w→w0
w − w0
w2 − w2
0
= lim
w→w0
1
w + w0
= 1
2w0
= 1
2
√
z0
,
we get
√· is differentiable and its derivative is
1
2
√· . �
Example 3.3.12 Define an analytic branch for
√
1 + √
z and justify that it is
analytic.
We know that
√
z is defined on �1 = C \ {(x, 0) : x ≤ 0} as an analytic
function by the unique value of
√
z, whose real part is positive. As the range
�
�
“book” — 2014/6/4 — 20:56 — page 154 — #32
�
�
�
�
�
�
154 Branch of a Multivalued Function
G of
√· is {z ∈ C : Re z > 0}, we get 1 + √
z ∈ G, ∀z ∈ �1. Furthermore,
G ⊆ �1, and we can define
√
1 + √
z using the above definition of
√· twice
as an analytic function.
RESULT 3.3.13 arccos(w) = ±i log
(
w + √
w2 − 1
)
Proof: Let z = arccos w. Then w = cos z = exp(iz) + exp(−iz)
2
. Hence, it
follows that
2w = exp(iz) + exp(−iz) ⇒ exp(i2z) − 2w exp(iz) + 1 = 0
which is a quadratic equation in exp(iz). Therefore,
exp(iz) = 2w ± √
4w2 − 4
2
= w ±
√
w2 − 1.
Therefore, iz = log(w ± √
w2 − 1) ⇒ z = −i log(w ± √
w2 − 1). As
(w +
√
w2 − 1) · (w −
√
w2 − 1) = w2 − (w2 − 1) = 1
arccos(w) = z = ±i log(w + √
w2 − 1) �
RESULT 3.3.14 arcsin(w) = −i log(iw ± √
1 − w2).
Proof: Let z = arcsin(w). Then w = sin(z) = exp(iz) − exp(−iz)
i2
. Hence, it
follows that
i2w = exp(iz) − exp(−iz) ⇒ exp(i2z) − i2w exp(iz) − 1 = 0
which is a quadratic equation in exp(iz). Therefore,
exp(iz) = i2w ± √−4w2 + 4
2
= iw ±
√
1 − w2.
Therefore, iz = log(iw ± √
1 − w2) ⇒ z = −i log(iw ± √
1 − w2). Thus,
arcsin w = −i log(iw ± √
1 − w2) �
RESULT 3.3.15 sinh−1 w = log
(
w ± √
w2 + 1
)
�
�
“book” — 2014/6/4 — 20:56 — page 155 — #33
�
�
�
�
�
�
Rational Functions and Multivalued Functions 155
Proof: If z = sinh−1 w, then
w = sinh z = exp(z) − exp(−z)
2
which implies that exp(2z) − 2w exp(z) − 1 = 0. Therefore,
exp(z) = 2w ± √
4w2 + 4
2
= w ±
√
w2 + 1
and hence, z = log
(
w ± √
w2 + 1
)
. �
RESULT 3.3.16 tanh−1 w = 1
2
log
(
1 + w
1 − w
)
.
If z = tanh−1 w, then w = tanh z = exp(z) − exp(−z)
exp(z) − exp(−z)
,
⇒ exp(z) − exp(−z) = w(exp(z) + exp(−z))
⇒ exp(2z) − 1 = w(exp(2z) + 1)
⇒ exp(2z)(1 − w) = 1 + w
⇒ exp(2z) = 1 + w
1 − w
.
Therefore, z = 1
2
log
(
1 + w
1 − w
)
. �
Exercise 3.3.17 Prove that cosh−1 w = log
(
w ± √
w2 − 1
)
and tan−1 w =
1
i2
log
(
1 + iw
1 − iw
)
.
Example 3.3.18 Find a branch for arccos.
Let �1 = C \ {(x, 0) : |x| ≥ 1}. Define arccos(w) = i log(w + √
w2 − 1)
with its real part lies between (−π , 0) and
√
w2 − 1 = i
√
1 − w2, ∀w ∈ �1.
To prove that this branch of arccos is analytic, we first show that 1 − w2 and
w+ √
w2 − 1 are never real and negative whenever w ∈ �1, so that from the
principal branch of log, arccos becomes analytic.
1 − w2 is real and less than or equal to 0 iff w2 is real and w2 ≥ 1
iff w is real and |w| ≥ 1
iff w � �1.
�
�
“book” — 2014/6/4 — 20:56 — page 156 — #34
�
�
�
�
�
�
156 Conformal Mapping
As (w + √
w2 − 1)(w − √
w2 − 1) = w2 − (w2 − 1) = 1, we have
(w + √
w2 − 1) is real iff (w − √
w2 − 1) is real
iff w and
√
w2 − 1 are real
iff w is real and |w| ≥ 1
iff w � �1.
If g(w) = w+√
w2 − 1, ∀w ∈ �1, then g is a continuous function on�1, and
hence, {w+ √
w2 − 1 : w ∈ �1} is a connected subset of C. As w+ √
w2 − 1
is never real on �1, {w + √
w2 − 1 : w ∈ �1} lies either in the upper half-
plane or in the lower half-plane. As i ∈ �1 and g(i) = i+i
√
2, which belongs
to the upper half-plane, {w+ √
w2 − 1 : w ∈ �1} should be in the upper half-
plane, and hence, we define log(w+ √
w2 − 1) as an analytic function on �1
whose imaginary part lies in (0,π ). Hence, arccos(w) = i log(w+ √
w2 − 1),
∀w ∈ �1 is analytic if its real part lies in (−π , 0).
Further, the derivative of arccos is obtained by
d
dw
(arccos(w)) = i
w +√w2 − 1
×
(
1 + 2w
2
√
w2 − 1
)
= i√
w2 − 1
= i
i
√
1 − w2
= 1√
1 − w2
.
Exercise 3.3.19 Define an analytic branch for arcsin.
Exercise 3.3.20 Define an analytic branch for n
√·. (Hint. Use the branch of
log and n
√
z = exp
(
1
n log(z)
)
.)
3.4 CONFORMAL MAPPING
Definition 3.4.1 (Curve)
If ϕ : [a, b] → C is a continuous function, then the image γ = ϕ([a, b]) of ϕ
is called a curve with initial point ϕ(a) and end point ϕ(b) in C.
We call ϕ(t), t ∈ [a, b] a parametric equation of the curve γ.
Definition 3.4.2 Let � be a region and let z0 ∈ �. A continuous function
f : �→ C is said to be a
1. conformal mapping of first type at z0 if for every curve γ with a para-
metric equation ϕ(t), t ∈ [a, b] such that ϕ(t0) = z0 and ϕ′(t0) � 0,
�
�
“book” — 2014/6/4 — 20:56 — page 157 — #35
�
�
�
�
�
�
Rational Functions and Multivalued Functions 157
we get ( f ◦ ϕ)′(t0) � 0 and
lim
t→t0
arg
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
ϕ(t) − ϕ(t0)
exists for a suitable branch of arg and is independent of the curve γ .
2. conformal mapping of second type at z0 if lim
z→z0
| f (z) − f (z0)|
|z − z0| exists
and is non-zero.
3. conformal mapping if f is conformal of first type at z0 as well as
conformal of second type at z0.
THEOREM 3.4.3 If f is differentiable at z0 and f ′(z0) � 0, then f is conformal
at z0.
Proof: By assumption, we have
lim
z→z0
f (z) − f (z0)
z − z0
� 0.
Let γ be any curve with a parametric equation ϕ(t), t ∈ [a, b] such that
ϕ(t0) = z0 and ϕ′(t0) � 0. By chain rule, we have
( f ◦ ϕ)′(t0) = f ′(z0)ϕ′(t0) � 0
and as f ′(z0) � 0, then arg is continuous on a region that contains f ′(z0) with
respect to a suitable branch (Lemma 3.3.8), and hence, we get
lim
t→t0
arg
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
ϕ(t) − ϕ(t0)
= arg lim
t→t0
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
ϕ(t) − ϕ(t0)
= arg lim
t→t0
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
t − t0
· t − t0
ϕ(t) − ϕ(t0)
= arg lim
t→t0
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
t − t0
· t − t0
ϕ(t) − ϕ(t0)
= arg
( f ◦ ϕ)′(t0)
ϕ′(t0)
= arg f ′(z0)
�
�
“book”— 2014/6/4 — 20:56 — page 158 — #36
�
�
�
�
�
�
158 Conformal Mapping
which is independent of the curve γ . Again using the hypothesis and the
continuity of the modulus function (Example 1.6.18), we get
lim
z→z0
| f (z) − f (z0)|
|z − z0| � 0.
Thus, f is conformal at z0. �
THEOREM 3.4.4 If f is conformal at z0, then f is differentiable at z0 and
f ′(z0) � 0.
Proof: By assumption, we have
1. for every curve γ with a parametric equation ϕ(t), t ∈ [a, b] such that
ϕ(t0) = z0 and ϕ′(t0) � 0, we get ( f ◦ ϕ)′(t0) � 0 and
lim
t→t0
arg
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
ϕ(t) − ϕ(t0)
= θ
for some θ ∈ R with respect to a suitable branch of arg and is
independent of the curve γ.
2. conformal of second type at z0. That is, lim
z→z0
| f (z) − f (z0)|
|z − z0| = A for
some A � 0.
To prove this theorem, we show that if
zn → z0 as n → ∞ with zn � z0, ∀n ∈ N
then
f (zn) − f (z0)
zn − z0
→ A exp(iθ ) as n → ∞.
First, we choose a curve γ with a parametric equation ϕ(t), t ∈ [a, b] with
the following properties:
1. ϕ(t0) = z0, for some t0 ∈ [a, b] and ϕ(tn) = zn, ∀n ∈ N for some
sequence (tn) of distinct points of [a, b] such that tn → t0 as n → ∞.
2. ϕ′(t0) � 0.
Now, using the assumptions and Lemma 3.3.8,
lim
n→∞
f (zn) − f (z0)
zn − z0
�
�
“book” — 2014/6/4 — 20:56 — page 159 — #37
�
�
�
�
�
�
Rational Functions and Multivalued Functions 159
= lim
n→∞
∣∣∣∣ f (zn) − f (z0)
zn − z0
∣∣∣∣ exp
(
iarg
(
f (zn) − f (z0)
zn − z0
))
= lim
n→∞
∣∣∣∣ f (zn) − f (z0)
zn − z0
∣∣∣∣ exp
(
i lim
n→∞ arg
(
f (ϕ(tn)) − f (ϕ((t0))
ϕ(tn) − ϕ(t0)
))
= A exp (iθ) .
Thus, f is differentiable at z0 and f ′(z0) = A exp (iθ) � 0. �
THEOREM 3.4.5 Let � be a region, z0 ∈ � and let f : �→ C be continuous
at z0.
1. If f is conformal of first type at z0 and fx, fy exist at z0 and they are
continuous, then f is differentiable at z0.
2. If f is conformal of second type at z0 and fx, fy exist at z0 and they are
continuous, then f or f is differentiable at z0.
Proof: Let γ be a curve with a parametric equation ϕ(t), t ∈ [a, b] such that
ϕ(t0) = z0 and ϕ′(t0) � 0. If ψ(t) = f (ϕ(t)), ∀t ∈ [a, b] and if x′(t0) and y′(t0)
are the real and imaginary parts of ϕ′(t0), respectively, thenψ is differentiable
at t0 and
ψ ′(t0) = fx(ϕ(t0))x′(t0) + fy(ϕ(t0))y′(t0)
= fx(z0)
(
ϕ′(t0) + ϕ′(t0)
2
)
+ fy(z0)
(
ϕ′(t0) − ϕ′(t0)
i2
)
= ϕ′(t0)
(
fx(z0) − ify(z0)
2
)
+ ϕ′(t0)
(
fx(z0) + ify(z0)
2
)
.
Thus, we have
ψ ′(t0)
ϕ′(t0)
=
(
fx(z0) − ify(z0)
2
)
+
(
fx(z0) + ify(z0)
2
)
ϕ′(t0)
ϕ′(t0)
. (3.3)
which represents an equation of a circle with centre
fx(z0) − ify(z0)
2
and radius
fx(z0) + ify(z0)
2
, as ϕ varies through the continuous function on [a, b] with
ϕ(t0) = z0 for some t0 and ϕ′(t0) � 0.
1. Assume that f is conformal of first type at z0.
Then, we have
arg
ψ ′(t0)
ϕ′(t0)
= arg
( f ◦ ϕ)′(t0)
ϕ(t0)
= lim
t→t0
arg
( f ◦ ϕ)(t) − ( f ◦ ϕ)(t0)
ϕ(t) − ϕ(t0)
�
�
“book” — 2014/6/4 — 20:56 — page 160 — #38
�
�
�
�
�
�
160 Conformal Mapping
which exists and independent of ϕ. Therefore, every point on the circle
represented by the equation (3.3) has same argument. This is possible
iff the radius of the circle is 0. That is, we have
fx(z0) + ify(z0)
2
= 0 ⇒ fx(z0) = −ify(z0).
Therefore, f satisfies C–R equations at z0 whence by using Theorem
3.1.48, we get f is differentiable at z0.
2. Assume that f is conformal of second type at z0.
By a similar argument, we get the absolute value of every point on the
circle given in the equation (3.3) is constant. This is possible iff the
radius of the circle is 0 or the centre of the circle is 0. That is, we have
fx(z0) + ify(z0)
2
= 0 or
fx(z0) − ify(z0)
2
= 0.
Just now we have seen that
fx(z0) + ify(z0)
2
= 0 implies that f is
differentiable at z0. If
fx(z0) − ify(z0)
2
= 0, then we have
fx(z0) = ify(z0) ⇒
(
f
)
x
(z0) = −i
(
f
)
y
(z0)
and hence, f satisfies the C–R equation at z0. Furthermore, as f
has continuous partial derivatives at z0, f also has continuous partial
derivatives at z0. Thus, f is differentiable at z0, by Theorem 2.2.28. �
Example 3.4.6 The function exp maps the rectangle {z ∈ C : 1 < Re z <
2,
π
6
< Im z < π
3 } onto the part of angular sector {w ∈ C : a < |w| <
b, π
6 < arg w < π
3 }.
Let z = x + iy. Under the exponential function,
1. the horizontal line segment {(x, π6 ) : 1 < x < 2} is mapped onto
{exp(x) exp(iπ6 ) : 1 < x < 2}, which is the line segment with slope π
6
with the positive real axis from |w| = exp(1) to |w| = exp(2).
2. similarly, the horizontal line segment {(x, π3 ) : 1 < x < 2} is mapped
onto {exp(x) exp(iπ3 ) : 1 < x < 2}, which is the line segment with
slope π
3 with the positive real axis from |w| = exp(1) to |w| = exp(2).
�
�
“book” — 2014/6/4 — 20:56 — page 161 — #39
�
�
�
�
�
�
Rational Functions and Multivalued Functions 161
3. the vertical line segment {(1, y) : π
6 < y < π
3 } is mapped onto
{exp(1) exp(iy) : π
6 < y < π
3 }, which is the part of the circular arc
with radius exp(1) with argument varies from π
6 to π
3 .
4. similarly, the vertical line segment {(2, y) : π
6 < y < π
3 } is mapped
onto {exp(2) exp(iy) : π6 < y < π
3 }, which is the part of the circular arc
with radius exp(2) with argument varies from π
6 to π
3 .
Thus, the image of the rectangle under exp is shown as in the following
diagram.
Exercise 3.4.7
1. Prove that the map f (z) = z2, ∀z ∈ C maps the first quadrant {(x, y) ∈
C : x > 0, y > 0} onto the upper half-plane {(x, y) ∈ C : y > 0}.
2. Prove that the map f (z) = z − 1
z + 1
, ∀z ∈ C \ {−1} maps the upper
half-plane {(x, y) ∈ C : y > 0} onto the unit circle {z ∈ C : |z| < 1}.
3.5 ELEMENTARY RIEMANN SURFACE
The functions z �→ z1/n, log and arccos, and so on are not single-valued
functions because z �→ zn, exp and cos are not injective. Therefore, in the
previous section, we have defined branches of each of these multivalued func-
tions by omitting all but one of its images at every point. However, Riemann
introduced an alternate concept, called elementary surfaces corresponding to
every multivalued function, using which the multivalued function becomes a
bijective single-valued function between a region of the complex plane onto
a suitable surface.
A Riemann surface corresponding to a non-injective function f , intu-
itively speaking, is a surface obtained by attaching domains (commonly
�
�
“book” — 2014/6/4 — 20:56 — page 162 — #40
�
�
�
�
�
�
162 Elementary Riemann Surface
called sheets) of all the different branches of the multivalued function f −1
so that f becomes a bijection from C onto the surface (or equivalently, f −1
can be defined as a single-valued function from the surface into C) without
losing any of the multivalued information.
The rigorous definition of elementary Riemann surface uses terminolo-
gies from topology, which are not within the scope of this book. For the
sake of completeness, here we provide the rigorous definition of a Riemann
surface.
Definition 3.5.1 A Riemann surface is a two-dimensional real analytic
manifold.1
Example 3.5.2 Construct the Riemann surface corresponding to z �→ zn for
some n ∈ N with n > 1.
We first note that z �→ zn is not injective; indeed, for each θ ∈ R, all of the
following pairwise distinct n points r exp
(
i
(
θ+2π
k
))
, k = 0, 1, 2, . . . , n − 1,
are mapped to the same value r exp (iθ). We also observe that the function
z �→ zn maps positive real axis onto itself. Although any sector{
z ∈ Z \ {0} : a < arg z < a + 2π
n
}
is mapped onto the same set C \ {(x, 0) : x ≥ 0}, for an arbitrary a ∈ R, we
distinguish the images of the sectors{
z ∈ Z \ {0} :
2kπ
n
< arg z <
2(k + 1)π
n
}
by denoting as
{w ∈ C \ {0} : 2kπ < arg w < 2(k + 1)π}
in n different complex planes, and we call them by sheet-k for every k ∈
0, 1, 2, . . . , n − 1. As the boundary arg z = 2(k+1)π
n in sheet-k and the bound-
ary arg z = 2(k+1)π
n in sheet-(k + 1) are common, by using these n sheets, we
construct the Riemann surface as follows:1. The edge arg z = 2(k+1)π
n at the bottom of the slit in sheet-k and arg
z = 2(k+1)π
n at the top of the slit in sheet-(k + 1) should be pasted, for
all k = 0, 1, 2, . . . , n − 2.
1A studious reader can refer to any book on topology to know the definitions of the
terminologies used in the above definition.
�
�
“book” — 2014/6/4 — 20:56 — page 163 — #41
�
�
�
�
�
�
Rational Functions and Multivalued Functions 163
2. The boundary arg z = 2π at the bottom of the slit in sheet-(n − 1)
should be pasted with the boundary arg z = 0 at the top of the slit in
sheet-0.
3. The origin at every sheet should be glued as a single point in the
surface.
Now, if arg z traverses in the z-plane from 0 to 2π through 2π
k , k =
1, 2, . . . , n−1, correspondingly, arg zn navigates from 0 to 2π in sheet-0, then
it moves to the successive sheets through the pasted edges arg z = 2(k+1)π
n ,
and finally, it comes back to the top edge of the slit in sheet-0. In other
words, we have shown geometrically that the function z �→ zn is a continuous
bijection from C onto the surface.
Example 3.5.3 Construct the Riemann surface associated with cos.
We observe that the non-negative imaginary axis {(0, y) : y ≥ 0} is
mapped into {(u, 0) : u ≥ 1} because
cos(0 + iy) = cos(0) cosh(y) − i sin(0) sinh(y) = cosh(y)
= exp(y) + exp(−y)
2
≥ 1.
In fact, we see that cos : {(0, y) : y ≥ 0} → {(u, 0) : u ≥ 1} is onto. Indeed,
for a given u > 1, using cosh(y) → +∞ as y → ∞, we find y > 0 such
that cosh(y) > u. Therefore, applying the intermediate value theorem, we can
find y0 ∈ (0, y) such that cosh(y0) = u. Being cosh an even function, we also
observe that {(0, y) : y ≤ 0} is mapped onto {(u, 0) : u ≥ 1}. By a similar
argument, one can verify that cos maps
• both of {(π , y) : y ≥ 0} and {(π , y) : y ≤ 0} onto {(u, 0) : u ≤ 1},
• {(x, y) : 0 < x < π , y > 0} onto the lower half-plane {(u, v) : v < 0},
• {(x, y) : 0 < x < π , y < 0} onto the upper half-plane {(u, v) : v > 0},
• {(x, y) : 0 < x < π , y = 0} onto the line segment {(u, 0) : |u| < 1}.
Now we claim that cos : {(x, y) : 0 < x < π} → C \ {(u, v) ∈ C : |u| ≥
0, v = 0} is a bijection. As, the injectivity of cos on {(x, y) : 0 < x < π}
is obvious, we only verify that surjectivity of cos. Let w ∈ C \ {(u, v) ∈ C :
|u| ≥ 0, v = 0} be arbitrary and let θ =
(
w ± √
w2 − 1
)
. If θ ∈ (0,π ),
then we take z = θ − i log
∣∣∣w ± √
w2 − 1
∣∣∣ so that z ∈ {(x, y) : 0 < x < π}
�
�
“book” — 2014/6/4 — 20:56 — page 164 — #42
�
�
�
�
�
�
164 Elementary Riemann Surface
and cos z = w. Otherwise, θ ∈ (π , 2π ). In this case, we choose the required
pre-image by z = 2π − θ − i log
∣∣∣w ± √
w2 − 1
∣∣∣.
By a similar argument, we obtain cos : {(x, y) : (k − 1)π < x < kπ} →
C \ {(u, v) ∈ C : |u| ≥ 0, v = 0} is a bijection, ∀k ∈ N, with the additional
information that
• the upper and lower half-strips are mapped onto the lower and upper
half-planes, respectively, when k is odd,
• the upper and lower half-strips are mapped onto themselves when k is
even.
Now we are ready to construct the Riemann surface associated with cos.
• We take countably infinite number of complex planes with two slits
from 1 to ∞ and −1 to ∞ and we denote them by
{w ∈ C\{0} : (2k+1)π � arg w∈ (2kπ , (2k+2)π )}∪{(u, 0)∈C : |u| ≤ 1},
called sheet-k, ∀k ∈ Z.
• Paste upper edge on the slit on the left side of sheet-k with lower edge
on the slit on the left side of sheet-(k + 1) and lower edge on the slit
on the left side of sheet-k with upper edge on the slit on the left side of
sheet-(k + 1).
• Paste upper edge on the slit on the right side of sheet-k with lower edge
on the slit on the right side of sheet-(k + 1) and lower edge on the slit
on the right side of sheet-k with upper edge on the slit on the right side
of sheet-(k + 1).
• By gluing the line segment {(u, 0) : |u| ≤ 1} in every sheet altogether,
identify it as a single line segment on the surface.
Then, cos becomes a continuous bijection from C onto the surface.
Exercise 3.5.4 Construct the Riemann surface corresponding to exp(z).
�
�
“book” — 2014/6/4 — 21:03 — page 165 — #1
�
�
�
�
�
�
4
Complex Integration
4.1 LINE INTEGRAL
In this section, first we recall some definitions and results (without proofs)
from the theory of Riemann integration of real-valued function on [a, b].
Definition 4.1.1 Let f : [a, b] → R be a bounded function, where −∞ <
a < b < +∞.
1. By a partition of [a, b], we mean a finite subset {t0, t1, t2, . . . , tn} with
a = t0 < t1 < t2 < · · · < tn = b.
2. Define U(P, f ) =
n∑
k=1
[
sup
t∈[tk−1, tk ]
f (t)
]
(tk − tk−1) and L(P, f ) =
n∑
k=1
[
inf
t∈[tk−1, tk ]
f (t)
]
(tk − tk−1).
3. f is Riemann integrable if inf
P
U(P, f ) = sup
P
L(P, f ) and this value is
denoted by
b∫
a
f (t) dt.
THEOREM 4.1.2 Let f : [a, b] → R be a bounded real-valued function. Then
f is Riemann integrable on [a, b] iff given ε > 0, there exists a partition P of
[a, b] such that U(P, f ) − L(P, f ) < ε.
THEOREM 4.1.3 If f , f1, and f2 are Riemann integrable functions on [a, b],
then
1.
b∫
a
( f1 + f2)(t) dt =
b∫
a
f1(t) dt +
b∫
a
f2(t) dt,
165
�
�
“book” — 2014/6/4 — 21:03 — page 166 — #2
�
�
�
�
�
�
166 Line Integral
2.
b∫
a
(αf )(t) dt = α
b∫
a
f (t) dt, ∀α ∈ R,
3.
b∫
a
f1(t) dt ≤
b∫
a
f2(t) dt, whenever f1 ≤ f2 on [a, b],
4.
b∫
a
f (t) dt =
b∫
c
f (t) dt +
c∫
a
f (t) dt, provided a < c < b.
RESULT 4.1.4 If f : [a, b] → R is bounded and continuous except at a finite
number of points of [a, b], then f is Riemann integrable.
RESULT 4.1.5 If f : [a, b] → R is Riemann integrable, then | f | is also
Riemann integrable over [a, b] and
∣∣∣∣∣
b∫
a
f (t) dt
∣∣∣∣∣ ≤
b∫
a
|f (t)| dt.
Definition 4.1.6 Let f = u + iv be a complex-valued continuous function on
[a, b], where u and v are real-valued continuous functions on [a, b]. We define
b∫
a
f (t) dt =
b∫
a
u(t) dt + i
b∫
a
v(t) dt,
where
b∫
a
u(t) dt and
b∫
a
v(t) dt are Riemann integrals of u and v, respectively, on
[a, b].
RESULT 4.1.7 If f : [a, b] → C, g : [a, b] → C are continuous and c ∈ C,
then
b∫
a
( f + g)(t) dt =
b∫
a
f (t) dt +
b∫
a
g(t) dt and
b∫
a
(cf )(t) dt = c
b∫
a
f (t) dt.
Proof: Let f = u1 + iv1, g = u2 + iv2 and c = α + iβ. Then
b∫
a
( f + g)(t) dt
=
b∫
a
(u1 + u2)(t) dt + i
b∫
a
(v1 + v2)(t) dt
�
�
“book” — 2014/6/4 — 21:03 — page 167 — #3
�
�
�
�
�
�
Complex Integration 167
=
⎛
⎝ b∫
a
u1(t) dt +
b∫
a
u2(t) dt
⎞
⎠+ i
⎛
⎝ b∫
a
v1(t) dt +
b∫
a
v2(t) dt
⎞
⎠
=
⎛
⎝ b∫
a
u1(t) dt + i
b∫
a
v1(t) dt
⎞
⎠+
⎛
⎝ b∫
a
u2(t) dt + i
b∫
a
v2(t) dt
⎞
⎠
=
b∫
a
f (t) dt +
b∫
a
g(t) dt,
b∫
a
(cf )(t) dt,
=
b∫
a
((α + iβ)(u + iv))(t)
=
b∫
a
(αu(t) − βv(t)) dt + i
b∫
a
(αv(t) + βu(t)) dt
= α
b∫
a
u(t) dt − β
b∫
a
v(t) dt + i
⎛
⎝α
b∫
a
v(t) dt + β
b∫
a
u(t) dt
⎞
⎠
= (α + iβ)
⎛
⎝ b∫
a
u(t) dt + i
b∫
a
v(t) dt
⎞
⎠
= c
b∫
a
f (t) dt.
Hence, the result follows. �
LEMMA 4.1.8 If f : [a, b] → C is a continuous function, then
∣∣∣∣∣
b∫
a
f (t) dt
∣∣∣∣∣ ≤
b∫
a
| f (t)| dt.
�
�
“book” — 2014/6/4 — 21:03 — page 168 — #4
�
�
�
�
�
�
168 Line Integral
Proof: If
b∫
a
f (t) dt = 0, then obviously the inequality follows. If
b∫
a
f (t) dt � 0,
then we can write it as
b∫
a
f (t) dt = reiθ . Then, using the previous theorem, we
get ∣∣∣∣∣∣
b∫
a
f (t) dt
∣∣∣∣∣∣ = r
= e−iθ
b∫
a
f (t) dt
=
b∫
a
e−iθ f (t) dt
= Re
⎛
⎝ b∫
a
e−iθ f (t) dt
⎞
⎠
=
b∫
a
Re (e−iθ f (t)) dt
≤
b∫
a
∣∣e−iθ f (t)
∣∣ dt
=
b∫
a
| f (t)| dt. �
RESULT 4.1.9 If f : [a, b] → C is Riemann integrable and | f (t)| ≤ M ,
∀t ∈ [a, b], then
b∫
a
| f (t)| dt ≤ M(b − a).
Proof of this result follows immediately from the previous Lemma.
RESULT 4.1.10 Let fn and f be complex-valued Riemann integrable func-
tions on [a, b], ∀n ∈ N. If fn → f uniformly on [a, b], then
b∫
a
fn(t) dt →
b∫
a
f (t) dt as n → ∞.
�
�
“book” — 2014/6/4 — 21:03 — page 169 — #5
�
�
�
�
�
�
Complex Integration 169
Proof: Let ε > 0 be given. Since fn → funiformly on [a, b], there exists
N ∈ N such that | fn(x) − f (x)| < ε/(b − a)∀ n ≥ N and ∀ x ∈ [a, b].
Now for n ≥ N ,∣∣∣∣∣∣
b∫
a
fn(x)dx −
b∫
a
f (x)dx
∣∣∣∣∣∣ ≤
b∫
a
| fn(x) − f (x)|dx <
ε
b − a
b∫
a
dx = ε.
Therefore,
b∫
a
fn(x) dx →
b∫
a
fdx as n → ∞. �
THEOREM 4.1.11 If f : [a, b] → C is continuous and g(t) =
t∫
a
f (τ ) dτ , ∀t ∈
[a, b], then g is differentiable on [a, b] and g′ = f.
Proof: Let t ∈ [a, b) be arbitrary. Since f is continuous at t, given ε > 0, there
exists δ > 0 such that [t, t+δ] ⊂ [a, b) and t ≤ s < t+δ ⇒ | f (s)− f (t)| < ε.
For 0 < h < δ, we have
∣∣∣∣g(t + h) − g(t)
h
− f (t)
∣∣∣∣ =
∣∣∣∣∣∣∣∣∣
t∫
a
f (τ ) dτ −
t+h∫
a
f (τ ) dτ −
t+h∫
t
f (t) dτ
h
∣∣∣∣∣∣∣∣∣
= 1
h
∣∣∣∣∣∣
t+h∫
t
( f (τ ) − f (t)) dτ
∣∣∣∣∣∣
≤ 1
h
t+h∫
t
| f (τ ) − f (t)| dτ
= 1
h
hε = ε, since |τ − t| < h < δ.
Similarly, we can prove that for every t ∈ (a, b], there exits δ > 0 such
that
∣∣∣∣g(t − h) − g(t)
−h
− f (t)
∣∣∣∣ < ε, whenever 0 < h < δ. Therefore, we have
proved that
lim
h→0
g(t + h) − g(t)
h
= f (t), ∀t ∈ [a, b].
Thus, g′ = f on [a, b]. �
�
�
“book” — 2014/6/4 — 21:03 — page 170 — #6
�
�
�
�
�
�
170 Line Integral
THEOREM 4.1.12 (Fundamental theorem of calculus)
If f : [a, b] → C is differentiable on [a, b] and f ′ is Riemann itegrable on
[a, b], then
b∫
a
f ′(t) dt = f (b) − f (a).
Proof: First, we prove this theorem for the case that f is a real-valued
Riemann integrable function on [a, b]. By Theorem 4.1.2, given ε > 0,
there exists a partition P = {t0, t1, t2, . . . , tn} of [a, b] such that U(P, f ′) −
L(P, f ′) < ε. By mean value theorem, there exists sk ∈ (tk−1, tk) such
that
f (tk) − f (tk−1) = f ′(sk)(tk − tk−1), ∀k = 1, 2, . . . , n. (4.1)
If Mk = sup
tk−1<t<tk
f ′(t) and mk = inf
tk−1<t<tk
f ′(t), ∀k = 1, 2, . . . , n, then we have
mk ≤ f ′(sk) ≤ Mk ,∀k = 1, 2, . . . , n, which implies that
L(P, f ′) =
n∑
k=1
mk(tk − tk−1)
≤
n∑
k=1
f ′(sk)(tk − tk−1)
≤
n∑
k=1
Mk(tk − tk−1)
= U(P, f ′).
On the other hand, by the definition of Riemann integral, we also have
L(P, f ′) ≤
b∫
a
f ′(t) dt ≤ U(P, f ′), and hence,
∣∣∣∣∣∣
b∫
a
f ′(t) dt −
n∑
k=1
f ′(sk)(tk − tk−1)
∣∣∣∣∣∣ ≤ U(P, f ) − L(P, f ).
Therefore, using equation (4.1), we get
∣∣∣∣∣∣
b∫
a
f ′(t) dt − f (a) + f (b)
∣∣∣∣∣∣ =
∣∣∣∣∣∣
b∫
a
f ′(t) dt −
n∑
k=1
[ f (tk) − f (tk−1)]
∣∣∣∣∣∣
�
�
“book” — 2014/6/4 — 21:03 — page 171 — #7
�
�
�
�
�
�
Complex Integration 171
=
∣∣∣∣∣∣
b∫
a
f ′(t) dt −
n∑
k=1
f ′(sk)(tk − tk−1)
∣∣∣∣∣∣
≤ U(P, f ) − L(P, f ) < ε.
Since ε > 0 is arbitrary, we get
b∫
a
f ′(t) dt = f (b) − f (a).
If f is a complex-valued function, let f = u + iv, where u and v are real-
valued Riemann integrable functions. Hence, using this theorem for u and v,
we obtain
b∫
a
f ′(t) dt =
b∫
a
u′(t) dt + i
b∫
a
v′(t)) dt
= (u(b) − u(a)) + i(v(b) − v(a))
= f (b) − f (a).
Hence, the theorem follows. �
Example 4.1.13 Find
2∫
0
[(1 + 9t2) + i(4t + 2)] dt.
2∫
0
[(1 + 9t2) + i(4t + 2)] dt =
2∫
0
(1 + 9t2) dt + i
2∫
0
(4t + 2) dt
=
[
t + 3t3
]2
0
+ i
[
2t2 + 2t
]2
0
= 26 + 12i.
Definition 4.1.14 (Piecewise smooth curve)
A curve γ with a parametric equation ϕ(t), t ∈ [a, b] is called a piecewise
smooth curve if ϕ is differentiable and ϕ′ is continuous for all but finite num-
ber of points {tj : 1 ≤ j ≤ n} of [a, b]. Furthermore, ϕ has both left limit and
right limit at each tj, 1 ≤ j ≤ n. Hereafter by a curve, we mean a piecewise
smooth curve.
Definition 4.1.15 A curve with a parametric equation ϕ(t), t ∈ [a, b] is said
to be
�
�
“book” — 2014/6/4 — 21:03 — page 172 — #8
�
�
�
�
�
�
172 Line Integral
1. a closed curve if ϕ(a) = ϕ(b).
2. a simple curve if ϕ is one-to-one on [a, b].
3. a simple closed curve if it is closed and φ(t) = φ(s) ⇒ either t = s or
{t, s} = {a, b}.
Definition 4.1.16 Let � be a region and let γ be a curve in �. If f is a
continuous function on �, then define the line integral
∫
γ
f (z) dz =
b∫
a
f (ϕ(t))ϕ′(t) dt,
where ϕ(τ ), τ ∈ [a, b] is a parametric equation of γ .
Definition 4.1.17 Let γ be a curve in a region�. If f is a continuous function
on � and g has a continuous partial derivatives on �, then define
∫
γ
f dg =
b∫
a
f
(
∂g
∂x
dx − ∂g
∂y
dy
)
. More explicitly, if ϕ(t), t ∈ [α,β] is the parametric
equation of γ , then
∫
γ
f dg =
β∫
α
f (φ(t))
(
∂g
∂x
(φ(t)) (Re ϕ)′(t) − ∂g
∂y
(φ(t)) (Im ϕ)′(t)
)
dt.
Example 4.1.18 Find
∫
γ
f (z) dz, where f (z) = z + 1, ∀z ∈ C and ϕ(t) =
t + it, t ∈ [0, 1].
∫
γ
f (z) dz =
1∫
0
f ((ϕ(t))ϕ′(t) dt
=
1∫
0
(ϕ(t) + 1)(1 + i) dt
=
1∫
0
((1 + t) + it)(1 + i) dt
=
1∫
0
[((1 + t) − t) + i(1 + t + t)] dt
�
�
“book” — 2014/6/4 — 21:03 — page 173 — #9
�
�
�
�
�
�
Complex Integration 173
=
1∫
0
(1 + i(2t + 1)) dt
=
1∫
0
dt + i
1∫
0
(2t + 1) dt
= 1 + i2.
Example 4.1.19 If f (z) = z2 and γ is the parabola given by 2t + i(t2 + 1)
t ∈ R, then find
∫
σ
f (z) dz, where σ is the part of γ from i to 2 + i2.
First, we note that the points i and 2 + i2 correspond to t = 0 and t = 1,
respectively, and hence, the parametric equation of σ is ϕ(t) = 2t + i(t2 + 1),
t ∈ [0, 1]. Now, for every t ∈ [0, 1], we have
ϕ′(t) = 2 + i2t
f (ϕ(t)) = (2t + i(t2 + 1))2
= (4t2 − (t2 + 1)2) + i2t(t2 + 1)
= (−t4 + 2t2 − 1) + i(2t3 + 2t)
f (ϕ(t))ϕ′(t) = 2(−t4 + 2t2 − 1) − 2t(2t3 + 2t)
+ i(4t3 + 4t − 2t5 + 4t3 − 2t)
= (−6t4 − 2) + i(−2t5 + 8t3 + 2t).∫
σ
f (z) dz =
1∫
0
f ((ϕ(t))ϕ′(t) dt
=
1∫
0
(−6t4 − 2) dt + i
1∫
0
(−2t5 + 8t3 + 2t) dt
=
(
−6
5
− 2
)
+ i
(
−1
3
+ 2 + 1
)
= −16
5
+ i
8
3
.
Example 4.1.20 Find
∫
γ
(z+ 3) dz, where γ is the circular arc given by ϕ(t) =
i + 2eit, t ∈ [0,π ].
�
�
“book” — 2014/6/4 — 21:03 — page 174 — #10
�
�
�
�
�
�
174 Line Integral
Solution:
∫
γ
(z + 3) dz =
π∫
0
[(i + 2eit) + 3]i2eit dt
= i2
π∫
0
[(3 + i)eit + 2ei2t)] dt
= i2
[
−(3 + i)ieit +−iei2t
]π
0
= −12 − i4.
Exercise 4.1.21
1. Evaluate
∫
γ
f (z) dz, where f and the parametric equation ‘ϕ(t),
t ∈ [a, b]’ of γ are given as follows:
(1) f (z) = z3, ϕ(t) = 1 + it, t ∈ [0, 1],
(2) f (z) = z, ϕ(t) = exp(iπ t), t ∈ [0, 1],
(3) f (z) = 1
z
, ϕ(t) = 1 + 2t + it2, t ∈ [0, 1],
(4) f (z) = z exp(z2), ϕ(t) = √
t, t ∈ [1, 4],
(5) f (z) = cos(z), ϕ(t) = t + it2, t ∈ [0, 1].
Answers: (1)
−5
4
; (2) iπ ; (3) log(3 + i); (4)
exp(1)
2
(exp(3) − 1);
(5) sin(1 + i).
2. Evaluate
∫
γ
f (z) dz, where f (z) = z2 + z, where γ is the part of the
parabola x = y2 from (0, 0) to (4, 2).
Answer:
2
3
(17 + i56).
3. Evaluate
∫
γ
f (z) dz, where f (z) = 1
z − a
, where γ is the part of the circle
with center a ∈ C and radius r > 0 from a + r to a + ir.
Answer: i
π
2
.
�
�
“book” — 2014/6/4 — 21:03 — page 175 — #11
�
�
�
�
�
�
Complex Integration 175
Definition 4.1.22 Let � be a region and let γ be a curve in �. If f is a
continuous function on �, then define
∫
γ
f (z) |dz| =
b∫
a
f (ϕ(t))|ϕ′(t)| dt, where
ϕ(t), t ∈ [a, b] is a parametric equation of γ .
THEOREM 4.1.23 If f is continuous on � and γ is a curve in �, then∣∣∣∣∣∫γ f (z) dz
∣∣∣∣∣ ≤ ∫
γ
| f (z)| |dz|.
Proof: Using Lemma 4.1.8, if ϕ(t), t ∈ [a, b] is a parametric equation of γ ,
then∣∣∣∣∣∣
∫
γ
f (z) dz
∣∣∣∣∣∣ =
∣∣∣∣∣∣
b∫
a
f (ϕ(t))ϕ′(t) dt
∣∣∣∣∣∣ ≤
b∫
a
| f (ϕ(t))ϕ′(t)| dt =
∫
γ
| f (z)| |dz|.
Hence, the theorem follows. �
Definition 4.1.24 If a curve γ is given by ϕ(t), t ∈ [a, b], then the opposite
curve −γ is defined by ψ(t) = ϕ(a + b − t), t ∈ [a, b].
Geometrically, −γ is obtained from γ just by changing its direction in the
opposite sense.
g −g
LEMMA 4.1.25 If f is continuous on � and γ is a curve in �, then∫
−γ
f (z) dz = − ∫
γ
f (z) dz.
Proof: If ϕ(t), t ∈ [a, b] is a parametric equation of γ and ψ(t) = ϕ(a +
b − t), t ∈ [a, b] is the parametric equation of −γ , then using the change of
variable s = a + b − t in the following, we get∫
−γ
f (z) dz =
∫ b
a
f (ψ(t))ψ ′(t) dt
=
∫ b
a
f (ϕ(a + b − t))ϕ′(a + b − t) (−dt)
�
�
“book” — 2014/6/4 — 21:03 — page 176 — #12
�
�
�
�
�
�176 Line Integral
=
∫ a
b
f (ϕ(s))ϕ′(s) ds
= −
∫ b
a
f (ϕ(s))ϕ′(s) ds
= −
∫
γ
f (z) dz.
Hence, the lemma follows. �
Definition 4.1.26 (Arc length)
If γ is curve with parametric equation z(t), t ∈ [a, b], then the length of
the curve is defined by L(γ ) = sup
P
|z(tk) − z(tk−1)|, where P varies over all
partitions of [a, b].
A curve with finite length is called a rectifiable curve.
THEOREM 4.1.27 If γ is a smooth curve with the parametric equation
z(τ ), τ ∈ [a, b], then it is rectifiable and L(γ ) = ∫ b
a |z′(t)| dt.
Proof: For each fixed t ∈ [a, b], let γa,t be the part of the curve γ with
parametric equation z(τ ) = u(τ ) + iv(τ ), τ ∈ [a, t], where u and v are
real-valued differentiable function on [a, b]. Define S : [a, b] → R by
S(t) = L(γa,t), ∀t ∈ [a, b]. We shall show that S is a differentiable function
and its S′(t) = |z′(t)|, ∀t ∈ [a, b]. Therefore, using the uniform continuity of
z′, given ε > 0, there exists δ > 0 such that
|τ − t| < δ ⇒ |z′(τ ) − z′(t)| < ε√
2
(4.2)
⇒ |u′(τ ) − u′(t)| < ε√
2
and |v′(τ ) − v′(t)| < ε√
2
.
Therefore, for |τ − t| < δ and |σ − t| < δ, we have∣∣ |(u′(τ ), v′(σ ))| − |z′(t)| ∣∣ = ∣∣ |(u′(τ ), v′(σ ))| − |(u′(t), v′(t))| ∣∣
≤ |(u′(τ ), v′(σ )) − (u′(t), v′(t))|
=
√
(u′(τ ) − u′(t))2 + (v′(σ ) − v′(t))2
<
√
ε2
2
+ ε2
2
= ε.
�
�
“book” — 2014/6/4 — 21:03 — page 177 — #13
�
�
�
�
�
�
Complex Integration 177
Hence, it follows that
|z′(t)| − ε < |(u′(τ ), v′(σ ))| < |z′(t)| + ε,∀σ , τ ∈ (t − δ, t + δ). (4.3)
Let 0 < s < δ and let P = {t0, t1, . . . , tn} be an arbitrary partition of [t, t + s].
Then, applying mean-value theorem (Theorem 2.1.18) for u and v, we get
n∑
k=1
|z(tk) − z(tk−1)| =
n∑
k=1
√
(u(tk) − u(tk−1))2 + (v(tk) − v(tk−1))2
=
n∑
k=1
(tk − tk−1)|(u′(xk), v′(yk))|, (4.4)
for some xk , yk ∈ (tk−1, tk).
Therefore,
s(|z′(t)| − ε) = (tn − t0)(|z′(t)| − ε)
=
n∑
k=1
(tk − tk−1)(|z′(t)| − ε)
<
n∑
k=1
(tk − tk−1)|(u′(xk), v′(yk))| (by using (4.3))
=
n∑
k=1
|z(tk) − z(tk−1)| (by using (4.4))
≤
n∑
k=1
(tk − tk−1)|z′(sk)|, for some sk ∈ (tk−1, tk)
(by using Theorem 2.1.19 for the function z.)
< (tn − t0)(|z′(t)| + ε)
(since |sk − t| < δ, by using (4.3))
= s(|z′(t)| + ε).
Hence, we proved that
s(|z′(t)| − ε) ≤ sup
P
n∑
k=1
|z(tk) − z(tk−1)| ≤ s(|z′(t)| + ε),
where P varies over all partitions of [t, t+ s]. Therefore, if γt, t+s is the part of
gamma, whose parametric equation is z(τ ), τ ∈ [t, t + s], then it follows that
�
�
“book” — 2014/6/4 — 21:03 — page 178 — #14
�
�
�
�
�
�
178 Line Integral
s(|z′(t)| − ε) ≤ L(γt, t+s) = (S(t + s) − S(t)) ≤ s(|z′(t)| + ε),
which implies that
∣∣∣∣S(t + s) − S(t)
s
− |z′(t)|
∣∣∣∣ < ε. Similarly, we can show
that if 0 < s < δ,
∣∣∣∣S(t − s) − S(t)
(−s)
− |z′(t)|
∣∣∣∣ < ε. That is, we have proved that
S′(t) = lim
s→0
S(t + s) − S(t)
s
= |z′(t)|. Therefore, by Theorem 4.1.11, we get
S(t) =
t∫
a
S′(τ ) dτ =
t∫
a
S′(τ ) dτ , ∀t ∈ [a, b].
In particular, L(γ ) = S(b) =
b∫
a
S′(τ ) dτ . �
LEMMA 4.1.28 Let f (z) be a continuous function on a region �. If there
exists an analytic function F on � such that F′ = f , then∫
γ
f (z) dz = F(ϕ(b)) − F(ϕ(a)),
for every curve γ in �, where ϕ(t), t ∈ [a, b] is a parametric equation of γ .
Proof: Define h(t) = F(ϕ(t)), ∀t ∈ [a, b], then h is piecewise differen-
tiable and h′(t) = F′(ϕ(t))ϕ′(t) = f (ϕ(t))ϕ′(t) for every t at which ϕ is
differentiable. By fundamental theorem of calculus (Theorem 4.1.12),
∫
γ
f (z) dz =
b∫
a
f (ϕ(t))ϕ′(t) dt
=
b∫
a
h′(t) dt
= h(b) − h(a)
= F(ϕ(b)) − F(ϕ(a)).
Hence, the lemma follows. �
COROLLARY 4.1.29 Let f (z) be a continuous function on a region�. If there
exists an analytic function F on � such that F′ = f , then
∫
γ
f (z) dz = 0, for
every closed curve γ in �.
�
�
“book” — 2014/6/4 — 21:03 — page 179 — #15
�
�
�
�
�
�
Complex Integration 179
Proof: Since γ is a closed curve, if ϕ(t), t ∈ [a, b] is a parametric equation
of γ , then ϕ(a) = ϕ(b). Hence, using the previous lemma, we get∫
γ
f (z) dz = F(ϕ(b)) − F(ϕ(a)) = 0.
Converse of the above corollary is also true, which follows. �
RESULT 4.1.30 If f is a continuous function on a region � such that∫
γ
f (z) dz = 0, for every closed curve γ in �, then f is the derivative of an
analytic function F on �.
Proof: Let w = (u, v) ∈ � be arbitrary. Then, for a given z0 = (x0, y0) ∈ �
and an ε > 0, choose δ > 0 such that B(z0, δ) ⊆ � and
| f (z) − f (z0)| < ε whenever |z − z0| < δ. (4.5)
Let Pz0 be the polygon joining w and z0 such that
1. the line segments of Pz0 are parallel to the coordinate axes.
2. the line segment incident with z0 is horizontal and completely con-
tained in B(z0, δ).
3. Pz0 ⊂ �.
Such a polygon exists by Theorem 1.4.31.
(x0, y0)
(x, y0)
(u, u)
�
�
“book” — 2014/6/4 — 21:03 — page 180 — #16
�
�
�
�
�
�
180 Line Integral
Define F(z0) = ∫
Pz0
f (z) dz. We claim that ∂F
∂x (x0, y0) = f (x0, y0). Let
(x, y0) be the point adjacent with (x0, y0) in the polygon Pz0 . As the parametric
equation of L(x, y0),(x0, y0) and L(x, y0),(x0+h, y0) are
ϕ(t) = t + iy0 = (t, iy0), t ∈ [x, x0]
and
ψ(t) = t + iy0 = (t, iy0), t ∈ [x, x0 + h],
respectively, for each h ∈ R such that |h| < δ,
F(x0 + h, y0) − F(x0, y0)
h
=
∫
Pz0+h
f (z) dz − ∫
Pz0
f (z) dz
h
=
∫
L(x, y0),(x0+h, y0)
f (z) dz − ∫
L(x, y0),(x0, y0)
f (z) dz
h
=
x0+h∫
x
f (t, y0) dt −
x0∫
x
f (t, y0) dt
h
= 1
h
x0+h∫
x0
f (t, y0) dt
= 1
h
h∫
0
f (x0 + s, y0) ds. (4.6)
Now, using the equation (4.6), we get∣∣∣∣F(x0 + h, y0) − F(x0, y0)
h
− f (x0, y0)
∣∣∣∣
=
∣∣∣∣∣∣
1
h
h∫
0
f (x0 + s, y0) ds − f (x0, y0)
∣∣∣∣∣∣
=
∣∣∣∣∣∣
1
h
h∫
0
( f (x0 + s, y0) − f (x0, y0)) ds
∣∣∣∣∣∣
≤ 1
|h|
h∫
0
| f (x0 + s, y0) − f (x0, y0)| |ds|
�
�
“book” — 2014/6/4 — 21:03 — page 181 — #17
�
�
�
�
�
�
Complex Integration 181
<
1
|h|
h∫
0
ε |ds|
= |h|ε
|h| = ε,
as |(x0 + s, y0) − (x0, y0)| = |t| < |h| < δ and by using equation (4.5). Thus,
our claim is proved. Next for every z0 ∈ �, we define G(z0) = ∫
Qz0
f (z) dz,
where Qz0 is a polygon joining w and z0 such that
1. the line segments of Qz0 are parallel to the coordinate axes.
2. the line incident with z0 is vertical and completely contained in B(z0, δ).
3. Qz0 ⊆ �.
Next we claim that ∂G
∂y (x0, y0) = if (x0, y0). We fix ε > 0, δ > 0 and h ∈ R as
before. As the parametric equation of L(x0, v),(x0, y0) is
ϕ(t) = x0 + it = (x0, t), t ∈ [v, y0],
by a similar argument, we get∣∣∣∣G(x0, y0 + h) − G(x0, y0)
h
− if (x0, y0)
∣∣∣∣
=
∣∣∣∣∣∣
1
h
h∫
0
f (x0, y0 + t) idt − if (x0, y0)
∣∣∣∣∣∣
≤ 1
|h|
h∫
0
| f (x0, y0 + t) − f (x0, y0)| |dt| < ε.
Since
F(z0) − G(z0) =
∫
Pz0
f (z) dz −
∫
Qz0
f (z) dz =
∫
Pz0∪(−Qz0 )
f (z) dz
and Pz0 ∪ (−Qz0 ) is a closed curve, by hypothesis, we get
∫
Pz0∪(−Qz0 )
f (z) dz =
0, and hence, F(z0) − G(z0) = 0. In other words, F = G on �. There-
fore, F has continuous partial derivatives, and they satisfy the C–R equation
∂F
∂x = −i ∂F
∂y (= f ). Thus, F is analytic on � (by Theorem 2.2.28) and
F′ = ∂F
∂x = f . �
�
�
“book” — 2014/6/4 — 21:03 — page 182 — #18
�
�
�
�
�
�
182 Winding Number and Cauchy’s Theorems
4.2 WINDING NUMBER AND CAUCHY’S THEOREMS
LEMMA 4.2.1 For a curve γ and a point z0 not on γ ,
1
i2π
∫
γ
dz
z − z0
is an
integer.
Proof: Let ϕ(t), t ∈ [a, b] be a parametric equation of the closed curve γ .
Then we have ϕ(a) = ϕ(b) and
1
i2π
∫
γ
dz
z − z0
= 1
i2π
b∫
a
ϕ′(t) dt
ϕ(t) − z0
.
If F(s) =
s∫
a
ϕ′(t) dt
ϕ(t)−z0
, ∀s ∈ [a, b], then F is a differentiable function on
[a, b] and F′(s) = ϕ′(s)
ϕ(s)−z0
, ∀s ∈ [a, b], by Theorem 4.1.11. To conclude
the theorem, we shall show that F(b) is an integral multiple of i2π . If
ψ(s) = exp(−F(s))(ϕ(s) − z0), ∀s ∈ [a, b], then
ψ ′(s) = − exp(−F(s))F′(s)(ϕ(s) − z0) + exp(−F(s))ϕ′(s)
= − exp(−F(s))ϕ′(s) + exp(−F(s))ϕ′(s) = 0,
we get that ψ is a constant function on [a, b]. Therefore,
exp(−F(a))(ϕ(a) − z0) = exp(−F(b))(ϕ(b) − z0),
and hence, exp(−F(a)) = exp(−F(b)) ⇒ exp( F(b)) = 1, as F(a) = 0. UsingLemma 2.4.20, it follows that F(b) = i2kπ for some k ∈ Z. �
Definition 4.2.2 Let γ be a closed curve and z0 be a point not on γ , then
we define the winding number of γ with respect to z0 or the index of z0 with
respect to γ , by WN(γ , z0) = 1
i2π
∫
γ
dz
z − z0
.
Definition 4.2.3 A region determined by a closed curve γ is defined by a
component of the complement of γ in the extended complex plane.
�
�
“book” — 2014/6/4 — 21:03 — page 183 — #19
�
�
�
�
�
�
Complex Integration 183
The above definition can be understood easily by the following diagram.
g Γ
R1
S1
S3
S2
S4R2
Remark 4.2.4: For a given closed curve, among the regions determined by
γ , there must be only one unbounded region determined by γ . That region is
treated as the region containing ∞ in C∞.
In the above diagrams, R2 and S4 are the unbounded regions determined by
γ and �, respectively.
THEOREM 4.2.5 Let γ be a closed curve in C and z0 ∈ C not on γ . Then,
1. WN(−γ , z0) = −WN(γ , z0),
2. WN(γ , z0) = WN(γ , w0) if z0 and w0 belong to a same region
determined by γ ,
3. WN(γ , z0) = 0 if z0 belongs to the unbounded region determined by γ .
Proof: Let z0 � γ .
1. Using Lemma 4.1.25, we get
WN(−γ , z0) = 1
i2π
∫
−γ
dz
z − z0
= − 1
i2π
∫
γ
dz
z − z0
= −WN(γ , z0).
2. Case 1. First, we show that if the line segment Lz0,w0 joining z0 and
w0 is completely contained in a same region determined by γ . We
know that log′
(
z − z0
z − w0
)
= 1
z − z0
− 1
z − w0
, provided the expression
z − z0
z − w0
belongs to the domain of log, where it is analytic. From Result
�
�
“book” — 2014/6/4 — 21:03 — page 184 — #20
�
�
�
�
�
�
184 Winding Number and Cauchy’s Theorems
3.3.9, we recall that log is analytic on � = C \ {(x, 0) : x ≤ 0}. As
z − z0
z − w0
is real and ≤ 0 ⇒ z − z0
z − w0
= −k for some k > 0
⇒ z = z0 + kw0
1 + k
⇒ z lies on Lz0,w0 .
Hence, log
(
z − z0
z − w0
)
is differentiable on the complement of Lz0,w0 .
Since Lz0,w0 does not intersect γ , we obtain that γ belongs to the
complement of Lz0,w0 . Thus, using Corollary 4.1.29, we get
∫
γ
(
1
z − z0
− 1
z − w0
)
dz =
∫
γ
log′
(
z − z0
z − w0
)
dz = 0.
Case 2. Next, let z0 and w0 be two points in the same region determined
by γ . Then, we can join z0 and w0 by a polygon
n∪
j=1
Laj−1,aj contained in
the same region determined by γ , where a0 = z0, an = w0, and Laj−1,aj
is the line segment joining ai−1 and ai. Applying case 1, repeatedly, we
get
WN(γ , z0) = WN(γ , a0) = WN(γ , a1) = · · · = WN(γ , an)
= WN(γ , w0).
3. As γ is the continuous image of [a, b], it is a compact subset of C,
and hence, it is bounded. Therefore, we choose M > 0 such that γ ⊆
B(0, M). Let z0 be the given point belonging to the unbounded region
determined by γ . Then for any w0 � B(0, M), z0 and w0 belong to the
same region determined by γ . Since
1
z − w0
is an analytic function in
B(0, M), using (2) and Cauchy’s theorem for simply connected region
(Theorem 4.2.14),1 we have
WN(γ , z0) = WN(γ , w0) = 1
i2π
∫
γ
dz
z − w0
= 0.
1Though Theorem 4.2.14 is proved later, and its proof does not depend on the present
theorem.
�
�
“book” — 2014/6/4 — 21:03 — page 185 — #21
�
�
�
�
�
�
Complex Integration 185
From the above properties. WN(γ , z0) is called the winding number of γ with
respect to z0. �
Example 4.2.6 If C is |ζ − a| = r, then WN(C, z) =
{
1 if |z − a| < r
0 if |z − a| > r
.
Since the parametric equation of the circle |ζ − a| = r is given by ζ =
a + r exp(iθ ), θ ∈ [0, 2π], we get
1
i2π
∫
|ζ−a|=r
dζ
ζ − a
= 1
i2π
2π∫
0
ir exp(iθ )dθ
r exp(iθ )
= 1.
If |z− a| < r, then a and z are lying in the same region determined by C, and
using Theorem 4.2.5(2), we get WN(C, z) = WN(C, a) = 1. If |z − a| > r,
then by using Theorem 4.2.5(3), we get WN(C, z) = 0.
There is an interesting theorem, namely, Jordan curve theorem, which
states that the complement of a simple closed curve has exactly to regions.
Although it can be realized geometrically, its proof is too lengthy. Hence, we
prefer to omit this theorem.
Definition 4.2.7 A region � in C is called a simply connected region if the
complement of � in the extended complex plane is also connected.
Geometrically, every closed curve in a simply connected region �
encloses only the points of �. We prove this statement rigorously through
the following theorem.
A simply connected region
THEOREM 4.2.8 Let � be a region in C. � is simply connected iff
WN(γ , z0) = 0, for every z0 � � and for every curve γ in �.
�
�
“book” — 2014/6/4 — 21:03 — page 186 — #22
�
�
�
�
�
�
186 Winding Number and Cauchy’s Theorems
Proof: Assume that � is simply connected, that is, complement of � in
C∞ is connected. Since γ ⊆ �, we have C∞\� ⊆ C∞\γ . How-
ever, C∞\γ is a disjoint union of the regions determined by γ and
∞ ∈ C∞\�. Hence, C∞\� is contained in the unbounded region de-
termined by γ . Hence, by Theorem 4.2.5(3), we get WN(γ , z0) = 0,
∀z0 � �.
Conversely, assume that WN(γ , z0) = 0, ∀z0 � �, for every closed curve
γ in �. Suppose, C∞\� is not connected. Then, C∞\� = A ∪ B, where
A � ∅, B � ∅, A ∩ B = ∅ and A and B are closed subsets of C∞\�.
Using ∞ ∈ C∞\�, without loss of generality, we assume that ∞ ∈ A and
let b ∈ B. Being C∞\B an open set containing ∞, it will be of the form
C∞\K for some compact subset K of C. (Cf. Definition 1.5.1.) Thus, B is a
compact subset of C, and hence, it is bounded. As A \ {∞} and B are disjoint
closed subsets of C with one of them is compact, if we let δ = inf{|x − y| :
x ∈ A \ {∞} and y ∈ B}, then δ > 0 (since A \ {∞} is closed in C and by
Theorem 1.4.41). Now, we cover the entire complex plane by a net consisting
of squares of diameter less than δ such that the point b lies at the centre of
a square. Let {Sα : α ∈ �} be the collection of all squares, which intersect
B. Being B is bounded, � is a finite set. Then, let σδ = ∑
α∈�
∂Sα , where
∂Sα is the boundary of the square Sα . After removing the common edges
with opposite directions present in σδ , it is a closed curve, and it does not
intersect B.
A
BΩ
sd
We claim that σδ ∩ A = ∅. Otherwise, there exists z0 ∈ σδ ∩ A. Then,
there exists a β ∈ � such that Sβ ∩ B � ∅ and z0 ∈ Sβ . Then, we can
choose w0 ∈ Sβ ∩ B, and hence, |z0 − w0| ≥ δ. However, z0, w0 ∈ Sβ ⇒
|z0 − w0| < δ, which is a contradiction. Hence, σδ does not intersect A.
Therefore, σδ ⊆ C\(A ∪ B) = �. From b ∈ B ⊆ (C∞\�), we have b � �.
Therefore, by assumption, WN(σδ , b) = 0. On the other hand, WN(σδ , b) =
�
�
“book” — 2014/6/4 — 21:03 — page 187 — #23
�
�
�
�
�
�
Complex Integration 187
∑
α∈�
WN(∂Sα , b) = 1, since WN(∂Sα , b) =
{
1 if α = α0
0 if α � α0,
where b ∈ Int Sα0 .
This is a contradiction. Therefore, C∞\� is connected. Thus, � is simply
connected. �
Definition 4.2.9 A region � is called a multiply connected region, if it is not
simply connected.
That is, the complement of � in C∞ has more than two components or
equivalently; there exists a closed curve γ in �, which encloses a point of
complement of �.
A multiply connected region
Definition 4.2.10 Let γ be a closed curve in a region �. We say that γ is
homologous to 0 in � if n(γ , a) = 0, ∀a � �. In this case, we write γ ∼ 0
in �.
The following result is an immediate consequence of Theorem 4.2.8.
RESULT 4.2.11 Let � be a region. Then, � is simply connected iff γ ∼ 0 in
�, ∀ closed curve γ in �.
In the following theorem, we mean a rectangle R by [a, b] × [c, d], for
some a, b, c, d ∈ R with a < b and c < d. We also use the notation ∂R to
denote the boundary of the rectangle R, which is a closed curve.
THEOREM 4.2.12 (Cauchy’s theorem for rectangle)
If f is analytic on a region � and R is a rectangle contained in �, then∫
∂R
f (z) dz = 0.
�
�
“book” — 2014/6/4 — 21:03 — page 188 — #24
�
�
�
�
�
�
188 Winding Number and Cauchy’s Theorems
Proof: Let I(R), D(R), and L(R) denote
∫
∂R
f (z) dz, the diameter of R, and
the perimeter of R, respectively. If R is subdivided into four equal rectangles
S1, S2,S3, and S4 of same size, then we have
I(R) =
4∑
k=1
I(Sk) ⇒ |I(R)| ≤
4∑
k=1
|I(Sk)|.
This implies that there exists at least one k ∈ {1, 2, 3, 4} such that |I(Sk)| ≥
1
4
|I(R)|. Denote that rectangle Sk by R1. Next, we subdivide R1 into four
equal subrectangles of same size as before and choose the one say R2
such that |I(R2)| ≥ 1
4
|I(R1)|. Proceeding further, we get a sequence of
rectangles R ⊇ R1 ⊇ R2 ⊇ · · · such that
|I(Rn+1)| ≥ 1
4
|I(Rn)| ⇒ |I(Rn)| ≥ 1
4n
|I(R)|. (4.7)
L(Rn+1) = 1
2
L(Rn) ⇒ L(Rn) = 1
2n
L(R). (4.8)
D(Rn+1) = 1
2
D(Rn) ⇒ D(Rn) = 1
2n
D(R). (4.9)
S4
S3 = R1
S1 S2
R2
R3
Since {Rn} is a decreasing sequence of compact sets with D(Rn) → 0 as
n → ∞, by Cantor’s intersection theorem (Theorem 2.4.33),
∞∩
n=1
Rn = {z0}, for some z0 ∈ R. (4.10)
As f is analytic at z0, given ε > 0, there exists δ > 0 such that
0 < |z − z0| < δ ⇒
∣∣∣∣ f (z) − f (z0)
z − z0
− f ′(z0)
∣∣∣∣ < ε
D(R)L(R)
.
�
�
“book” — 2014/6/4 — 21:03 — page 189 — #25
�
�
�
�
�
�
Complex Integration 189
Therefore, we have
0 < |z − z0| < δ ⇒ ∣∣ f (z) − f (z0) − (z − z0)f ′(z0)
∣∣ < ε
D(R)L(R)
|z − z0|.
Using equations (4.9) and (4.10), we choose m ∈ N such that D(Rm) < δ, and
hence, Rm ⊆ B(z0, δ). If f1(z) = 1 and f2(z) = z, ∀z ∈ C, then f1 and f2 are
the derivatives of the analytic functions F1 and F2 defined by F1(z) = z and
F2(z) = z2
2
, ∀z ∈ C, respectively. Therefore, by Corollary 4.1.29, we have∫
∂Rm
dz = 0 and
∫
∂Rm
z dz = 0. Now,
|I(R)| ≤ 4m|I(Rm)| (by equation (4.7))
= 4m
∣∣∣∣∣∣∣
∫
∂Rm
f (z) dz
∣∣∣∣∣∣∣
= 4m
∣∣∣∣∣∣∣
∫
∂Rm
[f (z) − f (z0) − (z − z0)f ′(z0)] dz
∣∣∣∣∣∣∣
≤ 4m
∫
∂Rm
| f (z) − f (z0) − (z − z0)f ′(z0)| |dz|
< 4m ε
D(R)L(R)
∫
∂Rm
|z − z0| |dz|
< 4m ε
D(R)L(R)
D(Rm)
∫
∂Rm
|dz|
= 4m ε
D(R)L(R)
D(Rm)L(Rm) (by Theorem 4.1.27)
= 4m ε
D(R)L(R)
4−mD(R)L(R) (by equations (4.8) and (4.9))
= ε.
As ε > 0 is arbitrary, we get |I(R)| = 0. �
THEOREM 4.2.13 Let � be a region, F ⊆ � be a finite set and let f be an
analytic function on �\F. If f is bounded in B(a, r)\{a}, for some ra > 0, for
every a ∈ F, then for every rectangle R such that ∂R ⊂ �\F,
∫
∂R
f (z) dz = 0.
�
�
“book” — 2014/6/4 — 21:03 — page 190 — #26
�
�
�
�
�
�
190 Winding Number and Cauchy’s Theorems
Proof: Let F∩R = {ak : k = 1, 2 . . . , m}. Then, ak ∈ Int R, ∀k = 1, 2, . . . , m.
By assumption, there exists rk > 0 such that B(ak , rk) ⊆ Int R, and
0 < |z − ak | < rk ⇒ | f (z)| ≤ Mk , for some Mk > 0, ∀1 ≤ k ≤ m.
For each k = 1, 2, . . . , m, we choose δk ∈ R such that 0 < δk <
min
{
ε
mMk
, rk
}
.
Find a subrectangle Sk of R such that ak ∈ Int Sk , and its circumference
L(Sk) of Sk is less than δk , ∀1 ≤ k ≤ m. If D(Sk) is the diameter of Sk , then
we get
z ∈ ∂Sk ⇒ z ∈ Sk ⇒ |z − ak | ≤ D(Sk) ≤ L(Sk) < δk . (4.11)
Now, extend all the line segments that are used to construct the rectangles Sk
up to the boundary of R so that the original rectangle R is subdivided into
finite number of rectangles.
Hence, by a similar argument used in the proof of the previous theorem, we
get ∫
∂R
f (z) dz =
m∑
k=1
∫
∂Sk
f (z) dz +
p∑
j=1
∫
∂Tj
f (z) dz,
where Tj’s are subrectangles of R not containing any ak . Applying the previ-
ous theorem, we get
∫
∂Tj
f (z) dz = 0, ∀1 ≤ j ≤ p. Now, for every 1 ≤ k ≤ m,
invoking equation(4.11), we get∣∣∣∣∣ ∫
∂Sk
f (z) dz
∣∣∣∣∣ ≤ ∫
∂Sk
| f (z)| |dz| ≤ ∫
∂Sk
| f (z)| |dz|
≤ ∫
∂Sk
Mk |dz| = MkL(Sk)
< Mkδk < Mk
ε
mMk= ε
m .
�
�
“book” — 2014/6/4 — 21:03 — page 191 — #27
�
�
�
�
�
�
Complex Integration 191
Therefore, ∣∣∣∣∣∣
∫
∂R
f (z) dz
∣∣∣∣∣∣ ≤
m∑
k=1
∣∣∣∣∣∣∣
∫
∂Sk
f (z) dz
∣∣∣∣∣∣∣ <
m∑
k=1
ε
m
= ε.
Thus,
∫
∂R
f (z) dz = 0. �
THEOREM 4.2.14 (Cauchy’s theorem for a simply connected region)
Let F be a finite2 subset of a simply connected region � and f be an analytic
function on �\F. If f is bounded in B(a, r)\{a}, for some ra > 0, for every
a ∈ F, then for every closed curve γ ⊂ �\F, we get
∫
γ
f (z) dz = 0.
Proof: First we show that there exists an analytic function � on � such that
�′ = f . This part of this theorem is almost similar to that of the proof of
Result 4.1.30. Let (a, b) ∈ �\F be arbitrarily fixed. For every (x0, y0) ∈ �\F,
let γ(x0,y0) ⊆ �\F be a polygon joining (a, b) and (x0, y0) consisting of the
line segments parallel to the coordinate axes and the line segment incident
with (x0, y0), which is horizontal (such a polygon exists by Theorem 1.4.31).
Furthermore, as F is a finite set, we can choose the polygon such that it does
not pass through any a ∈ F. Define �(x0, y0) = ∫
γ(x0,y0)
f (z) dz. First, we show
that the definition of �(x0, y0) is independent of the choice of the polygon
γ(x0,y0).
(a,b) (x
0
,y
0
)
If σ(x0,y0) is another polygon joining (a, b) and (x0, y0), then γ(x0,y0) ∪
(−σ(x0,y0)) is a finite union of boundary of rectangles. As � is simply con-
nected, all rectangles enclosed by γ(x0,y0) ∪ (−σ(x0,y0)) are contained in �.
2Note that empty set is a finite set, and hence, this theorem is also true for the case F = ∅.
�
�
“book” — 2014/6/4 — 21:03 — page 192 — #28
�
�
�
�
�
�
192 Winding Number and Cauchy’s Theorems
Hence, by Theorem 4.2.13, we get∫
γ(x0,y0)∪(−σ(x0,y0))
f (z) dz = 0 ⇒
∫
γ(x0,y0)
f (z) dz =
∫
σ(x0,y0)
f (z) dz.
Thus, � is well defined. Let ε > 0 be given. Using the continuity of f at
(x0, y0), choose δ > 0 such that
(x, y) ∈ B((x0, y0), δ) ⊂ �\F ⇒ | f (x, y) − f (x0, y0)| < ε.
Let h ∈ R with 0 < |h| < δ. Then, the line segment L(x0,y0),(x0+h,y0) ⊆ � and
is horizontal. Therefore, for |h| < δ, we have
�(x0 + h, y0) =
∫
γ(x0,y0)∪L(x0,y0),(x0+h,y0)
f (z) dz,
and hence, ∣∣∣∣�(x0 + h, y0) −�(x0, y0)
h
− f (x0, y0)
∣∣∣∣
=
∣∣∣∣∣∣∣
1
h
∫
L(x0,y0),(x0+h,y0)
f (z) dz − f (x0, y0)
∣∣∣∣∣∣∣
=
∣∣∣∣∣∣
1
h
h∫
0
f (x0 + t, y0) dt − f (x0, y0)
∣∣∣∣∣∣
≤ 1
|h|
h∫
0
| f (x0 + t, y0) − f (x0, y0)| |dt|
<
1
|h| |h|ε = ε.
Thus,
∂�
∂x
(x0, y0) = f (x0, y0).
Similarly, we can prove that
∂�
∂y
(x0, y0) = if (x0, y0). Furthermore, we get
∂�
∂x
(x0, y0) = −i
∂�
∂y
(x0, y0) = f (x0, y0) (the C–R equation), and
∂�
∂x
,
∂�
∂y
are continuous at (x0, y0). Therefore, by Theorem 2.2.28, we get that � is
�
�
“book” — 2014/6/4 — 21:03 — page 193 — #29
�
�
�
�
�
�
Complex Integration 193
differentiable at (x0, y0). As (x0, y0) ∈ �\F is arbitrary, we get that � is
analytic on � and �′ = ∂�
∂x
= f . Therefore, using Corollary 4.1.29, we get∫
γ
f (z) dz = 0, for all closed curve γ in �. �
THEOREM 4.2.15 (Cauchy–Goursat theorem)
Let γ be a simple closed curve and f be analytic inside and on γ , then∫
γ
f (z) dz = 0.
Proof: As f is analytic inside and on γ , for every point z on or inside γ , there
exists rz > 0 such that f is analytic on B(z, rz). If we put � as the union of
all these open balls B(z, rz) along with the region enclosed by γ , then � is a
simply connected region, γ ⊂ �, and f is analytic on �. Hence, by Cauchy’s
theorem for simply connected region, we get
∫
γ
f (z) dz = 0. �
4.3 CAUCHY’S INTEGRAL FORMULA
THEOREM 4.3.1 (Cauchy’s integral formula)
Let f be an analytic function on a simply connected region �. Then,
WN(γ , z0)f (z0) = 1
i2π
∫
γ
f (z)
z − z0
dz, for every closed curve γ in �\{z0}.
Proof: Let F(z) = f (z) − f (z0)
z − z0
, ∀z ∈ �\{z0}. Then, F is analytic on�\{z0}.
Since
lim
z→z0
F(z) = lim
z→z0
f (z) − f (z0)
z − z0
= f ′(z0),
there exists δ > 0 such that
0 < |z − z0| < δ ⇒
∣∣∣∣ f (z) − f (z0)
z − z0
− f ′(z0)
∣∣∣∣ < 1.
Hence, for every z ∈ B(z0, r)\{z0}, we have
| f (z)| =
∣∣∣∣ f (z) − f (z0)
z − z0
∣∣∣∣ ≤ 1 + | f ′(z0)| < +∞.
�
�
“book” — 2014/6/4 — 21:03 — page 194 — #30
�
�
�
�
�
�
194 Cauchy’s Integral Formula
Therefore, by Cauchy’s theorem for simply connected region, we get∫
γ
F(z) dz = 0. Hence, from the definition of WN(γ , z0), it follows that
1
i2π
∫
γ
f (z)
z − z0
dz − WN(γ , z0)f (z0) = 1
i2π
∫
γ
f (z) − f (z0)
z − z0
dz
= 1
i2π
∫
γ
F(z) dz = 0.
Hence, the theorem follows. �
THEOREM 4.3.2 Let γ be a curve in C and ϕ be a continuousfunction on γ .
If
Fn(z) =
∫
γ
φ(ζ )
(ζ − z)n
dζ ,
for every z � γ and n ∈ N, then Fn is differentiable on the complement of γ
and F′
n = nFn+1, ∀n ∈ N.
Proof: Let z0 � γ be arbitrary. First, we show that if
Gm,n(z) =
∫
γ
φ(ζ )
(ζ − z)m(ζ − z0)n
,∀z � γ and ∀m, n ∈ N,
then lim
z→z0
Gm,n(z) = Gm,n(z0). We choose δ > 0 such that B(z0, δ) ∩ γ = ∅.
Then, we have |ζ − z| ≥ δ
2
and |ζ − z0| ≥ δ, ∀z ∈ B
(
z0,
δ
2
)
and ∀ζ ∈ γ . If
M = ∫
γ
|φ(ζ )| |dζ |, then M <∞ and
|Gm,n(z) − Gm,n(z0)|
≤
∫
γ
∣∣∣∣ 1
(ζ − z)m(ζ − z0)n
− 1
(ζ − z0)m+n
∣∣∣∣ |φ(ζ )| |dζ |
=
∫
γ
1
|ζ − z0|n
∣∣∣∣ 1
(ζ − z)
− 1
(ζ − z0)
∣∣∣∣
×
∣∣∣∣∣
m−1∑
k=0
1
(ζ − z)m−1−k
1
(ζ − z0)k
∣∣∣∣∣ |φ(ζ )| |dζ |
�
�
“book” — 2014/6/4 — 21:03 — page 195 — #31
�
�
�
�
�
�
Complex Integration 195
(by using am − bm = (a − b)
m−1∑
k=0
am−1−kbk)
≤ |z − z0|
δn
m−1∑
k=0
∫
γ
|φ(ζ )| |dζ |
|ζ − z|m−k |ζ − z0|k+1
≤ |z − z0|
δn
m−1∑
k=0
∫
γ
2m−k
δm−kδk+1
|φ(ζ )| |dζ |
≤
M
m−1∑
k=0
2m−k
δm+n+1
|z − z0| → 0 as z → z0.
For z ∈ B
(
z0, δ2
)
, we have
Fn(z) − Fn(z0)
z − z0
= 1
z − z0
∫
γ
(
1
(ζ − z)n
− 1
(ζ − z0)n
)
φ(ζ ) dζ
= 1
z − z0
∫
γ
(
1
(ζ − z)
− 1
(ζ − z0)
) n−1∑
k=0
1
(ζ − z)n−1−k
1
(ζ − z0)k
φ(ζ ) dζ
= 1
z − z0
∫
γ
(
z − z0
(ζ − z)(ζ − z0)
) n−1∑
k=0
1
(ζ − z)n−1−k
1
(ζ − z0)k
φ(ζ ) dζ
=
n−1∑
k=0
∫
γ
1
(ζ − z)n−k
1
(ζ − z0)k+1
φ(ζ ) dζ
=
n−1∑
k=0
Gn−k,k+1(z) →
n−1∑
k=0
Gn−k,k+1(z0) =
n−1∑
k=0
∫
γ
φ(ζ )
(ζ − z0)n+1
dζ
= nFn+1(z0), as z → z0.
This completes the proof of the theorem. �
�
�
“book” — 2014/6/4 — 21:03 — page 196 — #32
�
�
�
�
�
�
196 Cauchy’s Integral Formula
COROLLARY 4.3.3 (Cauchy’s integral formula for derivatives)
Let f be an analytic function on a simply connected region � and z0 ∈ �.
Then for every n ∈ N, f (n) exists and
WN(γ , z0)f (n)(z0) = n!
i2π
∫
γ
f (z)
(z − z0)n+1
dz,
for every closed curve not passing through z0 in �, and for every n ∈ N.
Proof: Cauchy’s integral formula states that
WN(γ , z0)f (z0) = 1
i2π
∫
γ
f (ζ )
ζ − z0
dζ ,
where γ is the circle |z − z0| = r, where r > 0 is such that Cl B(z0, r) ⊆ �.
As f is continuous on γ , by previous theorem, we get that
1
i2π
∫
γ
f (ζ )
ζ − z
dζ is
analytic in B(z0, r), and for every z ∈ B(z0, r),
d
dz
⎛
⎝ 1
i2π
∫
γ
f (ζ )
ζ − z
dζ
⎞
⎠ = 1
i2π
∫
γ
f (ζ )
(ζ − z)2
dζ
d2
dz2
⎛
⎝ 1
i2π
∫
γ
f (ζ )
(ζ − z)
dζ
⎞
⎠ = 1 × 2
i2π
∫
γ
f (ζ )
(ζ − z)3
dζ
d3
dz3
⎛
⎝ 1
i2π
∫
γ
f (ζ )
(ζ − z)
dζ
⎞
⎠ = 1 × 2 × 3
i2π
∫
γ
f (ζ )
(ζ − z)4
dζ
...
dn
dzn
⎛
⎝ 1
i2π
∫
γ
f (ζ )
(ζ − z)
dζ
⎞
⎠ = n!
i2π
∫
γ
f (ζ )
(ζ − z)n+1
dζ .
Therefore, f (n)(z0) = n!
i2π
∫
γ
f (z)
(z − z0)n+1
dz, ∀n ∈ N. �
�
�
“book” — 2014/6/4 — 21:03 — page 197 — #33
�
�
�
�
�
�
Complex Integration 197
THEOREM 4.3.4 (Weierstrass theorem on convergence of sequence of
analytic functions)
If ( fn) is a sequence of analytic functions on a region � such that ( fn) con-
verges to f uniformly on every compact subset of �, then f is analytic and
( f ′n) converges to f ′ uniformly on every compact subset of �.
Proof: Let z ∈ � be arbitrary. Then, choose r > 0 such that Cl B(z, r) ⊆ �.
Then, by Cauchy’s integral formula, we have
fn(w) = 1
i2π
∫
|ζ |=r
fn(ζ )
ζ − w
dζ , ∀w ∈ B(z, r) ∀n ∈ N.
By assumption, we have fn → f uniformly on |ζ | = r as n → ∞. For all
w ∈ B(z, r), we have
f (w) = 1
i2π
∫
|ζ |=r
f (ζ )
ζ − w
dζ .
Since fn → f as n → ∞ uniformly on |ζ | = r and each fn is continuous,
we have f is continuous on |ζ | = r, by Theorem 1.6.31. Hence, by Theorem
4.3.2, we obtain that f is analytic on B(z, r). As z ∈ � is arbitrary, f is analytic
on �. Again by using ( fn) converges uniformly to f on |ζ | = r, given ε > 0,
there exists N ∈ N such that | fn(ζ ) − f (ζ )| < ε, ∀ζ with |ζ | = r, whenever
n ≥ N . By Cauchy’s integral formula for derivatives, we have
f ′n(w) = 1
i2π
∫
|ζ |=r
fn(ζ )
(ζ − w)2
dζ , ∀w ∈ B(z, r)
and
f ′(w) = 1
i2π
∫
|ζ |=r
f (ζ )
(ζ − w)2
dζ , ∀w ∈ B(z, r),
For all w ∈ Cl B
(
z,
r
2
)
, and for all n ≥ N ,
∣∣ f ′n(w) − f ′(w)
∣∣ ≤ 1
2π
∫
|ζ |=r
| fn(ζ ) − f (ζ )|
|ζ − w|2 |dζ | ≤ 1
2π
∫
|ζ |=r
ε
r2
4
|dζ | = 4ε
r
.
Therefore, for every z ∈ �, there exists ρz > 0 such that f ′n → f ′ uniformly
on Cl B(z, ρz). Let K be a compact subset of �. Then, {B(z, ρz) : z ∈ K} is
�
�
“book” — 2014/6/4 — 21:03 — page 198 — #34
�
�
�
�
�
�
198 Cauchy’s Integral Formula
a collection of open sets with K ⊆ ∪
z∈K
B(z, ρz). Since K is compact, there
exist z1, z2, . . . , zm ∈ K such that K ⊆ m∪
j=1
B(zj, ρzj ) ⊆
m∪
j=1
Cl B(zj, ρzj ). Since
f ′n → f ′ uniformly on Cl B(zj, ρzj ) as n → ∞, for a given ε > 0, there exists
Nj ∈ N such that
| f ′n(w) − f ′(w)| < ε, ∀w ∈ Cl B(zj, ρzj ), ∀n ≥ Nj, ∀1 ≤ j ≤ m.
If N0 = max{Nj : 1 ≤ j ≤ m}, then | f ′n(w) − f ′(w)| < ε, ∀w ∈ K, ∀n ≥ N0.
Thus, ( f ′n) converges to f ′ uniformly on every compact subset of �. �
The following algorithm is useful to evaluate integrals over a given simple
closed curve γ .
Algorithm 4.3.5 (To find
∫
γ
φ(z) dz by Cauchy’s integral formula)
Step 1: Let the given function be φ and find its poles. (If φ is a rational
function, the zeroes of the denominator of φ are the poles of φ.)
Step 2: Check which of the poles are lying outside γ .
Step 3: (a) If there is no pole lying inside γ , then
∫
γ
φ(z) dz = 0.
(b) If there is only one pole z0, which is enclosed by γ , then put
f (z) = (z − z0)mφ(z),
where m is the order of the pole z0 of φ. Then, we find∫
γ
φ(z) dz = ∫
γ
f (z)
(z − z0)m
dz using Cauchy’s integral formula as
follows.∫
γ
φ(z) dz =
∫
γ
f (z)
(z − z0)m
dz = i2π
(m − 1)!
f (m−1)(z0).
(c) If there are more than one pole, say z1, z2, . . . , zn, enclosed by γ ,
then put
f (z) = (z − z1)m1 (z − z2)m2 · · · (z − zn)mnφ(z),
where m1, m2, . . . , mn are the orders of the poles z1, z2, . . . , zn,
respectively.
�
�
“book” — 2014/6/4 — 21:03 — page 199 — #35
�
�
�
�
�
�
Complex Integration 199
(d) Split 1
(z−z1)m1 (z−z2)m2 ···(z−zn)mn into partial fractions as follows.
m1∑
k=1
A1,k
(z − z1)k
+
m2∑
k=1
A2,k
(z − z2)k
+ · · · +
mn∑
k=1
An,k
(z − zn)k
.
(Note that in the field of complex numbers, there is no non-
splittable factor. For example,
1
z2 + a2
should be written as
A
z + ia
+ B
z − ia
. Hence, in a partial fraction expansion, a general
term should be of the form
C
(z − a)p
.)
(e) Then write
∫
γ
φ(z) dz as
n∑
j=1
mj∑
k=1
Aj,k
∫
γ
f (z)
(z−zj)k dz.
(f) Doing Step 3(b) repeatedly, we can find the given integral.
Example 4.3.6 Evaluate the following integrals using Cauchy’s integral
formula:
(1)
∫
|z|=3
z3 + 1
z2 + 3z − 10
dz. (6)
∫
|z|=5
3z + 1
(z − 1)(z − 2)2
dz.
(2)
∫
|z−3|=1
z2 + 1
z − 1
dz. (7)
∫
|z−(1−i)|=3
exp(z)(3z2 + 4)
(z − 1)3
dz.
(3)
∫
γ
z4 − 4z − 6
z2 − 6z + 5
dz, where γ is (8)
∫
|z+1|=2
z2 + 5z + 6
(z2 + 1)(z − 1 + i)
dz.
the square with vertices (3, 3),
(−3, 3), (−3,−3), and (3,−3).
(9)
∫
|z|=4
exp(z)
z
(
z − π
2
)
(z − π) dz.
(4)
∫
|z|=5
2z + 3
z2 − 2z − 3
dz. (10)
∫
γ
z + i
(z + 4)(z−1−i)
dz, where γ
(5)
∫
|z−1|=4
z2 − 3
(z2 + 3z + 2)(z + 6)
dz. is the boundary of [0, 2]×[0, 3].
Solution:
(1) Let φ(z) = z3 + 1
z2 + 3z − 10
. As z2 + 3z − 10 = 0 ⇒ z = −5, z = 2, the
poles of φ are −5 and 2. We know that 2 is enclosed by |z| = 3, and
−5 is not enclosed by |z| = 3 because |2| = 2 < 3 and | − 5| = 5 > 3.
�
�
“book” — 2014/6/4 — 21:03 — page 200 — #36
�
�
�
�
�
�
200 Cauchy’s Integral Formula
We put f (z) = z3 + 1
z + 5
. Therefore,
∫
|z|=3
z3 + 1
z2 + 3z − 10
dz =
∫
|z|=3
f (z)
z − 2
dz = i2π f (2) = i2π
23 + 1
2 + 5
= 18iπ
7
.
(2) Let φ(z) = z2 + 1
z − 1
. Then, clearly 1 is the only pole of φ and is not
enclosed by |z − 3| = 1 (because |1 − 3| = 2 > 1.) Therefore,∫
|z−3|=1
z2 + 1
z − 1
dz = 0.
(3) Let φ(z) = z4 − 4z − 6
z2 − 6z + 5
. Since z2 − 6z + 5 = 0 ⇒ z = 5, z = 1,
the poles of φ are 5 and 1. Obviously, 1 is enclosed by γ and 5 is
not enclosed by γ (check geometrically).We put f (z) = z4 − 4z − 6
z − 5
.
Therefore,∫
γ
z4 − 4z − 6
z2 − 6z + 5
dz =
∫
γ
f (z)
z − 5
dz = i2π f (1) = i2π
1 − 4 − 6
1 − 5
= 9iπ
2
.
(4) Let φ(z) = 2z + 3
z2 − 2z − 3
. Now, z2 − 2z − 3 = 0 implies that z = −1,
and z = 3 are the poles of φ, and both are enclosed by |z| = 5, because
|− 1| = 1 < 5 and |3| = 3 < 5. Therefore, we put f (z) = 2z+ 3. Now,
we split
1
(z + 1)(z − 3)
into partial fractions as follows.
Let
1
(z + 1)(z − 3)
= A
z + 1
+ B
z − 3
⇒ 1 = A(z − 3) + B(z + 1).
Putting z = 3, we get 4B = 1 ⇒ B = 1
4
.
Putting z = −1, we get −4A = 1 ⇒ A = −1
4
.
Hence,
1
(z + 1)(z − 3)
= 1
4
(
1
z − 3
− 1
z + 1
)
. Therefore,
∫
|z|=5
2z + 3
z2 − 2z − 3
dz =
∫
|z|=5
f (z)
(z + 1)(z − 3)
dz
�
�
“book” — 2014/6/4 — 21:03 — page 201 — #37
�
�
�
�
�
�
Complex Integration 201
= 1
4
⎛
⎜⎝ ∫
|z|=5
f (z)
z − 3
dz −
∫
|z|=5
f (z)
z + 1
dz
⎞
⎟⎠
= i2π
4
( f (3) − f (−1)) = iπ
2
(9 − 1) = 4iπ .
(5) Let φ(z) = z2 − 3
(z2 + 3z + 2)(z + 6)
. As (z2+3z+2)(z+6) = 0, it follows
that z = −1, z = −2, and z = −6 are the poles of φ. Furthermore,
−1 and −2 are enclosed by |z − 1| = 4, and −6 is not enclosed by
|z − 1| = 4 because | − 1 − 1| = 2 < 4, | − 2 − 1| = 3 < 4, and
| − 6 − 1| = 7 > 4. Hence, we take f (z) = z2 − 3
z + 6
. Now, we split
1
(z + 1)(z + 2)
into partial fractions as follows.
1
(z + 1)(z + 2)
= A
z + 1
+ B
z + 2
⇒ 1 = A(z + 2) + B(z + 1).
Putting z = −2, we get −B = 1 ⇒ B = −1.
Putting z = −1, we get A = 1.
Hence,
1
(z + 1)(z + 2)
= 1
z + 1
− 1
z + 2
. Therefore,
∫
|z−1|=4
z2 − 3
(z2 + 3z + 2)(z + 6)
dz =
∫
|z−1|=4
f (z)
(z + 1)(z + 3)
dz
= −
∫
|z−1|=4
f (z)
z + 2
dz +
∫
|z−1|=4
f (z)
z + 1
dz
= −i2π ( f (−2) − f (−1))
= −i2π
(
(−2)2 − 3
−2 + 6
− (−1)2 − 3
−1 + 6
)
= − iπ
2
(
1
4
− 2
5
)
= −3iπ
40
.
(6) If φ(z) = 3z + 1
(z − 1)(z − 2)2
, then z = 1 and 2 are the poles of φ, and
both are enclosed by |z| = 5. Let f (z) = 3z + 1. Now, we split
�
�
“book” — 2014/6/4 — 21:03 — page 202 — #38
�
�
�
�
�
�
202 Cauchy’s Integral Formula
1
(z − 1)(z − 2)2
into partial fractions.
1
(z − 1)(z − 2)2
= A
z − 1
+ B
z − 2
+ C
(z − 2)2
⇒ 1 = A(z−2)2+B(z−1)(z−2)+C(z−1). Putting z = 1, z = 2, and
z = 0, we get, respectively, A = 1, C = 1, 1 = 4A+2B−C ⇒ B = −1.
Therefore,
1
(z − 1)(z − 2)2
= 1
z − 1
− 1
z − 2
+ 1
(z − 2)2
. Hence,
∫
|z|=5
3z + 1
(z − 1)(z − 2)2
dz =
∫
|z|=5
f (z)
(z − 1)(z − 2)2
dz
=
∫
|z|=5
f (z)
z − 1
dz −
∫
|z|=5
f (z)
z − 2
+
∫
|z|=5
f (z)
(z − 2)2
dz
= i2π
(
f (1) − f (2) + f ′(2)
)
= i2π (4 − 7 + 3) = 0.
(7) Let φ(z) = exp(z)(3z2 + 4)
(z − 1)3
. Then, 1 is the only pole of order 3 for φ,
which is enclosed by |z − (1 + i)| = 3, as |1 − (1 − i)| = 1 < 3. If
f (z) = exp(z)(3z2 + 4), then
f ′(z) = exp(z)(3z2 + 6z + 4),
f ′′(z) = exp(z)(3z2 + 12z + 10) ⇒ f ′′(1) = 25e.
Therefore,
∫
|z−(1+i)=3|
exp(z)(3z2 + 4)
(z − 1)3
dz = i2π
2!
f ′′(1) = 25π exp(1)i.
(8) Let φ(z) = z2 + 5z + 6
(z2 + 1)(z − 1 + i)
. Then, i,−i, and 1 − i are the poles
for φ, among which i and −i are enclosed by |z − 1| = 2, as
|i+1| = √
2 < 2, |− i+1| = √
2 < 2, and |1− i+1| = √
5 > 2. Let
f (z) = z2 + 5z + 6
z − 1 + i
. Now, we split
1
(z − i)(z + i)
into partial fractions
as follows.
1
(z − i)(z + i)
= A
z − i
+ B
z + i
⇒ 1 = A(z + i) + B(z − i).
�
�
“book” — 2014/6/4 — 21:03 — page 203 — #39
�
�
�
�
�
�
Complex Integration 203
Putting z = i and z = −i we get, respectively, A = 1
i2
and B = − 1
i2
.
Therefore,
∫
|z−1|=2
z2 + 5z + 6
(z2 + 1)(z − 1 + i)
dz
= 1
i2
⎛
⎜⎝ ∫
|z−1|=2
f (z)
z − i
dz −
∫
|z−1|=2
f (z)
z + i
dz
⎞
⎟⎠
= π ( f (i) − f (−i))
= π
(
i2 + i5 + 6
i − 1 + i
− i2 − i5 + 6
−i − 1 + i
)
= π
(
i5 + 5
i2 − 1
− 5 − i5
−1
)
= π (1 − i3 + 5 − i5) = π (6 − i8).
(9) Let φ(z) = exp(z)
z(z − π/2)(z − π )
. Then, 0,π , and π/2 are the poles for
φ, and all are enclosed by |z − 1| = 2, as |0| < 5, |π/2| < 5, and
|π | < 5. Let f (z) = exp(z). Now, we split
1
z(z − π/2)(z − π )
into
partial fractions as follows.
1
z(z − π/2)(z − π )
= A
z
+ B
z − π/2 + C
z − π ⇒ 1 =
A(z − π/2)(z − π ) + Bz(z − π ) + Cz(z − π/2).
Substituting z = 0, z = π/2, and z = π , we get, respec-
tively, A = 2
π2
, B = − 4
π2
, and C = 2
π2
. Therefore, we have
1
z(z − π/2)(z − π )
= 2
π2
(
1
z
− 2
z − π/2 + 1
z − π
)
. Hence,
∫
|z|=5
exp(z)
z(z − π/2)(z − π )
dz
�
�
“book” — 2014/6/4 — 21:03 — page 204 — #40
�
�
�
�
�
�
204 Cauchy’s Integral Formula
= 2
π2
⎛
⎜⎝ ∫
|z|=5
f (z)
z
dz − 2
∫
|z|=5
f (z)
z − π/2 dz +
∫
|z|=5
f (z)
z − π dz
⎞
⎟⎠
= i2π
2
π2
(f (0) − 2f (π/2) + f (π ))
= i4
π
(1 − 2 exp(π/2) + exp(π )).
(10) Let φ(z) = z + i
(z + 4)(z − 1 − i)
. Then, −4 and 1 + i are the poles for φ.
Since the given curve γ is the boundary of the rectangle with vertices
(0, 3), (2, 0), (2, 3), and (0, 3), 1 + i is enclosed by γ , and −4 is not
enclosed by γ . Therefore, let f (z) = z + i
z + 4
. Hence,
∫
γ
z + i
(z + 4)(z − 1 − i)
dz =
∫
γ
f (z)
z − 1 − i
dz
= i2π f (1+i) = i2π
1 + i2
5 + i
= π
13
(−9 + i7).
Exercise 4.3.7 Evaluate the following integrals using Cauchy’s integral
formula:
(1)
∫
|z|=2
exp(z)
(z + 1)(z − 3)2
dz.
(2)
∫
|z+i2|=2
dz
z2 + 1
dz.
(3)
∫
|z|=2
sin(z)
z + i
dz.
(4)
∫
|z|=1
exp(z)
(z − 2)3
dz.
�
�
“book” — 2014/6/4 — 21:03 — page 205 — #41
�
�
�
�
�
�
Complex Integration 205
(5)
∫
|z+ 1
2 |=1
z2 exp(z)
(z + 1)(z − 2)
dz.
(6)
∫
|z−1|=2
z + 1
z2 − 9z + 20
dz.
(7)
∫
|z|=3
z exp(z)
z − 1
dz.
(8)
∫
|z−(2+i)|=2
z
(z − (1 + i))
dz.
(9)
∫
|z|=4
exp(z)
(z + 1)(z − 3)2
dz.
(10)
∫
|z|=3
z2 + 1
(z2 − 1)(z − 2)
dz.
(11)
∫
|z|=2
z
(z − 1)2
dz.
(12)
∫
|z−i2|=4
2z + 3
(z − (1 − i))2
dz.
(13)
∫
|z|=4
ez
(z2 + 1)3
dz.
(14)
∫
|z−1|=4
z + i
(z − 1)2(z − 2)(z − 6)
dz.
(15)
∫
|z−2|=3
exp(z + 1)
(z + 6)4
dz.
�
�
“book” — 2014/6/4 — 21:03 — page 206 — #42
�
�
�
�
�
�
206 Cauchy’s Integral Formula
Answers: (1)
iπ exp(−1)
8
; (2) −π ; (3) 2π sinh(1); (4) 0; (5)
i2π exp(−1)
−3
;
(6) 0; (7) i2π exp(1); (8) 2π (−1 + i); (9)
iπ
8
(exp(−1) + 3 exp(3)); (10) i2π ;
(11) i2π ; (12) 4iπ ; (13)
π
4
(2 sin(1) − 3 cos(1); (14)
π
50
(1 − 6i); (15) 0
RESULT 4.3.8 (Cauchy’s estimate)
If f is analytic on a closed disk Cl B(a, r), then for each n ∈ N, and 0 < r < ε,
we have | f (n)(a)| ≤ Mrn!
rn
, where Mr = sup
|z|=r
| f (z)|.
Proof: Let γ be the circle |z − a| = r. By Cauchy’s integral formula for
derivatives, we get
∣∣∣ f (n)(a)
∣∣∣ =
∣∣∣∣∣∣
n!
i2π
∫
γ
f (z)
(z − a)n+1
dz
∣∣∣∣∣∣
≤ n!
2π
∫
γ
∣∣∣∣ f (z)
(z − a)n+1
∣∣∣∣ dz
≤ n!
2π
∫
γ
∣∣∣∣ Mr
rn+1
∣∣∣∣ dz
≤ Mrn!
2πrn+1
2πr = Mrn!
rn
.
Hence, the result follows. �
THEOREM 4.3.9 (Morera’s theorem)
Let f be a continuous function on � such that
∫
γ
f (z) dz = 0 for every closed
curve γ in �, then f is analytic on �.
Proof: As
∫
γ
f (z) dz = 0 for every closed curve γ in �, using Result 4.1.30,
there exists an analytic function F on� such that F′ = f . By Corollary 4.3.3,
F′ is analytic on �, and hence, f is analytic on �. �
Definition 4.3.10 An analytic function on C is called an entire function.
�
�
“book” — 2014/6/4 — 21:03 — page 207 — #43
�
�
�
�
�
�
Complex Integration 207
Clearly, every polynomial, exp, sin, and cos are entire functions. Obviously,
every constant function is a bounded entire function but the interesting fact
is that the converse of this statement is also true, which follows.
THEOREM 4.3.11 (Liouville’s theorem)
Every bounded entire function is constant.
Proof: Let f be a bounded entire function such that sup
z∈C
| f (z)| ≤ M for some
M > 0. For an arbitrary z ∈ C and for an arbitrary r > 0, from Cauchy’s
estimate, we get
| f ′(a)| ≤ Mr
r
, where Mr = sup
|z|=r
| f (z)|.
As Mr ≤ M , ∀r > 0, we have
| f ′(a)| ≤ M
r
→ 0 as r → ∞.
Thus, f ′(z) = 0, ∀z ∈ C. Hence, by Theorem 2.2.19, we get that f is a constant
function. �
THEOREM 4.3.12 (Fundamental theorem of algebra)
Every polynomial of degree n withcomplex coefficients has exactly n zeroes
in C including multiplicities.
Proof: First, we prove that every non-constant polynomial has a zero in C.
Let P be a non-constant polynomial. If P has no zero in C, then
1
P
is an entire
function, as P is an entire function (See Example 2.1.9). Now, we claim that
1
P
is bounded on C. If P(z) =
n∑
k=0
akzk , with an � 0, then for |z| ≥ 1, we have
|P(z)| ≥ ∣∣anzn
∣∣−
∣∣∣∣∣
n−1∑
k=0
akzk
∣∣∣∣∣
≥ ∣∣anzn
∣∣− n−1∑
k=0
∣∣∣akzk
∣∣∣
≥ ∣∣anzn
∣∣− |z|n−1
n−1∑
k=0
|ak |
�
�
“book” — 2014/6/4 — 21:03 — page 208 — #44
�
�
�
�
�
�
208 Cauchy’s Integral Formula
(since |z|k ≤ |z|n−1 if k ≤ n − 1)
≥ |z|n−1
[
|anz| −
n−1∑
k=0
|ak |
]
≥ |anz| −
n−1∑
k=0
|ak |
≥ 1 if |z| ≥ 1
|an|
[
1 +
n−1∑
k=0
|ak |
]
.
As
∣∣∣∣ 1
P
∣∣∣∣ is a continuous real-valued function on the compact set
{
z ∈ C : |z| ≤ max
{
1,
1
|an|
[
1 +
n−1∑
k=0
|ak |
]}}
,
there exists M > 0 such that
∣∣∣∣ 1
P(z)
∣∣∣∣ ≤ M on the compact set. Therefore,
∣∣∣∣ 1
P
∣∣∣∣ ≤ M + 1 <∞ on C,
and hence,
1
P
is a bounded entire function. Therefore, by Liouville’s theorem
(Theorem 4.3.11), we get that
1
P
is a constant function. This implies that P is
a constant function. This is a contradiction. Therefore, P has a zero in C. If
α1 is a zero of P, then using Lemma 3.1.3, we can write
P(z) = (z − α1)Q1(z), ∀z ∈ C.
Certainly, degree of Q1 is n − 1. If n − 1 > 0, then Q1 has a zero say α2, and
hence, we can write
P(z) = (z − α1)(z − α2)Q2(z), ∀z ∈ C.
Proceeding further, at the nth stage, we get
P(z) = (z − α1)(z − α2) · · · (z − αn)Qn(z), ∀z ∈ C,
where α1,α2, . . . ,αn are the zeroes of P, and Qn is a polynomial of degree 0,
which means that Qn is a constant. Therefore, P has exactly n zeroes in C. �
�
�
“book” — 2014/6/4 — 21:03 — page 209 — #45
�
�
�
�
�
�
Complex Integration 209
4.4 GENERAL VERSION OF CAUCHY’S THEOREM
THEOREM 4.4.1 (General version of Cauchy’s theorem)
If f is analytic on � and γ is a closed curve in � such that γ ∼ 0 in �, then∫
γ
f (z) dz = 0.
Proof: As γ is closed and bounded subset of C, it is compact. Since �c is
closed and �c ∩ γ = ∅, if
r = d(γ ,�c) = inf{|z − w| : z ∈ γ , w ∈ �c},
then r > 0, by Theorem 1.4.41. If z = φ(t), t ∈ [a, b] is the parametric
equation of γ , then φ is a continuous function on the compact set [a, b], and
hence, it is uniformly continuous on [a, b], by Theorem 1.6.26. Then, there
exists δ > 0 such that
t, s ∈ [a, b] with |t − s| < δ ⇒ |z(t) − z(s)| < r.
If {t0, t1, t2, . . . , tn} ⊆ [a, b] such that a = t0 < t1 < t2 < . . . < tn = b and
ti − ti−1 < δ, then let γk = φ([tk−1, tk]). Then, clearly,
γ = n∪
k=1
γk and γk ⊂ B(φ(tk), r), 1 ≤ k ≤ n.
Moreover, by the choice of r, we have B(φ(tk), r), which does not intersect�c,
as φ(tk) ∈ γ ⊂ �. Since φ(tk),φ(tk−1) ∈ γk ⊆ B(φ(tk), r), and B(φ(tk), r) is a
convex set,3 we can connect φ(tk) and φ(tk−1) by a polygon λk ⊆ B(φ(tk), r),
which consists of line segments parallel to coordinate axes. Hence, γk∪(−λk)
is a closed curve contained in the simply connected region B(φ(tk), r), ∀1 ≤
k ≤ n.
3A set C is said to be a convex set if for every pair of points of C, the line segment joining
the two points is contained in C.
�
�
“book” — 2014/6/4 — 21:03 — page 210 — #46
�
�
�
�
�
�
210 General Version of Cauchy’s Theorem
r
g
lf(tk)
f(tk−1)
Hence, by Cauchy’s theorem for simply connected region, we have∫
γk∪(−λk )
f (z) dz = 0 ⇒
∫
γk
f (z) dz =
∫
λk
f (z) dz, 1 ≤ k ≤ n.
Therefore, if λ = n∪
k=1
λk , then
∫
γ
f (z)dz =
∫
n∪
k=1
γk
f (z) dz =
∫
n∪
k=1
λk
f (z) dz =
∫
λ
f (z) dz. (4.12)
Now, we subdivide the entire complex plane into finite number of bounded
rectangles Bp, 1 ≤ p ≤ μ, and finite number of unbounded rectangles Uq,
1 ≤ q ≤ ν, by extending each line segment involved in the construction of λ
into straight lines. Now, for each p ∈ {1, 2, . . . ,μ} and q ∈ {1, 2, . . . , ν}, we
fix one point bp ∈ Int Bq and uq ∈ Int Uq. If we denote
μ∪
p=1
WN(λ, bp)∂Bp by
χ , then we claim that after removing the line segments, which are common
sides of two rectangles in χ , we get χ = λ. First, we note that for 1 ≤ j ≤ μ
and 1 ≤ l ≤ ν, we have
WN(χ , bj) = WN
(
μ∪
p=1
WN(λ, bp)∂Bp, bj
)
=
μ∑
p=1
WN(λ, bp)WN(∂Bp, bj)
�
�
“book” — 2014/6/4 — 21:03 — page 211 — #47
�
�
�
�
�
�
Complex Integration 211
= WN(λ, bj), since WN(∂Bp, bj) =
{
1 if p = j
0 if p � j
, (4.13)
and
WN(χ , ul) = WN
(
μ∪
p=1
WN(λ, bp)∂Bp, ul
)
=
μ∑
p=1
WN(λ, bp)WN(∂Bp, ul)
= 0, since WN(∂Bp, ul) = 0. (4.14)
Next, we claim that the multiplicity m of any line segment Lp, j in λ ∪ (−χ ),
which is common to two bounded rectangles Bp and Bj, is zero for some
1 ≤ p, j ≤ μ. Since λ ∪ (−χ ) ∪ (−m∂Bp) does not have the line Lp, j, the
points bp and bj belong to the same region determined by λ∪(−χ )∪(−m∂Bp).
Therefore,
WN(λ ∪ (−χ ) ∪ (−m∂Bp), bp) = WN(λ ∪ (−χ ) ∪ (−m∂Bp), bj).
Hence, by using equation(4.13), we get
WN(λ, bp) − WN(λ, bj) − m = WN(λ, bp) − WN(λ, bj) − 0 ⇒ m = 0.
Similarly, if Lp,q is the common side of a bounded rectangle Bp and an
unbounded rectangle Uq, then arguing as before by applying equation(4.14),
we get the multiplicity of the line segment Lp,q in λ ∪ (−χ ) as zero. Hence,
our claim holds.
By Cauchy’s theorem for rectangle, we have
∫
∂Bp
f (z) dz = 0, provided
Bp ⊆ �. If Bp � �, then there exists a ∈ Int Bp and a � �. Since γ is
homologous to zero in�, we get WN(γ , a) = 0. As γ ∪(−λ) does not enclose
a, we get
WN(γ ∪ (−λ), a) = 0 ⇒ WN(λ, a) = WN(γ , a) = 0.
Since a and bp belong to the same region determined by λ, we get
WN(λ, bp) = WN(λ, a) = 0.
This implies that for each p = 1, 2, . . . ,μ,
either
∫
∂Bp
f (z) dz = 0 or WN(λ, bp) = 0.
�
�
“book” — 2014/6/4 — 21:03 — page 212 — #48
�
�
�
�
�
�
212 Local Correspondence Theorem and Its Consequences
Hence, it follows that
∫
λ
f (z) dz = ∫
χ
f (z) dz =
μ∑
p=1
WN(λ, bp)
∫
∂Bp
f (z) dz = 0.
�
Definition 4.4.2 A cycle � is a finite union of closed curves
m∪
j=1
γj.
We define
∫
�
f (z) dz =
m∑
j=1
∫
γj
f (z) dz, so that we get WN(�, a) =
m∑
j=1
WN(γj, a).
Then, the general version of Cauchy’s theorem can be further extended to the
following form easily.
THEOREM 4.4.3 If f is analytic on a region � and � is a cycle, which is
homologous to zero in �, then
∫
�
f (z) dz = 0.
4.5 LOCAL CORRESPONDENCE THEOREM
AND ITS CONSEQUENCES
In this section, we shall use the following result, which will be proved in
Chapter 5. See Result 5.2.3.
RESULT 4.5.1 If f is an analytic function on a region�with a zero of order k
at z0 ∈ �, then there exists an analytic function g on � such that g is analytic
on �, f (z) = (z − z0)kg(z) on � and g(z0) � 0.
THEOREM 4.5.2 Let f be a non-zero analytic function on a region � with
zeroes z1, z2, . . . , zn, including multiplicities. If γ is a closed curve in � such
that γ ∼ 0 in � and it is not passing through any zj, then
1
i2π
∫
γ
f ′(z)
f (z)
dz =
n∑
j=1
WN(γ , zj).
Proof: Let �1 ⊆ � be a region such that γ and each region determined by
γ are contained in �1. We also assume that the only zeroes of f in �1 are
{zj : j = 1, 2, . . . , n}. Since zj’s are the zeroes of f in �, by Result 4.5.1, we
can write
f (z) = (z − z1)(z − z2) · · · (z − zn)g(z), ∀z ∈ �1,
�
�
“book” — 2014/6/4 — 21:03 — page 213 — #49
�
�
�
�
�
�
Complex Integration 213
where g is analytic on �1 and is nowhere zero on �1. Hence,
f ′(z)
f (z)
= 1
z − z1
+ 1
z − z2
+ · · · + 1
z − zn
+ g′(z)
g(z)
, ∀z on γ .
We note that g′
g is analytic on �1 because g and g′ are analytic and g is
nowhere zero on �1. Therefore,
1
i2π
∫
γ
f ′(z)
f (z)
dz =
n∑
j=1
1
i2π
∫
γ
dz
z − zj
+ 1
i2π
∫
γ
g′(z)
g(z)
dz =
n∑
j=1
WN(γ , zj),
since we have
∫
γ
g′(z)
g(z)
dz = 0, by general version of Cauchy’s theorem.
Note that if we take γ as a simple closed curve in the above theorem, then
1
i2π
∫
γ
f ′(z)
f (z)
dz counts the number of zeroes of f inside γ . �
THEOREM 4.5.3 (Local correspondence theorem)
Let f be a non-constantanalytic function on a region �, z0 ∈ � and f (z0) =
w0. For each w ∈ f (�), if gw(z) = f (z)−w0, ∀z ∈ � and the order of gw0 at z0
is n, then there exists r > 0 and ρ > 0 such that for each w ∈ B(w0, ρ)\{w0},
there exists exactly n distinct roots for gw in B(z0, r)\{z0}.
Proof: Since f is non-constant, gw0 is not identically zero, and hence, by the
principle of analytic continuation, it follows that z0 is isolated. Therefore,
there exists r > 0 such that
1. Cl B(z0, r) ⊆ �.
2. gw0 � 0 on B(z0, r)\{z0}, since z0 is an isolated zero of gw0 .
3. f ′ � 0 on B(z0, r)\{z0}. (This is possible because of the following rea-
son. Since f is non-constant, f ′ is not identically 0. If f ′(z0) = 0, then
z0 must be an isolated zero of f ′. Then, there exists r > 0. If f ′(z0) � 0,
using the continuity of f ′, we can find such an r > 0.)
Let γ be the circle |z − z0| = r. Since z0 � γ , we have w0 = f (z0) � f (γ ).
Hence, we can choose ρ > 0 such that B(w0, ρ) ∩ f (γ ) = ∅. Then, for each
w ∈ B(w0, ρ), we get that w and w0 lie in a same region determined by f (γ ).
Therefore, from a property of winding number (Thereom 4.2.5), it follows
that
WN( f (γ ), w) = WN( f (γ ), w0). (4.15)
�
�
“book” — 2014/6/4 — 21:03 — page 214 — #50
�
�
�
�
�
�
214 Local Correspondence Theorem and Its Consequences
For each w ∈ B(w0, ρ), if nw denotes the number of zeroes of gw inside γ
including the multiplicities and if Jw = {1, 2, . . . , nw}, then using Theorem
4.5.2 and using the change of variable ζ = f (z), we get
∑
j∈Jw
WN(γ , zj) = 1
i2π
∫
γ
g′w(z)
gw(z)
dz
= 1
i2π
∫
γ
f ′(z)
f (z) − w
dz
= 1
i2π
∫
f (γ )
dζ
ζ − w
= WN( f (γ ), w).
Since the only one zero of gw0 inside γ is z0 and its multiplicity is n, we have∑
j∈Jw0
WN(γ , zj) = n. Therefore, from equation (4.15), we get
n =
∑
j∈Jw0
WN(γ , zj) = WN( f (γ ), w0) = WN( f (γ ), w) =
∑
j∈Jw
WN(γ , zj).
In other words, the number of zeroes of gw inside B(z0, r) is n. Furthermore,
g′w = f ′ � 0 on B(z0, r)\{z0} implies that every zero of gw in B(z0, r) \ {z0}
is simple, and hence, for each w ∈ B(w0, ρ) \ {w0}, gw has exactly n distinct
roots inside B(z0, r)\{z0}. �
THEOREM 4.5.4 (Open mapping theorem)
Every non-constant analytic function on a region is an open mapping.
Proof: Let f be a non-constant analytic function on a region�. To prove that
f is an open mapping, we have to show that if U is open in �, then f (U)
is an open subset of C. Let U be an open subset of � and let w0 ∈ f (U).
Then, there exists z0 ∈ U such that w0 = f (z0). Choose ε > 0 such that
B(z0, ε) ⊆ U . By local correspondence theorem, there exists 0 < r < ε and
ρ > 0 such that for every w ∈ B(w0, ρ), there exists z ∈ B(z0, r) such that
f (z) = w. Therefore, we have proved that
B(w0, ρ) ⊆ f (B(z0, r)) ⊆ f (B(z0, ε)) ⊆ f (U).
Thus, f (U) is an open set. �
�
�
“book” — 2014/6/4 — 21:03 — page 215 — #51
�
�
�
�
�
�
Complex Integration 215
COROLLARY 4.5.5 If f : �1 → �2 is an analytic bijection, then f −1 : �2 →
�1 is also an analytic function and (f −1)′(w0) = 1
f ′(z0)
, where z0 = f −1(w0),
∀w0 ∈ �2.
Proof: First, we claim that f ′ is nowhere zero on �1. Suppose f ′(z0) = 0
for some z0 ∈ �1, then f (z) − f (z0) has at least two zeroes at z0. Therefore,
by local correspondence theorem, there exist r > 0 and ρ > 0 such that for
each w ∈ B(w0, ρ) \ {w0}, there exist at least two distinct zeroes for f (z) − w
from B(z0, r), which is a contradiction to the injectivity of f . Therefore, our
claim holds. If g is the inverse of f , then by open mapping theorem, g is a
continuous map. In fact, g is analytic; indeed, if w = f (z) and f (z0) = w0,
then w → w0 whenever z → z0, and hence,
lim
w→w0
g(w) − g(w0)
w − w0
= lim
z→z0
z − z0
f (z) − f (z0)
= lim
z→z0
1
f (z) − f (z0)
z − z0
= 1
f ′(z0)
.
Thus, the theorem follows. �
THEOREM 4.5.6 (Maximum modulus principle)
If f is a non-constant analytic function on a region �, then | f (z)| has no
maximum in �.
Proof: To prove this theorem, we show that for each z0 ∈ �, there exists
z1 ∈ � such that | f (z1)| > | f (z0)|. Since f is a non-constant analytic function
on�, then by open mapping theorem, f is an open map. Therefore, f (�) itself
is an open subset of C. If w0 = f (z0), then w0 ∈ f (�), and hence, there exists
ε > 0 such that B(w0, ε) ⊆ f (�). If |w0| < r < ε and if
w1 =
{
r exp(iarg w0) if w0 � 0
r if w0 = 0
,
then w1 ∈ B(w0, ε) ⊆ f (�) and |w1| > |w0|. Thus, there exists z1 ∈ � such
that f (z1) = w1, and hence, | f (z1)| = |w1| > |w0| = | f (z0)|. �
COROLLARY 4.5.7 [Minimum modulus principle]
Let f be a non-constant analytic function on a region �. If f is nowhere zero
on �, then |f | has no minimum in �.
�
�
“book” — 2014/6/4 — 21:03 — page 216 — #52
�
�
�
�
�
�
216 Local Correspondence Theorem and Its Consequences
Proof: Since f is nowhere zero on �,
1
f
is also an analytic function on �.
Hence, by maximum modulus principle,
∣∣∣∣1
f
∣∣∣∣ has no maximum in �. That is,
for every z ∈ �, there exists z1 ∈ � such that
∣∣∣∣ 1
f (z)
∣∣∣∣ <
∣∣∣∣ 1
f (z1)
∣∣∣∣, equivalently,
|f (z1)| < |f (z)|. Thus, |f | has no minimum on �. �
RESULT 4.5.8 (Schwarz lemma)
If f : B(0, 1) → Cl B(0, 1) is an analytic function such that f (0) = 0, then
1. | f (z)| ≤ |z| on B(0, 1) and | f ′(0)| ≤ 1.
2. If | f (z0)| = |z0| for some z0 � 0 or if | f ′(0)| = 1, then f (z) = exp(iθ )z,
on B(0, 1) for some θ ∈ R.
Proof: If g(z) = f (z)
z
for z � 0, then g is analytic on B(0, 1)\{0}, and 0 is a
removable singularity for g since
lim
z→0
g(z) = lim
z→0
f (z) − f (0)
z − 0
= f ′(0)
exists in C. Hence, if we define g(0) = f ′(0), then g becomes analytic on
B(0, 1), by Lemma 5.1.2. This implies that g is analytic on Cl B(0, r) for every
0 < r < 1. Since |g| is a continuous function on the compact set Cl B(0, r),
there exists zr with |zr| ≤ 1 such that |g(zr)| = sup
z∈Cl B(0,r)
|g(z)|. If g is a
constant function on B(0, 1), then we can choose this zr with |zr| = 1, and if
g is non-constant, then by maximum modulus principle, |g| does not attain its
maximum on |z| < r, and hence, |zr| = r. Since 0 < |zr| < 1, we have
sup
z∈Cl B(0,r)
|g(z)| = |g(zr)| =
∣∣∣∣ f (zr)
zr
∣∣∣∣ = | f (zr)|
r
≤ 1
r
,
and hence, by allowing r → 1, we get sup
|z|≤1
|g(z)| ≤ 1. Hence, by definition
of g, we get
∣∣∣ f (z)
z
∣∣∣ ≤ 1 on B(0, 1)\{0} and | f ′(0)| ≤ 1. Thus, | f (z)| ≤ |z| on
B(0, 1) and | f ′(0)| ≤ 1.
If | f (z0)| = |z0| for some z0 ∈ B(0, 1)\{0} or | f ′(0)| = 1, then we get
|g(ζ0)| = 1 for some ζ0(= z0 or 0) in B(0, 1). Hence, by maximum modulus
principle, we conclude that g is a constant say c. Since |g(ζ0)| = 1, we obtain
�
�
“book” — 2014/6/4 — 21:03 — page 217 — #53
�
�
�
�
�
�
Complex Integration 217
|c| = 1. Therefore, using Corollary 2.4.25, there exists θ ∈ R such that
c = exp(iθ ), and hence, f (z) = g(z) · z = cz, ∀z in B(0, 1). �
RESULT 4.5.9 (Generalized Schwarz lemma)
Let f be analytic on |z| < R such that | f (z)| ≤ S, ∀|z| < R. If |a| < R and
f (a) = b, then
∣∣∣∣S( f (z) − b)
S2 − bf (z)
∣∣∣∣ ≤
∣∣∣∣R(z − a)
R2 − az
∣∣∣∣.
Proof: Consider the linear fractional transform
w = T(z) = R(z − a)
R2 − az
, ∀w ∈ C∞.
By Example 3.2.26, we get T(B(0, R)) = B(0, 1).
Similarly, if ξ = �(ζ ) = S(ζ − b)
S2 − bζ
, ∀ζ ∈ C∞, then we have�(B(0, S)) =
B(0, 1). If F = (� ◦ f ◦ T−1), then F : B(0, 1) → B(0, 1) is analytic and
F(0) = �( f (T−1(0))) = �( f (a)) = �(b) = 0.
Hence, by Schwarz lemma, we get |F(w)| ≤ |w|, ∀w ∈ B(0, 1). If z ∈ B(0, R),
then |T(z)| < 1, and hence,
|F(T(z))| ≤ |T(z)| ⇒ |�( f (z))| ≤ |T(z)| ⇒
∣∣∣∣S( f (z) − b)
S2 − bf (z)
∣∣∣∣ ≤
∣∣∣∣R(z − a)
R2 − az
∣∣∣∣ .
Thus, the result follows. �
Now, we can see some applications of Schwarz lemma, which are used to
prove the uniqueness of Riemann mapping between a region and open unit
disc, in Result 7.4.5.
COROLLARY 4.5.10 If f : B(0, 1) → B(0, 1) is an analytic bijection, satis-
fying f (0) = 0, then f (z) = cz, ∀z ∈ B(0, 1) for some c ∈ C with |c| = 1.
Proof: Let f : B(0, 1) → B(0, 1) is the givenbijection. Then, by Corol-
lary 4.5.5, we get that f −1 : B(0, 1) → B(0, 1) is an analytic function and
(f −1)′(0) = 1
f ′(0)
. Applying Schwarz lemma to f and f −1, we have
|f ′(0)| ≤ 1 and
∣∣∣∣ 1
f ′(0)
∣∣∣∣ = |(f −1)′(0)| ≤ 1.
�
�
“book” — 2014/6/4 — 21:03 — page 218 — #54
�
�
�
�
�
�
218 Local Correspondence Theorem and Its Consequences
The second inequality implies that |f ′(0)| ≥ 1. Therefore, |f ′(0)| = 1. Then
using the condition for equality in Schwarz lemma, we get f (z) = cz, ∀z ∈ C
for some c ∈ C. �
COROLLARY 4.5.11 If f : B(0, 1) → B(0, 1) is an analytic bijection, then
f (z) = c
z − α
1 − αz
, ∀z ∈ B(0, 1) for some c ∈ C with |c| = 1, where α =
f −1(0) ∈ B(0, 1).
Proof: If Tα(z) = z − α
1 − αz
, ∀z ∈ C, then we claim that Tα : B(0, 1) → B(0, 1)
is an analytic bijection such that Tα(α) = 0 and T−1
α is also analytic.
If |z| = 1, then let z = exp(iθ ) for some θ ∈ R so that
|Tα(z)| =
∣∣∣∣ exp(iθ ) − α
1 − α exp(iθ )
∣∣∣∣ =
∣∣∣∣ exp(iθ ) − α
exp(iθ )(exp(−iθ ) − α)
∣∣∣∣ =
∣∣∣∣ exp(iθ ) − α
(exp(−iθ ) − α)
∣∣∣∣ = 1.
As every linear fractional transform maps circles to circles, if C is the
unit circle, then Tα(C) ⊆ C, which implies that Tα(C) = C. Since Tα
has a unique singularity 1
α
, which is not in Cl B(0, 1), Tα is an analytic
function on B(0, 1), and hence, Tα(B(0, 1)) is a connected subset of {z ∈
C : |z| � 1}. Therefore, either Tα(B(0, 1)) = B(0, 1) or Tα(B(0, 1)) =
{z ∈ C : |z| > 1}.
Since Tα(α) = 0 (by definition) and α ∈ B(0, 1), we get Tα(B(0, 1)) =
B(0, 1). As every linear fractional transform is a one-to-one map, Tα :
B(0, 1) → B(0, 1) is a bijection, satisfying Tα(α) = 0, and by Corollary
4.5.5, T−1
α is also an analytic map on B(0, 1) satisfying T−1
α (0) = α.
If F = f ◦ T−1
α , then F : B(0, 1) → B(0, 1) is an analytic bijection and
F(0) = (f ◦ T−1
α )(0) = f (α) = 0. Therefore, by the previous corollary, we
have F(w) = cw, ∀w ∈ C, for some c ∈ C with |c| = 1. Thus, for an arbitrary
z ∈ B(0, 1), Tα(z) ∈ B(0, 1) so that f (z) = f (T−1
α (Tα(z))) = F(Tα(z)) =
cTα(z) = c
z − α
1 − αz
. �
THEOREM 4.5.12 Let f be an analytic function on a simply connected region
�. If f is nowhere zero, then log( f ) can be suitably defined on � so that it is
analytic on �.
Proof: Using Theorem 4.2.8, we get that every closed curve γ in � is
homologous to zero in �. Since f is analytic and is nowhere zero, using
�
�
“book” — 2014/6/4 — 21:03 — page 219 — #55
�
�
�
�
�
�
Complex Integration 219
Theorem 4.5.5, we have
1
i2π
∫
γ
f ′(ζ )
f (ζ )
dζ = 0,
for every closed curve γ is �. Then by Result 4.1.30, there exists an analytic
function g on �, such that g′ = f ′
f
. If log( f ) is defined, then its derivative
must be
f ′
f
. Hence, we expect that log( f ) would be defined by g+ c for some
suitable constant c. We find c from
log( f ) = g + c ⇔ exp c = f exp(−g).
To conclude this proof, first we show that f exp(−g) is a constant. This can
be obtained since
d
dz
( f exp(−g)) = f ′ exp(−g) − f exp(−g)g′
= f ′ exp(−g) − f exp(−g)
f ′
f
= 0
and by using Theorem 2.2.19. Hence, let f exp(−g) = exp(c) for some con-
stant c, and this is possible since f exp(−g) is nowhere zero. Then, for a fixed
z0 ∈ �,
f (z0) exp(−g(z0)) = exp(c) ⇒ c = log( f (z0)) − g(z0),
where log( f (z0)) is any one of the infinite number of its values. Thus, if log( f )
is defined by g + log( f (z0)) − g(z0), then log( f ) is differentiable. �
COROLLARY 4.5.13 If f and� are as in the previous theorem, then for every
n ∈ N, there exists an analytic function g on � such that gn = f . In other
words, we can define n
√
f as an analytic function on �.
Proof: Using the previous theorem, we can define log( f ) as an analytic func-
tion on �. If we define g(z) = exp
(
1
n log( f (z)
)
, ∀z ∈ �, then clearly g is an
analytic function on � and
gn(z) =
[
exp
(
1
n
log( f (z) )
)]n
= exp (log( f (z)) = f (z), ∀z ∈ �,
and hence, g is the required analytic function such that g = n
√
f . �
�
�
“book” — 2014/6/4 — 21:03 — page 220 — #56
�
�
�
�
�
�
220 Local Correspondence Theorem and Its Consequences
THEOREM 4.5.14 (Hurwitz theorem)
Let ( fn) be a sequence of analytic functions such that fn is nowhere zero for
all n ∈ N. If fn → f uniformly on every compact subset of � as n → ∞,
then either f is identically zero or f is nowhere zero.
Proof: Assume that f is not identically zero. Let z ∈ � be arbitrary. Then,
we claim that there exists δ > 0 such that f � 0 on Cl B(z, δ)\{z}.
If f (z) � 0, then using the continuity of f at z, for ε = | f (z)|
2
, there exists
r > 0 such that
| f (w) − f (z)| < | f (z)|
2
whenever |w − z| < r.
If 0 < δ < r and w ∈ Cl B(z, δ), then
| f (z)| − | f (w)| ≤ | f (w) − f (z)| < | f (z)|
2
⇒ | f (w)| > | f (z)|
2
> 0.
Thus, our claim holds, in this case.
If f (z) = 0, then by Theorem 4.5.1, there exist m ∈ N and an analytic function
g on � such that
f (w) = (w − z)mg(w), ∀w ∈ � and g(z) � 0.
Then by the previous argument, there exists δ > 0 such that g � 0 on
Cl B(z, δ). This implies that f � 0 on 0 < |w − z| ≤ δ. Thus, our claim
holds.
Since fn → f uniformly on every compact subset of � as n → ∞, then
by Theorem 4.3.4, we get f ′n → f ′ uniformly on every compact subset of �
as n → ∞. By a similar technique employed in Theorem 1.3.9, we obtain
f ′n
fn
→ f ′
f
uniformly on |ζ − z| = δ, as n → ∞.
Indeed, since |f | is real, continuous, and nowhere zero on |ζ − z| = δ,
there exists μ > 0 such that μ = inf{|f (ζ )| : |ζ − z| = δ}. Let M1 and M2 be
positive real numbers such that
M1 > sup{|f (ζ )| : |ζ − z| = δ} and M2 > sup{|f ′(ζ )| : |ζ − z| = δ}.
Then, for a given ε > 0, there exist N1 and N2 ∈ N such that
|fn(ζ ) − f (ζ )| < min
{ |μ|
2
,
ε|μ|2
4M2
}
, ∀n ≥ N1
�
�
“book” — 2014/6/4 — 21:03 — page 221 — #57
�
�
�
�
�
�
Complex Integration 221
and
|f ′n(ζ ) − f ′(ζ )| < ε|μ|2
4M1
, ∀n ≥ N2.
Therefore, for n ≥ N1, we have |f | − |fn| ≤ |fn − f | < |μ|
2 , and hence,
|μ|
2
= |μ| − |μ|
2
≤ |f | − |μ|
2
< |fn|.
For every ζ ∈ C with |ζ − z| = δ and for every n ≥ max{N1, N2},∣∣∣ f ′n(ζ )
fn(ζ ) − f ′(ζ )
f (ζ )
∣∣∣
=
∣∣∣ f ′n(ζ )f (ζ )−fn(ζ )f ′(ζ )
fn(ζ )f (ζ )
∣∣∣
≤ 1
|fn(ζ )f (ζ )|
[∣∣f ′n(ζ )f (ζ ) − f (ζ )f ′(ζ )
∣∣+ ∣∣f (ζ )f ′(ζ ) − fn(ζ )f ′(ζ )
∣∣]
≤ 2
|μ|2
[
M1
∣∣f ′n(ζ ) − f ′(ζ )
∣∣+ M2 |f (ζ ) − fn(ζ )|]
≤ 2
|μ|2
[
M1
ε|μ|2
4M1
+ M2
ε|μ|2
4M2
]
= ε
2 + ε
2 = ε.
Therefore, by Theorem 4.5.2, we get that the number of zeroes of f in B(z, δ)
is equal to
1
i2π
∫
|ζ−z|=δ
f ′(ζ )
f (ζ )
dζ = lim
n→∞
1
i2π
∫
|ζ−z|=δ
f ′n(ζ )
fn(ζ )
dζ
= 1
i2π
∫
|ζ−z|=δ
(
lim
n→∞
f ′n(ζ )
fn(ζ )
)
dζ
= 0,
since each fn has no zeroes. Therefore, f � 0 on B(z, δ). In particular, f (z) � 0.
Using that z ∈ � is arbitrary, we get that f is nowhere zero on �. �
This page is intentionally left blank
�
�
“book” — 2014/6/4 — 21:09 — page 223 — #1
�
�
�
�
�
�
5
Series Developments
and Infinite Products
5.1 TAYLOR SERIES AND LAURENT SERIES
Definition 5.1.1 Let f be an analytic function on �\{a}. Then, the point a is
said to be a removable singularity of f if lim
z→a
f (z) exists in C.
LEMMA 5.1.2 Let f be an analytic function on �\{a} for some a ∈ �. Then,
the point a is a removable singularity of f if and only if there exists an analytic
function g on � such that f (z) = g(z), ∀z ∈ �\{a}.
Proof: Assume that there exists an analytic function g on � such that
f (z) = g(z), ∀z ∈ �\{a}. Then, g is continuous at a. Therefore, lim
z→a
f (z) =
lim
z→a
g(z) = g(a), which exists in C.
Conversely, assume that a is a removable singularity. We choose r > 0
such that B(a, r) is contained in �. Then, for each z ∈ B(a, r)\{a}, we define
gz(ζ ) = f (ζ ) − f (z)
ζ − z
, ∀ζ ∈ �\{a, z}.
Then, gz is analytic on �\{a, z},
lim
ζ→z
gz(ζ ) = lim
ζ→z
f (ζ ) − f (z)
ζ − z
= f ′(z)
and
lim
ζ→a
gz(ζ ) = lim
ζ→a
f (ζ ) − f (z)
ζ − z
=
lim
ζ→a
f (ζ ) − f (z)
a − z
.
223
�
�
“book” — 2014/6/4 — 21:09 — page 224 — #2
�
�
�
�
�
�
224 Taylor Series and LaurentSeries
Therefore, gz is bounded on some neighbourhoods of a and z. Using Cauchy’s
theorem for open disk with exceptional points, if 0 < s < r and 0 < |z−a| <
s, then we get∫
|ζ−a|=s
gz(ζ ) dζ = 0 ⇒ 1
i2π
∫
|ζ−a|=s
f (ζ )
ζ − z
dζ = f (z)
i2π
∫
|ζ−a|=s
dζ
ζ − z
= f (z).
Since f is continuous on |z − a| = s, if f1(z) = 1
i2π
∫
|ζ−a|=s
f (ζ )
ζ − z
dζ , then f1
is analytic on B(a, r) (including at a) and f1(z) = f (z),∀z ∈ B(a, r). Hence, if
g : �→ C is defined by
g(z) =
{
f (z) if z � a
f1(a) if z = a
,
then g is analytic on � and g(z) = f (z), ∀z ∈ �\{a}. �
THEOREM 5.1.3 (Taylor’s theorem)
Let f be an analytic function on � and z0 ∈ �. Then, for every n ∈ N, there
exists an analytic function φn on � such that
f (z) =
n−1∑
k=1
f (k)(z0)
k!
(z − z0)k + φn(z)(z − z0)n, ∀z ∈ �.
Furthermore,
φn(z) = 1
i2π
∫
C
f (ζ )
(ζ − z)(ζ − z0)n
, dζ , ∀z inside C,
where C is a circle |z − z0| = r contained in �.
Proof: If we define
F(z) = f (z) − f (z0)
z − z0
, ∀z ∈ �\{z0},
then F is analytic on �\{z0} and z0 is a removable singularity for F, as
lim
z→z0
F(z) = lim
z→z0
f (z) − f (z0)
z − z0
= f ′(z0).
�
�
“book” — 2014/6/4 — 21:09 — page 225 — #3
�
�
�
�
�
�
Series Developments and Infinite Products 225
Then, by Lemma 5.1.2, there exists an analytic function φ1 on � such that
φ1(z) = F(z), ∀z ∈ �\{z0}. Hence, for every z ∈ �\{z0}, clearly we have
f (z)−f (z0) = φ1(z)(z−z0) and f (z0)−f (z0) = 0 = φ1(z0)(z0−z0). Therefore,
f (z) = f (z0) + φ1(z)(z − z0), ∀z ∈ �.
Applying the same technique, we can write φ1(z) = φ1(z0) + φ2(z)(z − z0)
for some analytic function φ2 on �. Proceeding like this, at the nth stage, we
get φn−1(z) = φn−1(z0) + φn(z)(z − z0) for some analytic function φn on �.
Therefore, for every z ∈ �,
f (z) = f (z0) + φ1(z)(z − z0)
= f (z0) + (φ1(z0) + φ2(z)(z − z0)) (z − z0)
= f (z0) + φ1(z0)(z − z0) + φ2(z)(z − z0)2
= f (z0) + φ1(z0)(z − z0) + φ2(z0)(z − z0)2 + φ3(z)(z − z0)3
...
= f (z0) +
n−1∑
k=1
φk(z0)(z − z0)k + φn(z)(z − z0)n.
Differentiating n times, we get
f (n)(z) =
n∑
j=0
nCj
dj
dzj
((z − z0)n)φ(n−j)
n (z), ∀z ∈ �.
For each j = 0, 1, 2, . . . , n − 1, dj
dz j ((z − z0)n) has the factor (z − z0), and
dn
dzn ((z − z0)n) = n!, we get f (n)(z0) = n!φn(z0). Hence, φn(z0) = f (n)(z0)
n!
,
∀n ∈ N. Therefore, we have
f (z) = f (z0) +
n−1∑
k=1
f (k)(z0)
k!
(z − z0)k + φn(z)(z − z0)n, ∀z ∈ �.
Next, we choose r > 0 such that ClB(z0, r) ⊆ �. If C is the circle with centre
z0 and radius r > 0, then by Cauchy’s integral formula, we have
φn(z) = 1
i2π
∫
C
φn(ζ )
ζ − z
dζ , ∀z ∈ B(z0, r).
�
�
“book” — 2014/6/4 — 21:09 — page 226 — #4
�
�
�
�
�
�
226 Taylor Series and Laurent Series
Since
φn(z) = 1
(z − z0)n
(
f (z) −
n−1∑
k=0
f (k)(z0)
k!
(z − z0)k
)
,
we have
φn(z) = 1
i2π
∫
C
[
f (ζ )
(ζ − z0)n(ζ − z)
−
n−1∑
k=0
f (k)(z0)
k!(ζ − z0)n−k(ζ − z)
]
dζ . (5.1)
Now, we claim that
∫
C
dζ
(ζ − z0)j(ζ − z)
= 0, ∀j ∈ N and ∀z ∈ B(z0, r). For
each j ∈ R and z ∈ B(z0, r), define
Fj,z(w) =
∫
C
dζ
(ζ − w)j(ζ − z)
, ∀w ∈ B(z0, r).
Applying partial fraction technique, we get
1
(ζ − w)(ζ − z)
= 1
z − w
(
1
ζ − z
− 1
ζ − w
)
,
and hence, using Example 4.2.6, for every w ∈ B(z0, r), we have
F1,z(w) =
∫
C
dζ
(ζ − w)(ζ − z)
= 1
z − w
∫
C
(
1
ζ − z
− 1
ζ − w
)
dζ
= i2π
z − w
(WN(C, z) − WN(C, w)) = 1 − 1 = 0.
Using Theorem 4.3.2, we get
Fj,z(w) =
F(1)
j−1,z(w)
j
=
F(2)
j−2,z(w)
j(j − 1)
=
F(3)
j−3,z(w)
j(j − 1)(j − 2)
= · · · = F(j−1)
1,z (w)
j!
= 0,
for all j > 1. Thus, Fj,z(w) = 0, ∀j > 0 and ∀w ∈ B(z0, r). In particular,∫
C
dζ
(ζ − z0)j(ζ − z)
= Fj,z(z0) = 0, ∀j = 1, 2, 3, . . . , n − 1.
�
�
“book” — 2014/6/4 — 21:09 — page 227 — #5
�
�
�
�
�
�
Series Developments and Infinite Products 227
Using this observation in equation (5.1), we get
φn(z) = 1
i2π
∫
C
f (ζ )
(ζ − z0)n(ζ − z)
dζ , ∀z ∈ B(z0, r).
Hence, the theorem follows. �
COROLLARY 5.1.4 Let P be a non-constant polynomial and z0 ∈ C. If m ∈ N
such that P(k)(z0) = 0, ∀k = 0, 1, 2 . . . , m − 1 and P(m)(z0) � 0, then m is the
order of the zero z0 for P.
Proof: By Taylor’s theorem, we have P(z) = (z − z0)mφ(z) for some analytic
function φ on C such that φ(z0) � 0. Then, φ(z) = P(z)
(z−z0)m , ∀z ∈ C, which is
a rational function and has no pole in C. Hence, using Lemma 3.1.16, we get
φ, which is a polynomial. Therefore, m is the order of the zero of P at z0. �
THEOREM 5.1.5 (Infinite Taylor series)
Let f be analytic in a region � and z0 ∈ �. Then, we have
f (z) =
∞∑
k=0
f (k)(z0)
k!
(z − z0)k , ∀z ∈ B(z0, r),
where r is a positive real number such that ClB(z0, r) ⊆ �.
Proof: For every n ∈ N, from Taylor theorem, we get
f (z) =
n−1∑
k=0
f (k)(z0)
k!
(z − z0)k + φn(z)(z − z0)n, ∀z ∈ �,
where
φn(z) = 1
i2π
∫
C
f (ζ )
(ζ − z)(ζ − z0)n
dζ , ∀z ∈ B(z0, r),
where C is the circle |ζ − z0| = r. If M = sup
ζ∈C
| f (ζ )|, then for z ∈ B(z0, r),
∣∣∣∣∣ f (z) −
n−1∑
k=0
f (k)(z0)
k!
(z − z0)k
∣∣∣∣∣ = |z − z0|n|φn(z)|
≤ |z − z0|n
2π
∫
C
| f (ζ )|
|ζ − z||ζ − z0|n |dζ |
�
�
“book” — 2014/6/4 — 21:09 — page 228 — #6
�
�
�
�
�
�
228 Taylor Series and Laurent Series
≤ |z − z0|n
2π
∫
C
M
rn(r − |z − z0|) |dζ |
≤ Mr
r − |z − z0|
( |z − z0|
r
)n
→ 0 as n→∞
(as
∫
C
|dζ | = 2πr).
Thus, f (z) = lim
n→∞
n−1∑
k=0
f (k)(z0)
k!
(z− z0)k =
∞∑
k=0
f (k)(z0)
k!
(z− z0)k , ∀z ∈ B(z0, r).
�
Remark 5.1.6: At this juncture, it should be noted that finite Taylor series
expansion f (z) =
n−1∑
k=0
f (k)(z0)
k!
(z− z0)k +φn(z)(z− z0)n of an analytic function
f on � is valid on the entire region �, but the infinite Taylor series expan-
sion
∞∑
k=0
f (k)(z0)
k!
(z − z0)k is valid only in a largest possible neighbourhood
B(z0, r) ⊆ �.
Definition 5.1.7 Let the power series
∞∑
n=0
an(z − a)n and
∞∑
n=0
bn(z − a)n con-
verge on B(a, R). Then, the Cauchy product of the two power series is defined
by
∞∑
n=0
(
n∑
k=0
akbn−k
)
(z − a)n, ∀z ∈ B(a, R).
THEOREM 5.1.8 If f (z) =
∞∑
n=0
an(z − a)n and g(z) =
∞∑
n=0
bn(z − a)n,∀z ∈
B(a, R), then ( f · g)(z) =
∞∑
n=0
(
n∑
k=0
akbn−k
)
(z − a)n, ∀z ∈ B(a, R).
Proof: We know that f and g are analytic on B(a, R), and hence, f · g is also
analytic on B(a, R). By Taylor’s theorem, we have
( f · g)(z) =
∞∑
n=0
( f · g)(n)(a)
n!
(z − a)n, ∀z ∈ B(a, R).
�
�
“book” — 2014/6/4 — 21:09 — page 229 — #7
�
�
�
�
�
�
Series Developments and Infinite Products 229
Now using Leibniz rule (Result 2.1.13) and Corollary 2.3.4, we have
( f · g)(n)(a)
n!
= 1
n!
n∑
k=0
nCkf (k)(a) · g(n−k)(a)
= 1
n!
n∑
k=0
nCkf (k)(a) · g(n−k)(a)
= 1
n!
n∑
k=0
n!
k!(n − k)!
(akk!) · (bn−k(n − k)!)
=
n∑
k=0
akbn−k .
Hence, for every z ∈ B(a, R), using Corollary 2.3.5, we get ( f · g)(z) =
∞∑
n=0
( f · g)(n)(a)
n!
(z − a)n. �
COROLLARY 5.1.9 If
∞∑
n=0
an(z− a)n and
∞∑
n=0
bn(z− a)n converges on B(a, R),
then the Cauchy product of the two power series converge on B(a, R).
The proof of this corollary follows immediately from the previous
theorem.
Definition 5.1.10 Let
∞∑
n=0
an(z − a)n and 0 �
∞∑
n=0
bn(z − a)n converge ∀z ∈
B(a, R). Then, the quotient
∞∑
n=0
dn(z − a)n =
∞∑
n=0
an(z − a)n
∞∑
n=0
bn(z − a)n
of the power series is defined as the Taylor series of
f
g
, where
f (z) =
∞∑
n=0
an(z − a)n and 0 � g(z) =
∞∑
n=0
bn(z − a)n, ∀z ∈ B(a, R).
�
�
“book” — 2014/6/4 — 21:09 — page 230 — #8
�
�
�
�
�
�
230 Taylor Series and Laurent Series
Finding the coefficients of the Taylor series of f
g in terms of an and bn are not
easy as that of f · g. However, the first few coefficients of the Taylor series of
f
g can be calculated successively.
In the following sequel, we use the notation ANs
r(a) to denote the annulus
{z ∈ C : r < |z − a| < s}.
Definition 5.1.11 We say that
∞∑
n=−∞
an is convergent if
∞∑
n=0
an and
∞∑
n=1
a−n
are convergent; in this case, we write
∞∑
n=−∞
an =
∞∑
n=1
a−n +
∞∑
n=0
an.
THEOREM 5.1.12 For a given
∞∑
n=−∞
an(z − z0)n, if r = lim sup
n→∞
|a−n|1/n, s =
lim supn→∞
|an|1/n in [−∞,+∞], and r < s, then
1.
∞∑
n=−∞
anzn converges on r < |z − z0| < s,
2.
∞∑
n=−∞
anzn converges uniformly on r1 ≤ |z − z0| ≤ s1, if r < r1 <
s1 < s,
3.
∞∑
n=−∞
anzn diverges on |z − z0| < r or |z − z0| > s.
Proof of this theorem follows immediately from Abel’s theorem on conver-
gence of power series (Theorem 2.3.2).
THEOREM 5.1.13 (Laurent’s Theorem)
Let f be analytic on an annulus ANR2
R1
(z0) = {z ∈ C : R1 < |z − z0| < R2}.
Then, f can be written as
f (z) =
∞∑
k=−∞
ak(z − z0)k , ∀z ∈ ANR2
R1
(z0).
Proof: Choose positive real numbers r1 and r2 such that R1 < r1 < r2 < R2.
If Cj is the circle with centre z0 and radius rj, for j = 1, 2, then we claim that
C2 ∪ (−C1) ∼ 0 in ANR2
R1
(z0). Let w � ANR2
R1
(z0).
�
�
“book” — 2014/6/4 — 21:09 — page 231 — #9
�
�
�
�
�
�
Series Developments and Infinite Products 231
Case 1: |w − z0| > R2.
Then, |w − z0| > r1 and |w − z0| > r2, which implies WN(C2 ∪
(−C1), w) = WN(C2, w) − WN(C1, w) = 0.
Case 2: |w − z0| < R1. In this case, |w − z0| < r1, |w − z0| < r2, and hence,
WN(C2 ∪ (−C1), w) = WN(C2, w) − WN(C1, w) = 1 − 1 = 0.
Hence, our claim holds. Then, by the general version of Cauchy’s integral
formula, we have
WN(C2 ∪ (−C1), z)f (z) = 1
i2π
∫
C2∪(−C1)
f (ζ )
ζ − z
dζ ∀z � C2 ∪ (−C1).
In particular, as WN(C2 ∪ (−C1), z) = 1 whenever r1 < |z − z0| < r2, we
have
f (z) = 1
i2π
∫
C2∪(−C1)
f (ζ )
ζ − z
dζ = 1
i2π
∫
C2
f (ζ )
ζ − z
dζ − 1
i2π
∫
C1
f (ζ )
ζ − z
dζ .
Since
|ζ − z0|
|z − z0| = r1
|z − z0| < 1, ∀ζ ∈ C1, we get
1
ζ − z
= 1
(ζ − z0) − (z − z0)
=
( −1
z − z0
)(
1
1 − ζ−z0
z−z0
)
=
( −1
z − z0
) ∞∑
k=0
(
ζ − z0
z − z0
)k
.
We note that the power series on the right-hand side converges uniformly on
C1. Similarly, using
|z − z0|
|ζ − z0| < 1, ∀ζ ∈ C2, we get
1
ζ − z
= 1
(ζ − z0) − (z − z0)
=
(
1
ζ − z0
)(
1
1 − z−z0
ζ−z0
)
=
(
1
ζ − z0
) ∞∑
k=0
(
z − z0
ζ − z0
)k
,
and the series uniformly converges on C2, ∀z ∈ ANr2
r1
(z0).
Therefore, we get
f (z) =
∞∑
k=0
(z − z0)k
i2π
∫
C2
f (ζ )
(ζ − z0)k+1
dζ
�
�
“book” — 2014/6/4 — 21:09 — page 232 — #10
�
�
�
�
�
�
232 Taylor Series and Laurent Series
+
∞∑
k=0
(z − z0)−k−1
i2π
∫
C1
f (ζ )
(ζ − z0)−k
dζ
=
∞∑
k=0
(z − z0)k
i2π
∫
C2
f (ζ )
(ζ − z0)k+1
dζ
+
∞∑
j=1
(z − z0)−j
i2π
∫
C1
f (ζ )
(ζ − z0)−j+1
dζ
=
∞∑
k=−∞
ak(z − z0)k ,
where ak =
⎧⎪⎪⎨
⎪⎪⎩
1
i2π
∫
C2
f (ζ )
(ζ − z0)k+1
dζ if k ≥ 0
1
i2π
∫
C1
f (ζ )
(ζ − z0)k+1
dζ if k < 0
, ∀k ∈ Z. �
RESULT 5.1.14 (Uniqueness)
If f (z) =
∞∑
k=−∞
ck(z − z0)k , ∀z ∈ ANs
r(z0) and if the Laurent series of f is
f (z) =
∞∑
k=−∞
ak(z − z0)k , ∀z ∈ ANs
r(z0), then ck = ak , ∀k ∈ Z.
Proof: Let n be a non-negative integer. Then, by definition, we have
an = 1
i2π
∫
C2
f (ζ )
(ζ − z0)n+1
dζ
= 1
i2π
∫
C2
∞∑
k=−∞
ck(ζ − z0)k
(ζ − z0)n+1
dζ
=
∞∑
k=−∞
ck
1
i2π
∫
C2
(ζ − z0)k−n−1 dζ
(as the series converges uniformly on C2)
�
�
“book” — 2014/6/4 — 21:09 — page 233 — #11
�
�
�
�
�
�
Series Developments and Infinite Products 233
=
∞∑
k=−∞
ck
1
i2π
2π∫
0
(r2 exp(it))k−n−1 ir2 exp(it) dt
(as C2 is given by ζ (t) = z0 + r2 exp(it), t ∈ [0, 2π])
=
∞∑
k=−∞
ckirk−n
2
1
i2π
2π∫
0
exp(i(k − n)t) dt
= cn,
since
1
2π
2π∫
0
exp(i(k − n)t) dt =
{
1 k = n
0 k � n
. By a similar argument, we get
an = cn for every negative integer also. Hence, the result follows. �
Algorithm 5.1.15 (To find Laurent series of rational functions)
Algorithm to find Laurent series
Step 1. If the given rational function is
P(z)
Q(z)
, where P(z), Q(z) are polyno-
mials, then find the partial fraction expansion of
1
Q(z)
if Q(z) has
more than one root. In this case, a typical summand in the partial
fraction expansion is of the form
A
(z ± a)m
for some A, a ∈ C.
Step 2(a). To expand
1
(z ± a)m
, in |z − z0| < R for some R > 0 and for some
a ∈ C with R < |b|, where b = −(z0 ± a).
Then write
1
(z ± a)m
= 1
((z − z0) − b)m
= 1
|b|m
1(
1 − (z−z0)
b
)m
= 1
|b|m
1
m!
∞∑
k=m
k(k − 1) · · · (k − m + 1)
(
(z − z0)
b
)k−m
.
Step 2(b). To expand
1
(z ± a)
, in |z − z0| > R for some R > 0 and for some
a ∈ C with |b| < R, where b = −(z0 ± a).
�
�
“book” — 2014/6/4 — 21:09 — page 234 — #12
�
�
�
�
�
�
234 Taylor Series and Laurent Series
Then write
1
(z ± a)m
= 1
((z − z0) − b)m
= 1
(z − z0)m
1(
1 − b
(z−z0)
)m
= 1
(z − z0)m
1
m!
∞∑
k=m
k(k − 1) · · · (k−m+1)
(
b
(z − z0)
)k−m
.
Step 3. Finally, replace all the summands by the corresponding series and
simplify.
From the following examples, one can understand the algorithm still better.
Example 5.1.16 Find the series development of the following functions in
the specified regions:
1.
1
z2 − 5z + 4
in (i) 0 < |z| < 1; (ii) 1 < |z| < 4; (iii) |z| > 4.
2.
z + 1
z2 − z − 6
in (i) 0 < |z| < 2; (ii) 2 < |z| < 3; (iii) |z| > 3.
3.
z
z2 − 8z + 15
in (i) 0 < |z−2| < 1; (ii) 1 < |z−2| < 3; (iii) |z−2| > 3.
4.
z + 5
z2 − 6z + 8
in 1 < |z + 1| < 3.
Solution:
1. Applying partial fraction technique, we write
1
z2 − 5z + 4
= 1/3
z − 4
− 1/3
z − 1
.
(a) If 0 < |z| < 1, then we have1
1/3
z − 4
= − 1
12
(
1 − z
4
) = − 1
12
∞∑
k=0
( z
4
)k
(5.2)
1Here note that |z| < 1 < 4 so we have taken −4 and −1 as the common factors in equations
(5.2) and (5.3), respectively.
�
�
“book” — 2014/6/4 — 21:09 — page 235 — #13
�
�
�
�
�
�
Series Developments and Infinite Products 235
and
−1/3
z − 1
= 1
3 (1 − z)
= 1
3
∞∑
k=0
zk . (5.3)
Thus, we get
1
z2 − 5z + 4
= − 1
12
∞∑
k=0
( z
4
)k + 1
3
∞∑
k=0
zk
= 1
3
∞∑
k=0
(
1 − 1
4k+1
)
zk , 0 < |z| < 1.
(b) If 1 < |z| < 4, then we have2
1/3
z − 4
= − 1
12
(
1 − z
4
) = − 1
12
∞∑
k=0
( z
4
)k
(5.4)
and
−1/3
z − 1
= − 1
3z
(
1 − 1
z
) = − 1
3z
∞∑
k=0
(
1
z
)k
= −1
3
∞∑
k=0
(
1
z
)k+1
.
(5.5)
Thus, we get
1
z2 − 5z + 4
= − 1
12
∞∑
k=0
( z
4
)k − 1
3
∞∑
k=0
(
1
z
)k+1
= −1
3
[ ∞∑
k=0
1
4k+1
zk +
∞∑
k=1
z−k
]
, 1 < |z| < 4.
(c) If |z| > 4, then we have
1/3
z − 4
= 1
3z
(
1 − 4
z
) = 1
3z
∞∑
k=0
(
4
z
)k
= 1
3
∞∑
k=0
4k
zk+1
2Here note that 1 < |z| < 4 so we have taken −4 and z as the common factors in equations
(5.4) and (5.5), respectively.
�
�
“book” — 2014/6/4 — 21:09 — page 236 — #14
�
�
�
�
�
�
236 Taylor Series and Laurent Series
and
−1/3
z − 1
= − 1
3z
(
1 − 1
z
) = − 1
3z
∞∑
k=0
(
1
z
)k
= −1
3
∞∑
k=0
(
1
z
)k+1
.
Thus, we get
1
z2 − 5z + 4
= 1
3
∞∑
k=0
4k
zk+1
− 1
3
∞∑
k=0
(
1
z
)k+1
= −1
3
∞∑
k=1
(4k − 1)z−(k+1), 0 < |z| < 1.
2. Applying partial fraction technique, we write
1
z2 − z − 6
= 1/5
z − 3
− 1/5
z + 2
.
(a) If 0 < |z| < 2, then we have
z + 1
5 (z − 3)
= − z + 1
15
(
1 − z
3
)
= − z + 1
15
∞∑
k=0
( z
3
)k
= − 1
15
[ ∞∑
k=0
3
( z
3
)k+1 +
∞∑
k=0
( z
3
)k
]
= − 1
15
[ ∞∑
k=0
3
( z
3
)k+1 + 1 +
∞∑
k=0
( z
3
)k+1
]
= − 1
15
+ 4
15
∞∑
k=0
( z
3
)k+1
and
−(z + 1)
5(z + 2)
= −(z + 1)
10
(
1 + z
2
)
= − z + 1
10
∞∑
k=0
(
z
−2
)k
�
�
“book” — 2014/6/4 — 21:09 — page 237 — #15
�
�
�
�
�
�
Series Developments and Infinite Products 237
= − 1
10
[ ∞∑
k=0
(−2)
(
z
−2
)k+1
+
∞∑
k=0
(
z
−2
)k
]
= − 1
10
[ ∞∑
k=0
(−2)
(
z
−2
)k+1
+ 1 +
∞∑
k=0
(
z
−2
)k+1
]
= −1
10
+ 1
10
∞∑
k=0
(
z
−2
)k+1
.
Thus, we get
z + 1
z2 − z − 6
= − 1
15
+ 4
15
∞∑
k=0
( z
3
)k+1 + −1
10
+ 1
10
∞∑
k=0
(
z
−2
)k+1
= −1
6
+
∞∑
k=1
[
4
15
3−k + 1
10
(−2)k
]
zk , 0 < |z| < 2.
(b) If 2 < |z| < 3, then using (i), we have
z + 1
5 (z − 3)
= − z + 1
15
(
1 − z
3
) = − 1
15
+ 4
15
∞∑
k=0
( z
3
)k+1
and
− z + 1
5(z + 2)
= − z + 1
5z
(
1 + 2
z
)
= − z + 1
5z
∞∑
k=0
(
−2
z
)k
= −1
5
[ ∞∑
k=0
(
−2
z
)k
− 1
2
∞∑
k=0
(
−2
z
)k+1
]
= −1
5
[
1 +
∞∑
k=0
(
−2
z
)k+1
− 1
2
∞∑
k=0
(
−2
z
)k+1
]
= −1
5
+ 1
10
∞∑
k=0
(
−2
z
)k+1
.
�
�
“book” — 2014/6/4 — 21:09 — page 238 — #16
�
�
�
�
�
�
238 Taylor Seriesand Laurent Series
Thus, we get
z + 1
z2 − z − 6
= − 1
15
+ 4
15
∞∑
k=0
( z
3
)k+1− 1
5
+ 1
10
∞∑
k=0
(
−2
z
)k+1
= − 4
15
+ 4
15
∞∑
k=1
3−kzk
+ 1
10
∞∑
k=1
(−2)kz−k , 2 < |z| < 3.
(c) If |z| > 3, then we have
z + 1
5(z − 3)
= z + 1
5z
(
1 − 3
z
)
= z + 1
5z
∞∑
k=0
(
3
z
)k
= 1
5
[ ∞∑
k=0
(
3
z
)k
+
∞∑
k=0
1
3
(
3
z
)k+1
]
= 1
5
+
[
1 +
∞∑
k=1
(
3
z
)k
+
∞∑
k=1
1
3
(
3
z
)k
]
= 1
5
+ 4
15
∞∑
k=1
(
3
z
)k
,
and from case (ii), we have
− z + 1
5(z + 2)
= − z + 1
5z
(
1 + 2
z
) = −1
5
+ 1
10
∞∑
k=0
(
−2
z
)k+1
.
Thus, we get
z + 1
z2 − z − 6
= 4
15
∞∑
k=1
3kz−k + 1
10
∞∑
k=1
(−2)kz−k
=
∞∑
k=1
(
4
5
3k−1 + 1
10
(−2)k
)
z−k, |z| > 3.
�
�
“book” — 2014/6/4 — 21:09 — page 239 — #17
�
�
�
�
�
�
Series Developments and Infinite Products 239
3. Applying partial fraction technique, we write
1
z2 − 8z + 15
= 1/2
z − 5
− 1/2
z − 3
.
(a) If 0 < |z − 2| < 1, then we have
z
2(z − 5)
= z
2(z − 2 − 3)
= z − 2 + 2
−6
1(
1 − z−2
3
)
= −
(
z − 2
6
+ 1
3
)
1(
1 − z−2
3
)
= −
(
z − 2
6
+ 1
3
) ∞∑
k=0
(
z − 2
3
)k
= −1
2
∞∑
k=0
(
z − 2
3
)k+1
− 1
3
∞∑
k=0
(
z − 2
3
)k
= −1
2
∞∑
k=1
(
z − 2
3
)k
− 1
3
− 1
3
∞∑
k=1
(
z − 2
3
)k
= −1
3
− 5
6
∞∑
k=1
3−k(z − 2)k
and
− z
2(z − 3)
= − z
2(z − 2 − 1)
= z − 2 + 2
2
1
(1 − (z − 2))
=
(
z − 2
2
+ 1
)
1
(1 − (z − 2))
=
(
z − 2
2
+ 1
) ∞∑
k=0
(z − 2)k
= 1
2
∞∑
k=0
(z − 2)k+1 +
∞∑
k=0
(z − 2)k
= 1
2
∞∑
k=1
(z − 2)k + 1 +
∞∑
k=1
(z − 2)k
�
�
“book” — 2014/6/4 — 21:09 — page 240 — #18
�
�
�
�
�
�
240 Taylor Series and Laurent Series
= 1 + 3
2
∞∑
k=1
(z − 2)k .
Therefore,
1
z2 − 8z + 15
= −1
3
− 5
6
∞∑
k=1
3−k(z − 2)k + 1
+ 3
2
∞∑
k=1
(z − 2)k
= 2
3
+
∞∑
k=1
(
3
2
− 5
6
3−k
)
(z − 2)k .
(b) If 1 < |z − 2| < 3, then we have
z
2(z − 5)
= z
2(z − 2 − 3)
= z − 2 + 2
−6
1(
1 − z−2
3
)
= −
(
z − 2
6
+ 1
3
)
1(
1 − z−2
3
)
= −
(
z − 2
6
+ 1
3
) ∞∑
k=0
(
z − 2
3
)k
= −1
2
∞∑
k=0
(
z − 2
3
)k+1
− 1
3
∞∑
k=0
(
z − 2
3
)k
= −1
2
∞∑
k=1
(
z − 2
3
)k
− 1
3
− 1
3
∞∑
k=1
(
z − 2
3
)k
= −1
3
− 5
6
∞∑
k=1
3−k(z − 2)k
and
− z
2(z − 3)
= − z
2(z − 2 − 1)
= − z − 2 + 2
2(z − 2)
1(
1 − 1
z−2
)
�
�
“book” — 2014/6/4 — 21:09 — page 241 — #19
�
�
�
�
�
�
Series Developments and Infinite Products 241
= −
(
1
2
− 1
z − 2
)
1(
1 − 1
z−2
)
= −1
2
∞∑
k=0
(z − 2)−k +
∞∑
k=1
(z − 2)−k
= −1
2
+ 1
2
∞∑
k=1
(z − 2)−k .
Therefore,
1
z2 − 8z + 15
= −1
3
− 5
6
∞∑
k=1
3−k(z − 2)k − 1
2
+ 1
2
∞∑
k=1
(z − 2)−k
= −5
6
− 5
6
∞∑
k=1
3−k(z − 2)k + 1
2
∞∑
k=1
(z − 2)−k .
(c) If |z − 2| > 3, then we have
z
2(z − 5)
= z
2(z − 2 − 3)
= z − 2 + 2
2(z − 2)
1(
1 − 3
z−2
)
=
(
1
2
+ 1
z − 2
)
1(
1 − 3
z−2
)
=
(
1
2
+ 1
z − 2
) ∞∑
k=0
(
3
z − 2
)k
= 1
2
∞∑
k=0
(
3
z − 2
)k
+ 1
3(z − 2)
∞∑
k=0
(
3
z − 2
)k
= 1
2
+ 1
2
∞∑
k=1
(
3
z − 2
)k
+ 1
3
∞∑
k=1
(
3
z − 2
)k
= 1
2
+ 5
6
∞∑
k=1
3k(z − 2)−k
�
�
“book” — 2014/6/4 — 21:09 — page 242 — #20
�
�
�
�
�
�
242 Taylor Series and Laurent Series
and
− z
2(z − 3)
= − z
2(z − 2 − 1)
= − z − 2 + 2
2(z − 2)
1(
1 − 1
z−2
)
= −
(
1
2
− 1
z − 2
)
1(
1 − 1
z−2
)
= −1
2
+ 1
2
∞∑
k=1
(z − 2)−k .
Therefore,
1
z2 − 8z + 15
= 1
2
+ 5
6
∞∑
k=1
3k(z − 2)−k − 1
2
+ 1
2
∞∑
k=1
(z − 2)−k
= 5
6
∞∑
k=1
3−k(z − 2)k + 1
2
∞∑
k=1
(z − 2)−k .
4. Applying partial fraction technique, we write
1
z2 − 6z + 8
= 1/2
z − 4
− 1/2
z − 2
.
(a) If 1 < |z + 1| < 3, then we have
z + 5
2(z − 4)
= z + 5
2(z + 1 − 5)
= − z + 5
10
1(
1 − z + 1
5
)
= − z + 1 + 4
10
∞∑
k=0
(
z + 1
5
)k
= − 1
10
[
5
∞∑
k=0
(
z + 1
5
)k+1
+ 4
∞∑
k=0
(
z + 1
5
)k
]
= −1
2
∞∑
k=0
(
z + 1
5
)k+1
− 2
5
∞∑
k=0
(
z + 1
5
)k
�
�
“book” — 2014/6/4 — 21:09 — page 243 — #21
�
�
�
�
�
�
Series Developments and Infinite Products 243
= −1
2
∞∑
k=1
(
z + 1
5
)k
− 2
5
− 2
5
∞∑
k=1
(
z + 1
5
)k
= −2
5
− 7
10
∞∑
k=1
5−k(z + 1)k
and
− z + 5
2(z + 2)
= − z + 5
2(z + 1 + 1)
= − z + 1 + 4
2(z + 1)
(
1 +
(
1
z + 1
))
= −
(
1
2
+ 2
z + 1
) ∞∑
k=0
( −1
z + 1
)k
= −1
2
∞∑
k=0
( −1
z + 1
)k
+ 2
∞∑
k=0
( −1
z + 1
)k+1
= −1
2
− 1
2
∞∑
k=1
( −1
z + 1
)k
+ 2
∞∑
k=1
( −1
z + 1
)k
= −1
2
+ 3
2
∞∑
k=1
(−1)k (z + 1)−k .
Thus, we get
1
z2 − 5z + 4
= −1
2
− 7
10
∞∑
k=1
5−k(z + 1)k + 2
− 1
2
∞∑
k=1
(−1)k (z + 1)−k
= −1 − 7
10
∞∑
k=1
5−k(z + 1)k
+ 3
2
∞∑
k=1
(−1)k (z + 1)−k ,∀z with 1<|z + 1|<3.
�
�
“book” — 2014/6/4 — 21:09 — page 244 — #22
�
�
�
�
�
�
244 Zeroes, Poles, and Singularities
Exercise 5.1.17 Find the Laurent series expansions of the given functions in
the specified regions.
1. f (z) = 1
z2(1 − z)
in 0 < |z| < 1.
2. f (z) = 1
1 + z
in (i) 0 < |z − i| < √
2, (ii) |z − i| > √
2.
3. f (z) = 1
z2 − 5z + 6
in 0 < |z| < 2.
4. f (z) = (1 − z)3
z − 2
in |z − 1| > 1.
5. f (z) = 1
(z − 1)2(z − 3)
in (i) 0 < |z − 1| < 2, (ii) 0 < |z − 3| < 2.
Answers:
1.
∞∑
n=−2
1
zn
.
2. (i)
∞∑
n=0
(−1)n
(
z − i
1 + i
)n
, (ii)
∞∑
n=1
(−1)n+1
(
1 + i
z − i
)n
.
3.
∞∑
n=1
(
1
2n
− 1
3n
)
zn−1.
4. −(z − 1)3
∞∑
n=1
1
(z − 1)n
.
5. (i) − 1
2(z − 1)2
∞∑
n=0
(
z − 1
2
)n
, (ii)
1
4(z − 3)
∞∑
n=1
n
(
−
[
z − 3
2
])n−1
.
5.2 ZEROES, POLES, AND SINGULARITIES
Definition 5.2.1 If f is an analytic function on a region � and f (z0) = 0 for
some z0 ∈ �, then z0 is called a zero of f . If there exists k ∈ N, such that
f (z0) = f ′(z0) = f ′′(z0) = · · · = f (k−1)(z0) = 0 and f (k)(z0) � 0,
then k is called the order of the zero of f at z0.
�
�
“book” — 2014/6/4 — 21:09 — page 245 — #23
�
�
�
�
�
�
Series Developments and Infinite Products 245
THEOREM 5.2.2 Let f be an analytic function, which is not identically zero
on a region �. If f (z0) = 0 for some z0 ∈ �, then order of zero of f at z0
exists.
Proof: Let A = {
z ∈ � : f (z) = 0 and f (k)(z) = 0, ∀k ∈ N
}
and B = �\A.
That is, B = {
z ∈ � : f (z) � 0 or f (k)(z) � 0 for some k ∈ N
}
. Then clearly,
� = A ∪ B and A ∩ B = ∅. Since f (k) is analytic (see Corollary 4.3.3), f (k)
is continuous for every k = 0, 1, 2, 3, . . . . Therefore,
{
z ∈ � : f (k)(z) = 0
}
=
(
f (k)
)−1
({0})
is a closed set, and hence,
{
z ∈ � : f (k)(z) � 0
}
is an open set for every k =
0, 1, 2, 3, . . . . Therefore, B = ∞∪
k=0
{
z ∈ � : f (k)(z) � 0
}
is an open set.
Next, we show that A is also an open set. If a ∈ A is arbitrary, then for every
n ∈ N, by Taylor’s theorem, we have
f (z) =
n−1∑
k=0
f (k)(a)
k!
(z − a)k + φn(z)(z − a)n = φn(z)(z − a)n, ∀z ∈ �,
where
φn(z) = 1
i2π
∫
|ζ−a|=r
f (ζ )
(ζ − a)n(ζ − z)
dζ , ∀z inside C,
where r > 0 is such that ClB(a, r) ⊆ �. If M = sup
ζ∈C
| f (ζ )|, then M <∞ and
for every z ∈ B(a, r),
| f (z)| ≤ |z − a|n
2π
∫
|ζ−a|=r
| f (ζ )|
|ζ − a|n|ζ − z| |dζ |
≤ M |z − a|n
2π
∫
|ζ−a|=r
|dζ |
rn(r − |z − a|)
(as |ζ − a| ≤ |z − ζ | + |z − a| ⇒ r − |z − a| ≤ |z − ζ |)
≤ M |z − a|n
2π
2πr
rn(r − |z − a|)
≤
( |z − a|
r
)n Mr
r − |z − a| → 0 as n → ∞.
�
�
“book” — 2014/6/4 — 21:09 — page 246 — #24
�
�
�
�
�
�
246 Zeroes, Poles, and Singularities
Therefore, f = 0 on B(a, r), and hence, f (k)(z) = 0, for all k = 0, 1, 2, 3, . . . .
Thus, B(a, r) ⊆ A. Given that z0 ∈ � such that f (z0) = 0. Suppose f (k)(z0) =
0, ∀k ∈ N, then A � ∅. By using the connectedness of �, we get
B = ∅ ⇒ � = A ⇒ f = 0 on �,
which is a contradiction. Hence, there exists a least positive integer k such
that f (k)(z0) � 0. In other words, order of the zero at z0 exists for f . �
RESULT 5.2.3 If f is an analytic function on a region�with a zero of order k
at z0 ∈ �, then there exists an analytic function g on � such that g is analytic
on �, f (z) = (z − z0)kg(z) on � and g(z0) � 0.
Proof: By Taylor’s theorem, there exists an analytic function g on � such
that
f (z) =
k−1∑
j=0
f (j)(z0)
j!
(z− z0) j + g(z)(z − z0)k = g(z)(z − z0)k , ∀z ∈ �.
Suppose g(z0) = 0, then using
f (k)(z) =
k∑
ν=0
kCνg(k−ν)(z)
dν
dzν
(
(z − a)k
)
and
dν
dzν
(
(z − a)k
)∣∣∣∣
z=a
=
{
0 ν < k
k! ν = k
,
we obtain f (k)(a) = g(a)k! = 0, which contradicts the fact that the order of
the zero of f at z0 is k. Thus, g(a) � 0. �
THEOREM 5.2.4 (Principle of analytic continuation)
Let f and g be analytic functions on a region �. If {z ∈ � : f (z) = g(z)} has
a limit point in �, then f (z) = g(z), ∀z ∈ �.
Proof: Since f and g are analytic, clearly f −g is an analytic function. If z0 is
a limit point of {z ∈ � : f (z) = g(z)}, then there exists a sequence (zn) from
�\{z0} such that
f (zn) − g(zn) = 0, ∀n ∈ N and zn → z0 as n → ∞.
�
�
“book” — 2014/6/4 — 21:09 — page 247 — #25
�
�
�
�
�
�
Series Developments and Infinite Products 247
Therefore, ( f − g)(z0) = lim
n→∞( f − g)(zn) = 0. If f (w) � g(w) for some
w ∈ �, then f − g is not identically zero. Then, by the previous result, there
exists k ∈ N such that
( f − g)(z) = (z − z0)kφ(z), ∀z ∈ B(z0, r),
for some analytic function φ on � such that φ(z0) � 0. Since φ is continuous
at z0, for
|φ(z0)|
2
> 0, we find 0 < δ < r such that
|z − z0| < δ ⇒ |φ(z) − φ(z0)| < |φ(z0)|
2
⇒ |φ(z0)| − |φ(z)| ≤ |φ(z) − φ(z0)| < |φ(z0)|
2
⇒ |φ(z)| ≥ |φ(z0)|
2
.
Therefore, φ is nowhere zero on B(z0, δ). Since zn → z0 as n → ∞, for this
δ > 0, we find an m ∈ N such that
|zm − z0| < δ ⇒ zm ∈ B(z0, δ)
⇒ (zm − z0)kφ(zm) = ( f − g)(zm) = 0
⇒ φ(zm) = 0.
which is a contradiction. Hence, f (z) = g(z), ∀z ∈ �. �
COROLLARY 5.2.5 If f is a non-constant analytic function on a region �,
then f cannot have uncountable number of distinct zeroes.
Proof: Let Z( f ) = {z ∈ � : f (z) = 0}. By principle of analytic continuation,
Z( f ) has no limit point in �. Hence, for every z ∈ Z( f ), there exists rz > 0
such that B(z, rz)∩Z( f ) = {z}. From Theorem 2.4.34, there exists a countable
collection B = {Bn : n ∈ N} of open balls such that for every z ∈ Z( f ), there
exists Bnz ∈ B such that z ∈ Bnz ⊆ B(z, rz). Therefore, it follows that
{z} ⊆ Bnz ∩ Z( f ) ⊆ B(z, rz) ∩ Z( f ) = {z},
and hence, Bnz ∩ Z( f ) = {z}, ∀z ∈ Z( f ). If we define
φ : Z( f ) → N by φ(z) = nz, ∀z ∈ Z( f ),
where nz ∈ N is the least positive integer such that z ∈ Bnz ⊆ B(z, rz), then
claim that φ is one-to-one. If z, w ∈ Z( f ) be such that φ(z) = φ(w), then
Bnz = Bnw ⇒ {z} = Bnz ∩ Z( f ) = Bnw ∩ Z( f ) = {w} ⇒ z = w.
Thus, φ is an injection. Hence, Z( f ) is at most countable. �
�
�
“book” — 2014/6/4 — 21:09 — page 248 — #26
�
�
�
�
�
�
248 Zeroes, Poles, and Singularities
LEMMA 5.2.6 Let f be a non-zero analytic function on a region � and γ be
a closed curve not passing through any zero of f . Then, the number of zeroes
of f enclosed by γ is finite.
Proof: We note that the regions enclosed by a closed curve are bounded sets.
Suppose, there are infinitely many zeroes of f , which are enclosed by γ , then
the set {z ∈ � : f (z) = 0} has a limit point, by Bolzano–Weierstrass property
(Theorem 1.4.47). Hence, by the principle of analytic continuation (Theorem
5.2.4), it would follow that f is identically zero. Hence, there must be only
finitely many zeroes of f enclosed by γ . �
PROBLEM 5.2.7 If f is an entire function such that |f | = 1 on {z ∈ C : |z| =
1}, then f = czn, ∀z ∈ C, for some constant c ∈ C with |c| = 1 and for some
n ∈ {0, 1, 2, . . .}.
Solution: If f is a constant function, then the proof is obvious. Hence,
assume that f is not a constant. Let z1, z2, . . . , zn be the zeroes of f includ-
ing multiplicities on Cl B(0, 1). (Suppose f has infinite number of zeroes in
Cl B(0, 1), being a compact set, then by Theorem 1.4.47, the zeroes of f can
have limit point in Cl B(0, 1). Thus, by the principle of analytic continuation
(Theorem 5.2.4), f becomes identically zero on C, which is not possible.)
Moreover, note that |zk | < 1, ∀k ∈ {1, 2, . . . , n}, as |zk | ≤ 1 and |f (z)| = 1 on
|z| = 1.
Consider the rational function R(z) =
n∏
k=1
z − zk
1 − zkz
, whose finite zeroes
are z1, z2, . . . , zn and its finite poles are
1
zk
, k = 1, 2, 3, . . . , n, with zk � 0.
If F(z) = f (z)
R(z)
, then after removing the common zeroes z1, z2, . . . , zn (which
are removable singularities for F), F becomes an entire function with zeroes
at the poles of R. Since
∣∣∣∣ 1
zk
∣∣∣∣ > 1, F is nowhere zero on Cl B(0, 1). As |F(z)|
is continuous real-valued function on the compact set Cl B(0, 1), there exist
w1, w2 ∈ Cl B(0, 1) such that
|F(w1)| ≤ |F(z)| ≤ |F(w2)|, ∀z ∈ Cl B(0, 1).
Hence, applying maximum and minimum principle for analytic functions, we
see |w1| = |w2| = 1, and hence, by hypothesis,
1 = |F(w1)| ≤ |F(z)| ≤ |F(w2)| = 1 on Cl B(0, 1).
�
�
“book” — 2014/6/4 — 21:09 — page 249 — #27
�
�
�
�
�
�
Series Developments and Infinite Products 249
In particular, maximum of |F| is attained at all points of B(0, 1); hence, again
by using maximum modulus principle, we get F = c on Cl B(0, 1) for some
c ∈ C with |c| = 1. Then, by principle of analytic continuation, we get F = c
on C \
{
1
zk
: k = 1, 2, . . . n with zk � 0
}
. Therefore,
f (z) = c
n∏
k=1
z − zk
1 − zkz
,∀z ∈ C \
{
1
zk
: k = 1, 2, . . . n with zk � 0
}
.
If zk � 0 for some k, then f should have a pole at zk , which is a contradiction
to the assumption on f . Hence, zk = 0, ∀k = 1, 2, 3, . . . , n. In other words,
f (z) = czn for all z ∈ C.
Definition 5.2.8 Let f be a complex-valued function on a region � and a ∈
�. We say that a is a singularity of f if f is not differentiable at a. A singularity
a of f is called
1. a removable singularity if lim
z→a
f (z) exists in C.
2. a pole if lim
z→a
f (z) = ∞.
3. an essential singularity if lim
z→a
f (z) does not exist in C∞. That is, a is
neither a removable singularity nor a pole.
4. an isolated singularity if there exists r > 0 such that f is analytic in
B(a, r) \ {a}.
RESULT 5.2.9 Let f be an analytic function on �\{a}. Then, the following
statements are equivalent:
1. a is a removable singularity for f ,
2. f is bounded on B(a, r)\{a} for some r > 0,
3. lim
z→a
(z − a)f (z) = 0,
4. There exists an analytic function φ on � such that φ(z) = f (z), ∀z ∈
�\{a}.
Proof:
(1) ⇒ (2) Since a is a removable singularity of f , we have lim
z→a
f (z) = 	 for
some 	 ∈ C. Therefore, for ε = 1, there exists r > 0 such that
0 < |z − a| < r ⇒ | f (z) − 	| < 1.
Therefore, | f (z)| ≤ 1 + |	|, ∀z ∈ B(a, r)\{a}. Hence, (2) follows.
�
�
“book” — 2014/6/4 — 21:09 — page 250 — #28
�
�
�
�
�
�
250 Zeroes, Poles, and Singularities
(2) ⇒ (3) By assumption, there exists r > 0 and M > 0 such that
| f (z)| ≤ M , ∀z ∈ B(a, r)\{a}.
For a given ε > 0, choose δ = min
{ ε
M
, r
}
so that δ > 0 and
0 < |z − a| < δ ⇒ |z − a| | f (z)| ≤ δM ≤ ε
M
M = ε.
Therefore, (3) holds.
(3) ⇒ (4) Assume that lim
z→a
(z − a)f (z) = 0. Choose r > 0 such that
ClB(a, r) ⊆ �. Define
φ(z) =
⎧⎪⎨
⎪⎩
∫
|ζ−a|=r
f (ζ )
ζ − a
dζ , ∀z ∈ B(a, r)
f (z) ∀z ∈ �\B(a, r)
.
Then, by Theorem 4.3.2 and by Cauchy’s integral formula (Theorem
4.3.1), we get φ, which is analytic on � and φ(z) = f (z), ∀z ∈ �\{a}.
(4) ⇒ (1) In particular, φ is continuous at a, and hence,
φ(a) = lim
z→a
φ(z) = lim
z→a
f (z).
Thus, proof (1) follows.
Hence, the result follows.
Definition 5.2.10 Let f be a complex-valued function on a region�. f is said
to be a meromorphic function if for each z ∈ �, either f is differentiable at z
or f has a pole at z.
RESULT 5.2.11 Let f be an analytic function on a region � except at a pole
a. Then, there exist δ > 0 and p ∈ N such that
f (z) = (z − a)−pψ(z), ∀z ∈ B(a, δ)\{a},
where ψ is analytic on B(a, δ) and ψ(a) � 0.
Proof: Since lim
z→a
f (z) = ∞, there exists ε > 0 such that B(a, ε) ⊆ � and
0 < |z − a| < ε ⇒ | f (z)| > 1.
�
�
“book” — 2014/6/4 — 21:09 — page 251 — #29
�
�
�
�
�
�
Series Developments and Infinite Products 251
Hence, f (z) � 0, ∀z ∈ B(a, ε)\{a}. If
g(z) = 1
f (z)
, ∀z ∈ B(a, ε)\{a},
then g is analytic and nowherezero on B(a, ε)\{a}. Since
lim
z→a
g(z) = lim
z→a
1
f (z)
= 0,
a is a removable singularity for g. Therefore, if we define
g(a) = lim
z→a
g(z) = 1
f (z)
= 0,
then g becomes analytic on B(a, ε), and hence, there exist p ∈ N and an
analytic function ϕ on B(a, ε) such that
g(z) = (z − a)pϕ(z), ∀z ∈ B(a, ε) and ϕ(a) � 0.
Since ϕ is non-zero at a and g is nowhere zero on B(a, δ) \ {a}, it follows that
ϕ is nowhere zero on B(a, r). If we define
ψ(z) = 1
ϕ(z)
, ∀z ∈ B(a, δ),
then ψ is a nowhere zero analytic function on B(a, δ), and
f (z) = (z − a)−pψ(z), ∀z ∈ B(a, δ).
Hence, the result follows. �
COROLLARY 5.2.12 If f is a meromorphic function on �, then f has at most
countable number of poles.
Proof: From Result 5.2.11, every pole of f is isolated. Hence, by a simi-
lar argument employed in the proof of Corollary 5.2.5, we can prove this
corollary. �
Definition 5.2.13 If a is a pole of f , then the order of the pole at a is defined
by the natural number p such that
f (z) = (z − a)−pψ(z), ∀z ∈ B(a, δ)\{a},
for some δ > 0 and for some analytic function ψ on B(a, δ) with ψ(a) � 0.
If f is analytic on � except at an isolated singularity a then the nature of
singularity at a can be described easily from the Laurent series expansion of
f in B(a, r)\{a} as follows, where r > 0 is such that ClB(a, r) ⊆ �.
�
�
“book” — 2014/6/4 — 21:09 — page 252 — #30
�
�
�
�
�
�
252 Zeroes, Poles, and Singularities
RESULT 5.2.14 If f has an isolated singularity at a and
f (z) =
∞∑
n=−∞
cn(z − a)n, in B(a, r)\{a},
for some r > 0, then a is
1. a removable singularity iff cn = 0, ∀n < 0.
2. a pole iff there exists m ∈ N such that c−m � 0 and cn = 0, ∀n < −m.
Furthermore, this m is the order of the pole of f at a.
3. an essential singularity iff cn � 0 for infinitely many n < 0.
Proof: Let a be an isolated singularity of f (z) =
∞∑
n=−∞
cn(z − a)n in 0 <
|z − a| < r.
1. Assume that a is a removable singularity. Suppose there exists at least
one k < 0 such that ck � 0, then
lim
z→a
f (z) = lim
z→a
∞∑
n=−∞
cn(z − a)n
does not exist in C, which is a contradiction. Hence, cn = 0, ∀n < 0.
Conversely, assume that f (z) =
∞∑
n=0
cn(z − a)n. Then, lim
z→a
f (z) = f (a)
exists in C. Hence, f has a removable singularity at a.
2. Assume that a is a pole of order m. Then, using Theorem 5.2.11, we
get
f (z) = (z − a)−mψ(z), ∀z ∈ B(a, r)\{a},
whereψ is analytic on B(a, δ) andψ(a) � 0. Asψ is analytic on B(a, δ),
we have the Taylor series expansion
ψ(z) =
∞∑
n=0
bn(z − a)n, ∀z ∈ B(a, r).
Hence, we have
f (z) = (z − a)−m
∞∑
n=0
bn(z − a)n =
∞∑
n=0
bn(z − a)n−m, ∀z ∈ B(a, r)\{a}.
�
�
“book” — 2014/6/4 — 21:09 — page 253 — #31
�
�
�
�
�
�
Series Developments and Infinite Products 253
Since the coefficient of (z − a)−m is b0 = ψ(a) � 0, the proof of this
part follows. Conversely, assume that c−m � 0 and cn = 0, ∀n < −m
for some m ∈ N. Then, let
ψ(z) =
{
(z − a)mf (z), if z � a
c−m if z = a,
∀z ∈ B(a, r).
Since
ψ(z) = (z − a)m
∞∑
n=−m
cn(z − a)n
=
∞∑
n=−m
cn(z − a)n+m
=
∞∑
n=0
cn−m(z − a)n, |z − a| < r
and the power series
∞∑
n=0
cn−m(z − a)n is analytic inside B(a, r), we get
that ψ is an analytic function in B(a, r) with
f (z) = (z − a)−mψ(z), ∀z ∈ B(a, r)\{a} and ψ(a) = c−m � 0.
Hence, f has a pole at a of order m.
3. a is essential singularity iff a is neither a removable singularity nor a
pole iff cn � 0 for some n < 0 and for every m ∈ N, there exists n ∈ Z
such that n < −m and cn � 0 iff cn � 0 for infinitely many n < 0.
Hence, the result follows. �
RESULT 5.2.15 Let f be an analytic function on a region � except at an
isolated singularity a.
1. If lim
z→a
|z − a|r| f (z)| = 0 for some r ∈ R
or
2. if lim
z→a
|z − a|s| f (z)| = ∞ for some s ∈ R,
then there exists an integer m such that
lim
z→a
|z − a| j| f (z)| = 0, ∀j > m and lim
z→a
|z − a|k | f (z)| = ∞, ∀k < m.
�
�
“book” — 2014/6/4 — 21:09 — page 254 — #32
�
�
�
�
�
�
254 Zeroes, Poles, and Singularities
Proof: Let (1) hold. Then, choose an integer n ≥ r. Now,
lim
z→a
|z − a|n+1| f (z)| = lim
z→a
|z − a|n−r+1 lim
z→a
|z − a|r| f (z)| = 0.
Hence, if φ(z) = (z − a)nf (z), ∀z ∈ �\{a}, then φ is an analytic function
except at a, and it has removable singularity at a and
|φ(a)| = lim
z→a
|φ(z)|
= lim
z→a
|z − a|n| f (z)|
= lim
z→a
|z − a|n−r lim
z→a
|z − a|r| f (z)|
= 0.
Hence, φ(a) = 0. Since φ is not identically zero3 using Result 5.2.3, we have
φ(z) = (z − a)νφ1(z), ∀z ∈ B(a, r) ⊆ �,
for some ν ∈ N, for some r > 0 and for an analytic function φ1 on B(a, r)
with φ1(a) � 0. Now, let m = n − ν. Then,
lim
z→a
|z − a|m| f (z)| = lim
z→a
|φ1(z)| = |φ1(a)|,
which is neither 0 nor ∞. Hence, for j > m,
lim
z→a
|z − a| j| f (z)| = lim
z→a
|z − a| j−m lim
z→a
|z − a|m| f (z)| = 0 · |φ1(a)| = 0,
and for k < m,
lim
z→a
|z − a|k | f (z)| = lim
z→a
|z − a|k−m lim
z→a
|z − a|m| f (z)| = ∞ · |φ1(a)| = ∞.
Next, assume that (2) holds. Then, choose an integer l such that l < s. Then,
lim
z→a
|z − a|l| f (z)| = lim
z→a
|z − a|l−s lim
z→a
|z − a|s| f (z)| = ∞,
and hence, a is a pole for ψ , where
ψ(z) = (z − a)lf (z), ∀z ∈ �\{a}.
Hence, by using Result 5.2.11, there exist a negative integer λ, δ > 0 and an
analytic function ψ1 on B(a, δ) ⊆ � such that
ψ(z) = (z − a)λψ1(z), ∀z ∈ B(a, δ)\{a}.
3As f has a singularity, f should not be identically zero, and hence, φ is not identically zero.
�
�
“book” — 2014/6/4 — 21:09 — page 255 — #33
�
�
�
�
�
�
Series Developments and Infinite Products 255
Therefore,
f (z) = (z − a)−lψ(z) = (z − a)λ−lψ1(z), ∀z ∈ B(a, δ) \ {a}.
If m = l − λ, then
lim
z→a
|z − a|m| f (z)| = lim
z→a
|ψ1(z)| = |ψ1(a)|,
which is neither 0 nor ∞. Thus, by a similar argument, we get the result.
Note that if there exist two integers m1 and m2 such that
lim
z→a
|z − a| j| f (z)| = 0, ∀j > mi
and
lim
z→a
|z − a|k | f (z)| = ∞, ∀k < mi, for i = 1, 2,
then obviously, m1 = m2. �
Definition 5.2.16 Let f be an analytic function on a region � except at an
isolated singularity a. If there exists an integer m such that lim
z→a
|z−a|j| f (z)| =
0, ∀j > m and lim
z→a
|z − a|k | f (z)| = ∞, ∀ k < m, then m is called the
algebraic order of f at a. It is easy to see that algebraic order of f at a is m iff
lim
z→a
|z − a|m| f (z)| exists and is neither 0 nor ∞.
The following lemma characterizes the singularity in terms of algebraic
order.
LEMMA 5.2.17 Let f be an analytic function on a region � except at an
isolated singularity a, and the algebraic order of f at a is m. Then,
1. m < 0 iff a is a removable singularity of f and f (a) = 0,
2. m = 0 iff a is a removable singularity of f and f (a) � 0,
3. m > 0 iff a is a pole of f .
Proof: Since m is the algebraic order of f at a, we have
lim
z→a
|z − a| j| f (z)| = 0, ∀j > m and lim
z→a
|z − a|k | f (z)| = ∞, ∀k < m.
1. m < 0 ⇒ lim
z→a
|z − a|0| f (z)| = 0. Hence, a is a removable singularity
of f and f (a) = 0. Conversely, assume that a is a removable singularity
�
�
“book” — 2014/6/4 — 21:09 — page 256 — #34
�
�
�
�
�
�
256 Zeroes, Poles, and Singularities
of f and f (a) = 0. Then, f (z) = (z− a)ng(z) in a neighbourood of a for
some n ∈ N and for some analytic function g on the neighbourhood of
a such that g(a) � 0. Therefore,
lim
z→a
|z − a|−n| f (z)| = lim
z→a
| f (z)| = |g(a)|,
which is neither 0 nor ∞; hence, −n is the algebraic order of f at a.
2. m = 0 ⇒ lim
z→a
|z − a|0| f (z)| � 0. Hence, a is a removable singularity
of f and f (a) � 0. Conversely, assume that a is a removable singularity
of f and f (a) � 0. Then, lim
z→a
|z − a|0| f (z)| = | f (a)|, which is neither 0
nor ∞, and hence, 0 is the algebraic order of f at a.
3. If m > 0, then lim
z→a
|z−a|0| f (z)| = ∞ ⇒ f has a pole at a. Conversely,
assume that a is a pole for f . Then, we have f (z) = (z − a)−ng(z) in a
neighbourhood of a for some n ∈ N and for an analytic function g on
the neighbourhood of a such that g(a) � 0. Therefore,
lim
z→a
|z − a|n| f (z)| = lim
z→a
|g(z)| = |g(a)|,
whichis neither 0 nor ∞. Hence, the algebraic order of f at a
is n > 0. �
COROLLARY 5.2.18 If f is an analytic function on a region � except at an
isolated singularity a, then the algebraic order of f at a does not exist iff a is
an essential singularity of f .
Proof: The algebraic order m of f at a exists iff m < 0 or m = 0 or m > 0 iff
a is a removable singularity or a pole iff a is not an essential singularity of f .
Thus, the corollary follows. �
THEOREM 5.2.19 (Weierstrass theorem for essential singularity)
Let f be an analytic function on a region �\{a}. If f has an essential singu-
larity at a, then given c ∈ C, given ε > 0, and given δ > 0, there exists
z ∈ B(a, δ) such that | f (z) − c| < ε.
Proof: Suppose that this theorem is not true. That is, there exist c ∈ C, ε > 0
and δ > 0 such that
|z − a| < δ ⇒ | f (z) − c| ≥ ε.
�
�
“book” — 2014/6/4 — 21:09 — page 257 — #35
�
�
�
�
�
�
Series Developments and Infinite Products 257
Clearly,
lim
z→a
|z − a|−1| f (z) − c| ≥ lim
z→a
|z − a|−1ε = ∞.
Hence, by Result 5.2.15, there exists r > 0, such that
lim
z→a
|z − a|r| f (z) − c| = 0.
Therefore,
lim
z→a
|z − a|r| f (z)| ≤ lim
z→a
|z − a|r| f (z) − c| + lim
z→a
|z − a|r|c| = 0.
It follows that f has algebraic order at a. Then, a is a pole of f or a removable
singularity of f . This is a contradiction to the assumption that a is an essential
singularity of f . Hence, the theorem follows. �
Exercise 5.2.20 Prove that the algebraic order of f at a is the least integer
m ∈ Z such that cm � 0 in the Laurent series expansion of f (z) =
∞∑
k=−∞
ck(z−
z0)k , 0 < |z − z0| < δ for some δ > 0.
5.3 PARTIAL FRACTION OF ENTIRE FUNCTIONS
Definition 5.3.1 Let f be a meromorphic function on C. If {an : n ∈ N} is
the set of all poles of f , then the following representation of f is called the
partial fraction expansion of f .
f (z) =
∞∑
n=−∞
(
Pn
(
1
z − an
)
− Qn(z)
)
+ φ(z), ∀z ∈ C\{an : n ∈ N},
where Pn is a polynomial without constant term, Qn is a polynomial ∀n ∈ N,
and φ is an entire function.
In the partial fraction of f , Pn
(
1
z − an
)
is called the singular part of f at an.
THEOREM 5.3.2 (Mittag-Leffler theorem)
Let {an} be a sequence of complex numbers such that an → ∞ as n → ∞ and
let {Pn} be an arbitrary sequence of polynomials without constant term. Then,
there exists a meromorphic function such that whose poles are precisely {an :
n ∈ N} and the singular part of the function at an is Pn
(
1
z − an
)
.
�
�
“book” — 2014/6/4 — 21:09 — page 258 — #36
�
�
�
�
�
�
258 Partial Fraction of Entire Functions
Proof: We shall construct a meromorphic function f such that
f (z) =
∞∑
n=0
(
Pn
(
1
z − an
)
− Qn(z)
)
, ∀z ∈ C\{an : n = 0, 1, 2, . . .},
for some suitable polynomial Qn such that the above series converges
uniformly on every compact subset of C\{an : n = 0, 1, 2, . . .}.
Let M > 0 be arbitrary. Since an → ∞ as n → ∞, we choose N ∈ N
such that 4|an| > M , ∀n ≥ N . For n ≥ N , if gn(z) = Pn
(
1
z − an
)
, ∀z ∈
B(0, |an|), then gn is analytic on B(0, |an|). Then, the finite Taylor series
expansion of gn is given by
gn(z) = Pn
(
1
z − an
)
=
mn∑
k=0
ckzk + φmn+1(z)zmn+1, ∀z ∈ B(0, |an|),
where mn ∈ N is such that
2mn−n > Mn = sup
|z|≤ |an|
2
∣∣∣∣Pn
(
1
z − an
)∣∣∣∣ < +∞
and
φmn+1(z) = 1
i2π
∫
|ζ |= |an|
2
Pn
(
1
z − an
)
ζmn+1(ζ − z)
dζ , ∀z ∈ B
(
0,
|an|
2
)
.
Let Qn(z) =
mn∑
k=0
ckzk , ∀z ∈ C. Now for z ∈ B
(
0, |an|
4
)
,
∣∣∣∣Pn
(
1
z − an
)
− Qn(z)
∣∣∣∣ =
∣∣∣φmn+1(z)zmn+1
∣∣∣
≤ |z|mn+1 Mn
2π
∫
|ζ |= |an|
2
|dζ |( |an|
2
)mn+1 |ζ − z|
≤ |z|mn+1 Mn
2π
∫
|ζ |= |an|
2
|dζ |( |an|
2
)mn+1 ( |an|
2 − |an|
4
)
�
�
“book” — 2014/6/4 — 21:09 — page 259 — #37
�
�
�
�
�
�
Series Developments and Infinite Products 259
≤
( |an|
4
)mn+1 Mn
2π
2π
|an|
2
(
2
|an|
)mn+1 4
|an|
= 2−mn Mn < 2−n.
Hence, by comparison test,
∞∑
n=N
(
Pn
(
1
z − an
)
− Qn(z)
)
converges uni-
formly on |z| ≤ M . As every compact subset of C\{an : n =
0, 1, 2, . . .} is contained in B (0, M) for some M > 0, we conclude that
∞∑
n=N
(
Pn
(
1
z − an
)
− Qn(z)
)
converges uniformly on every compact subset
of C\{an : n = 0, 1, 2, . . .}. Thus, there exists a meromorphic function
f (z) =
∞∑
n=0
(
Pn
(
1
z − an
)
− Qn(z)
)
,+φ(z),∀z ∈ C\{an : n = 0, 1, 2, . . .},
whose singular part at an is Pn
(
1
z − an
)
, ∀n = 0, 1, 2, . . ., where φ is an
arbitrary entire function. �
Example 5.3.3 Prove the following:
1.
π2
sin2(πz)
=
∞∑
k=−∞
1
(z − k)2
,∀z ∈ C\Z.
2. π cot(πz) = lim
n→∞
n∑
k=−n
1
z − k
,∀z ∈ C\Z.
3.
π
sin(πz)
= lim
n→∞
n∑
k=−n
(−1)k
z − k
, ∀z ∈ C\Z.
Let f (z) = π2
sin2(πz)
, ∀z ∈ C\N. Clearly, f has double pole at every integer.
Therefore, the Laurent series expansion of f around 0 is of the form
f (z) = A2
z2
+ A1
z
+
∞∑
n=0
Anzn, 0 < |z| < δ for some δ > 0,
and hence, the singular part of f at 0 is
A2
z2
+ A1
z
, where
A−2 = lim
z→0
(
z2 π2
sin2(πz)
)
= 1 and A−1 = 0,
�
�
“book” — 2014/6/4 — 21:09 — page 260 — #38
�
�
�
�
�
�
260 Partial Fraction of Entire Functions
as f is an even function, and hence, all odd coefficients in the Laurent series
expansion of f around 0 are zeroes. Thus, the singular part of f at 0 is
1
z2
.
Since sin2(π (z−n)) = sin2(πz), the singular part of sin2(πz) at n is
1
(z − n)2
.
Therefore, by Mittag-Leffler theorem, we have
f (z) =
∞∑
n=−∞
1
(z − n)2
+ φ(z), ∀z ∈ C\N,
for some entire function φ. (As
∞∑
n=−∞
1
(z − n)2
itself is convergent, we have
taken Qn(z) = 0, ∀z ∈ C and n ∈ N.) We claim that φ is identically zero.
First, we note that as
π2
sin2(πz)
and
∞∑
n=−∞
1
(z − n)2
are periodic functions of
period 1, φ is also a periodic function of period 1. Now, writing z = x + iy,
we get
| sin(πz)|2 = | sin(π (x + iy))|2
= | sin(πx) cosh(πy) + i sinh(πy) cos(πx)|2
= sin2(πx) cosh2(πy) + sinh2(πy) cos2(πx)
= (1 − cos2(πx)) cosh2(πy) + sinh2(πy) cos2(πx)
= cosh2(πy) − cos2(πx)(cosh2(πy) − sinh2(πy))
= cosh2(πy) − cos2(πx) → ∞ as |y| → +∞ uniformly in x.
Therefore,
g(x + iy) → 0 as |y| → +∞ uniformly on C\Z. (5.6)
In particular, g is bounded in the strip {(x, y) : 0 ≤ x ≤ 1, y ∈ R}. Since
g is a periodic function of period 1, g is bounded on C. Thus, by Liouville’s
theorem (Theorem 4.3.11), g is a constant function. Moreover, using equation
(5.6), we conclude that g is identically 0. Thus,
π2
sin2(πz)
=
∞∑
k=−∞
1
(z − k)2
= lim
n→∞
n∑
k=−n
1
(z − k)2
.
Integrating the above equation, we get
π2
(− cot(πz)
π
)
= lim
n→∞
n∑
k=−n
( −1
z − k
)
,
�
�
“book” — 2014/6/4 — 21:09 — page 261 — #39
�
�
�
�
�
�
Series Developments and Infinite Products 261
which implies the required series representation of π cot(πz). Next, consider
lim
n→∞
n∑
k=−n
(−1)k
z − k
= lim
j→∞
(2j+1)∑
k=−(2j+1)
(−1)k
z − k
= lim
j→∞
⎛
⎝ j∑
k=−j
1
z − 2k
−
j∑
k=−j−1
1
z − (2k + 1)
⎞
⎠
= 1
2
lim
j→∞
⎛
⎝ j∑
k=−j
1
z
2 − k
−
j∑
k=−j−1
1
z−1
2 − k
⎞
⎠
= π
2
(
cot
(πz
2
)
− cot
(
π (z − 1)
2
))
= π
2
⎛
⎜⎜⎝
cos
(πz
2
)
sin
(πz
2
) −
cos
(
π (z − 1)
2
)
sin
(
π (z − 1)
2
)
⎞
⎟⎟⎠
= π
2
⎛
⎜⎝cos
(πz
2
)
sin
(πz
2
) +
sin
(πz
2
)
cos
(πz
2
)
⎞
⎟⎠
= π
2
1
sin
(πz
2
)
cos
(πz
2
)
= π
sin(πz)
.
PROBLEM 5.3.4 Find the partial fraction expansion for sec and deduce the
Gregory–Leibniz–Madhava series, which is given by
∞∑
n=0
(−1)n+1
2n + 1
.
Solution:
From Example 5.3.3(3), we have
π
cos
(
π
2 − πz
) = π
sin(πz)
=
∞∑
n=−∞
(−1)n
z − n
.
By substituting w = π
2 − πz, we get
π
cos(w)
=
∞∑
n=−∞
(−1)n
−w
π
+ 1
2 − n
,
�
�
“book” — 2014/6/4 — 21:09 — page 262 — #40
�
�
�
�
�
�
262 Partial Fraction of Entire Functions
which implies that
sec(w) = −
∞∑
n=−∞
(−1)n
w +
(
n − 1
2
)
π
= −
∞∑
n=1
(−1)n
w +
(
n − 1
2
)
π
−
∞∑
n=0
(−1)n
w +
(
−n − 1
2
)
= −
∞∑
n=1
(−1)n
w +
(
n − 1
2
)
π
+
∞∑
m=1
(−1)m
w −
(
m − 1
2
)
(using the change of variable m = n + 1 in the second sum)
=
∞∑
n=1
(−1)n
⎡⎣− 1
w +
(
n − 1
2
)
π
+ 1
w −
(
n − 1
2
)
π
⎤
⎦
=
∞∑
n=1
(−1)n
2
(
n − 1
2
)
π
w2 −
(
n − 1
2
)2
π2
= 4
∞∑
n=1
(−1)n (2n − 1)π
4w2 − (2n − 1)2π2
.
Substituting w = 0 in the above series representation of sec(w), we get
1 = 4
∞∑
n=1
(−1)n+1
(2n − 1)π
⇒
∞∑
n=1
(−1)n+1
(2n − 1)
= π
4
.
Exercise 5.3.5
1. Prove that π tan
(
πz
2
) = 4z
∞∑
n=1
1
(2n − 1)2 − 4z2
. Hint: Use the identity
tan
(
πz
2
) = cot
(
πz
2
) − 2 cot(πz) and the partial fraction expansion of
π cot(πz).
2. Obtain the value of Gregory–Leibniz–Madhava series from the partial
fraction expansions of πz cot(πz) and
π
sin(πz)
. Hint: Substitute z = 1
4
in both expansions.
PROBLEM 5.3.6 Prove that
∞∑
k=1
1
k2
= π2
6
.
�
�
“book” — 2014/6/4 — 21:09 — page 263 — #41
�
�
�
�
�
�
Series Developments and Infinite Products 263
Solution:
From Example 5.3.3, we have
π2
sin2(πz)
=
∞∑
k=−∞
1
(z − k)2
=
∑
0�k∈Z
1
k2
1(
1 − z
k
)2 + 1
z2
=
∑
0�k∈Z
1
k2
∞∑
m=1
m
( z
k
)m−1 + 1
z2
. (5.7)
Since the Laurent series of
1
sin2(πz)
is given by
1
sin2(πz)
=
⎡
⎢⎢⎢⎣ 1
πz
1
1 −
(
(πz)2
3!
+ (πz)4
5!
− · · ·
)
⎤
⎥⎥⎥⎦
2
=
[
1
πz
∞∑
n=0
(
(πz)2
3
+ (πz)4
5!
− · · ·
)n
]2
=
[
1
πz
∞∑
n=0
(
(πz)2
3!
+ (πz)4
5!
− · · ·
)n
]2
= 1
(πz)2
+ 2
3!
+
∞∑
m=1
Amzm,
for some Am ∈ C, m ∈ N. Hence,
π2
sin2(πz)
= 1
z2
+ 2π2
3!
+
∞∑
m=1
Amzm. (5.8)
Equating the constant terms of the expansions of
π2
sin2(πz)
in equations
(5.7) and (5.8), we get
2π2
6
=
∑
0�k∈Z
1
k2
= 2
∞∑
k=1
1
k2
⇒
∞∑
k=1
1
k2
= π2
6
.
�
�
“book” — 2014/6/4 — 21:09 — page 264 — #42
�
�
�
�
�
�
264 Infinite Product
5.4 INFINITE PRODUCT
Definition 5.4.1 (Rigorous definition) A sequence (ζk) of non-zero complex
numbers is said to be multiplyable if
(
n∏
k=1
ζk
)
converges in C\{0}. Then, the
product of (ζk) is denoted by
n∏
k=1
ζk .
Although the above definition is rigorous, it is customary to define the
infinite product of complex numbers as follows.
Definition 5.4.2 (Customary definition)
n∏
k=1
ζk is said to be convergent if
lim
n→∞
(
n∏
k=1
ζk
)
exists in C\{0}.
Note that according to this definition,
∞∏
k=1
1
2
does not converge, as
n∏
k=1
1
2
=
1
2n
→ 0 as n → ∞.
LEMMA 5.4.3 If
n∏
k=1
ζk converges, then ζk → 1 as k → ∞.
Proof: Let �n =
n∏
k=1
ζk , ∀n ∈ N. Then, by assumption, we have lim
n→∞�n =
A for some A ∈ C\{0}. As ζk = �k
�k−1
, ∀k > 1, allowing k → ∞, we get
lim
k→∞
ζk = A
A
= 1. �
LEMMA 5.4.4
∞∏
k=1
ζk converges iff
∞∑
k=1
log(ζk) converges, for a suitable
branch of log.
Proof: Using Corollary 2.4.25, for every ζk , there exists unique θk ∈ (0, 2π ]
such that arg ζk = θk . Since {θk : k ∈ N} is countable, there exists θ ∈ (0, 2π )
such that arg ζk � θ , ∀k ∈ N. Now, we define a branch of log as follows. Let
�θ = C\ ({0} ∪ {r exp(iθ ) : r > 0}) and define
log z = ln |z| + i arg z with arg z ∈ (θ − 2π , θ ),
�
�
“book” — 2014/6/4 — 21:09 — page 265 — #43
�
�
�
�
�
�
Series Developments and Infinite Products 265
so that �θ contains positive real axis and arg 1 = 0. Then, as in the proof
of Result 3.3.9, we can prove that log is analytic on �θ . Now, assume that
∞∑
k=1
log(ζk) converges. Then, using the continuity of exp, we get
exp
( ∞∑
k=1
log(ζk)
)
= exp
(
lim
n→∞
n∑
k=1
log(ζk)
)
= lim
n→∞ exp
(
n∑
k=1
log(ζk)
)
= lim
n→∞
n∏
k=1
exp (log(ζk))
= lim
n→∞
n∏
k=1
ζk .
Therefore,
∞∏
k=1
ζk converges.
Conversely, assume that
∞∏
k=1
ζk converges to A. If �n =
n∏
k=1
ζk , ∀n ∈ N, then
we have
log
(
�n
A
)
=
n∑
k=1
log(ζk) − log A + i2πνn, where νn ∈ Z, ∀n ∈ N.
Using the continuity of log at 1, we get log
(
�n
A
)
→ log(1) = 0 as n → ∞.
Therefore,
0 = lim
n→∞
[
log
(
�n+1
A
)
− log
(
�n
A
)]
= lim
n→∞
[
n+1∑
k=1
log(ζk) −
n∑
k=1
log(ζk) + i2π (νn+1 − νn)
]
= lim
n→∞ [log(ζn+1) + i2π (νn+1 − νn)]
= lim
n→∞ i2π (νn+1 − νn) (by Lemma 5.4.3).
Using
�n
A
→ 1 and ζn → 1 as n → ∞, we obtain
arg
(
�n+1
A
)
− arg
(
�n
A
)
→ 0 and arg ζn+1 → 0 as n → ∞.
�
�
“book” — 2014/6/4 — 21:09 — page 266 — #44
�
�
�
�
�
�
266 Infinite Product
Since
(νn+1 − νn)2π = arg
(
�n+1
A
)
− arg
(
�n
A
)
+ arg (1 + ζn+1),
we get νn+1 − νn → 0 as n → ∞. Therefore, by definition of a convergent
sequence, for ε = 1
2
> 0, there exists N ∈ N such that |νn+1 − νn| < 1
2
.
Since νn+1 − νn is an integer, for each n ∈ N, we have
νn = νn+1 = ν(say), ∀n ≥ N .
Therefore, for n ≥ N , we get
n∑
k=1
log(ζk) = log
(
�n
A
)
+ log A − i2πν → log A − i2πν as n → ∞.
Thus,
∞∑
k=1
log(ζk) converges. �
It is interesting to note that we can use any branch of log which is analytic
on the positive real axis, in particular the principal logarithm.
Indeed, ζk → 1 iff log(ζk) → 0 implies that both
∞∏
k=1
ζk and
∞∑
k=1
log(ζk)
do not converge whenever ζk �→ 1 (or equivalently log(ζk) �→ 0). In the other
case, as ζk → 1 as k → ∞, we can very well define principal logarithm
of ζk by log(ζk) = In(|ζk |) + i arg(ζk) where ζk ∈ (−π ,π ]. However, we
can find N ∈ N such that ζk ∈ C\{(x, 0) : x ≤ 0} ∀k ≥ N , using ζk → 1 as
k → ∞. This is sufficient to get the proof of the previous lemma for principal
logarithm.
Definition 5.4.5 An infinite product
∞∏
k=1
ζk is said to be absolutely convergent
if
∞∑
k=1
log(ζk) converges absolutely.
As in the case of infinite series, we can define the concept of rearrange-
ment of an infinite product as follows.
Definition 5.4.6 If f : N → N is a bijection, then
∞∏
k=1
ζf (k) is called an
rearrangement of
∞∏
k=1
ζk .
�
�
“book” — 2014/6/4 — 21:09 — page 267 — #45
�
�
�
�
�
�
Series Developments and Infinite Products 267
RESULT 5.4.7 If
∞∏
k=1
ζk converges absolutely, then every rearrangement of
∞∏
k=1
ζk converges.
Proof: By definition,
∞∏
k=1
ζk converges absolutely means that
∞∑
k=1
log(ζk) con-
verges absolutely. Then, using Result 1.3.29, we get every rearrangement of
∞∑
k=1
log(ζk) converges, and hence, every rearrangement of
∞∏
k=1
ζk converges.
�
THEOREM 5.4.8
∞∏
k=1
ζk converges absolutely iff
∞∑
k=1
(ζk − 1) converges
absolutely.
Proof: First, we note that if
∞∏
k=1
ζk converges absolutely or
∞∑
k=1
(ζk − 1) con-
verges absolutely, then by Lemma 5.4.3 or by Result 1.3.27, we have ζk → 1
as k → ∞.
We claim that
∞∑
k=1
(ζk − 1) converges absolutely iff
∞∑
k=1
log(ζk) converges
absolutely. We note that Taylor series expansion of log(ζk) around 1 is given
by
log(ζk) =
∞∑
m=0
log(m)(1)
m!
(ζk − 1)m =
∞∑
m=1
(−1)m+1
m
(ζk − 1)m.
Therefore,
lim
k→∞
∣∣∣∣ log(ζk)
ζk − 1
∣∣∣∣ = lim
k→∞
∣∣∣∣∣
∞∑
m=1
(−1)m+1
m
(ζk − 1)m−1
∣∣∣∣∣ = 1.
Therefore, for a given ε > 0, there exists N ∈ N such that
k ≥ N ⇒
∣∣∣∣ log(ζk)
ζk − 1
− 1
∣∣∣∣ < ε
⇒ 1 − ε <
∣∣∣∣ log(ζk)
ζk − 1
∣∣∣∣ < 1 + ε
⇒ (1 − ε)(ζk − 1) < | log(ζk)| < (1 + ε)(ζk − 1).
Therefore, by comparison test (Theorem 1.3.30), our claim holds.
�
�
“book” — 2014/6/4 — 21:09 — page 268 — #46
�
�
�
�
�
�
268 Infinite Product
Thus, by definition of absolute convergence of infinite product and by our
claim, we have
∞∏
k=1
ζk converges absolutely iff
∞∑
k=1
log(ζk) converges absolutely
iff
∞∑
k=1
(ζk − 1) converges absolutely. �
PROBLEM 5.4.9 Prove that
∞∏
n=2
(
1 − 1
n2
)
converges and find its value.
Solution:
Let Pm =
m∏
n=2
(
1 − 1
n2
)
, ∀m ∈ N with m ≥ 2. Therefore,
Pm =
m∏
n=2
(
1 − 1
n
)(
1 + 1
n
)
=
m∏
n=2
n − 1
n
n + 1
n
=
(
1
2
3
2
) (
2
3
4
3
)
· · ·
(
m − 1
m
m + 1
m
)
= 1
2
m + 1
m
→ 1
2
� 0 as m → ∞.
Hence,
∞∏
n=2
(
1 − 1
n2
)
converges to
1
2
.
PROBLEM 5.4.10 Prove that
∞∏
n=1
(1 + z2n
) = 1
1 − z
, when |z| < 1.
Solution:
If Pm(z) =
m−1∏
n=1
(1 + z2n
), ∀m ∈ N, then we claim that Pm(z) =
2m−1∑
k=0
zk ,
∀m ∈ N. For m = 1, our claim is true. Assume that our claim is true for some
m ∈ N, then for
Pm+1(z) =
m∏
n=1
(1 + z2n
)
= Pm(z) (1 + z2m
)
�
�
“book” — 2014/6/4 — 21:09 — page 269 — #47
�
�
�
�
�
�
Series Developments and Infinite Products269
=
(
2m−1∑
k=0
zk
)
(1 + z2m
) (by assumption)
=
2m−1∑
k=0
zk +
2m−1∑
k=0
zk+2m
=
2m−1∑
k=0
zk +
2m+1−1∑
j=2m
z j (using the change of variable j = k + 2m)
=
2m+1−1∑
j=0
z j.
Hence, by induction, our claim holds. Since |z| < 1,
lim
m→∞ Pm(z) = lim
m→∞
2m−1∑
k=0
zk = 1
1 − z
using Example 1.3.31.
PROBLEM 5.4.11 Prove that
∞∏
n=1
(
1 + z
n
)
e
−z
n converges absolutely and
uniformly on every compact subset of C.
Solution:
To prove this statement, we shall obtain that
∞∑
n=1
log
([
1 + z
n
]
e
−z
n
)
converges
absolutely and uniformly on {z ∈ C : |z| ≤ R}, ∀R > 0 with respect to
the principal branch of log. As principal logarithm is analytic at 1, we can
expand the Taylor series expansion of log(1+w) = −
∞∑
k=1
1
k
wk , ∀w ∈ B(0, 1).
Therefore, for each |z| ≤ R and each n > R, we have∣∣∣log
([
1 + z
n
]
e
−z
n
)∣∣∣ =
∣∣∣log
(
1 − z
n
)
− z
n
∣∣∣
=
∣∣∣∣∣
∞∑
k=1
(−1)k
k
( z
n
)k − z
n
∣∣∣∣∣
=
∣∣∣∣∣
∞∑
k=2
(−1)k
k
( z
n
)k
∣∣∣∣∣
�
�
“book” — 2014/6/4 — 21:09 — page 270 — #48
�
�
�
�
�
�
270 Infinite Product
≤
∞∑
k=2
1
k
(
R
n
)k
≤ 1
2
R2
n2
∞∑
k=0
(
R
n
)k
≤ 1
2
R2
n2
⎛
⎜⎝ 1
1 − R
n
⎞
⎟⎠
= R2
2n(n − R)
,
which is the nth term of a convergent series of real numbers (i.e., this
term is independent of z). Hence, by comparison test,
∞∑
n=1
log
([
1 + z
n
]
e
−z
n
)
converges absolutely and uniformly on {z ∈ C : |z| ≤ R}.
Exercise 5.4.12 Prove the following:
1.
∞∏
n=2
(
1 + 2n + 1
n2 − 1
)
= 1
3
.
2.
∞∏
n=2
(
1 − 2n + 1
n(n + 2)
)
= 3.
3.
∞∏
n=2
(
1 + (−1)n
n
)
= 1.
Definition 5.4.13 Let f be an entire function with zeroes {an : n ∈ N}. Then,
the representation
f (z) = zm exp(g(z))
∞∏
n=1
(
1 − z
ak
)
exp(Pn(z)), ∀z ∈ C
is called the canonical product representation of f , where g is an entire
function, Pn are some polynomials, and m ∈ {0, 1, 2, 3, . . .}.
�
�
“book” — 2014/6/4 — 21:09 — page 271 — #49
�
�
�
�
�
�
Series Developments and Infinite Products 271
THEOREM 5.4.14 (Weierstrass theorem on existence of entire functions
with prescribed zeroes)
Let {an} be a sequence of complex numbers such that an → ∞ as n → ∞.
Then, there exists an entire function f whose zeroes are {an : n ∈ N}.
Proof:
Case (i) Let f be an entire function without any zeroes in C. In view of
Theorem 4.5.12, we observe that if f is an entire function and
has no zeroes, then there exists an entire function g such that
g(z) = log( f (z)), ∀z ∈ C and g′(z) = f ′(z)
f (z)
, ∀z ∈ C. Now, we
claim that f exp(−g) is a constant function on C.
d
dz
(f (z) exp(−g(z))) = f ′(z) exp(−g(z)) − exp(−g(z))g′(z)f (z)
= f ′(z) exp(−g(z)) − exp(−g(z))
f ′(z)
f (z)
f (z)
= f ′(z) exp(−g(z)) − exp(−g(z))f ′(z)
= 0.
Hence, f exp(−g) = k for some 0 � k ∈ C. Then, choose c ∈ C
such that exp(c) = k. Therefore, f (z) exp(−g(z)) = exp(c), ∀z ∈ C.
Thus, g + c is an entire function such that f = exp(g + c).
Case (ii) If f has finite number of zeroes, say for example, and
a1, a2, a3, . . . , an ∈ C\{0} including multiplicities and 0 is a zero
of order m (where m ≥ 0), then applying Result 5.2.3 repeatedly,
we get an entire function h, which has no zeroes in C, such that
f (z) = zm(z − a1)(z − a2) · · · (z − an)h(z)
= zm
n∏
k=1
(
1 − z
ak
)
× (−1)n
n∏
k=1
ak × h(z), ∀z ∈ C.
Therefore, by Case (i), there exists an entire function g such that
(−1)n
n∏
k=1
akh(z) = exp(g(z)), ∀z ∈ C.
�
�
“book” — 2014/6/4 — 21:09 — page 272 — #50
�
�
�
�
�
�
272 Infinite Product
Thus, we can write
f (z) = zm exp(g(z))
n∏
k=1
(
1 − z
ak
)
, ∀z ∈ C.
Case (iii) Let f has countably infinite number of zeroes {an : n ∈ N} such
that an → ∞ as n → ∞. In this case, with the motivation from
Case (ii), we shall find an entire function f such that
f (z) = zm exp(g(z))
∞∏
n=1
(
1 − z
ak
)
, ∀z ∈ C.
However, the convergence of
∞∏
n=1
(
1 − z
ak
)
is not guaranteed
for an arbitrary sequence of complex numbers with an →
∞ as n → ∞. Thus, we have to find a suitable sequence
of polynomials (Pn) such that
∞∏
n=1
(
1 − z
ak
)
exp(Pn(z)) con-
verges. We know that
∞∏
n=1
(
1 − z
ak
)
exp(Pn(z)) converges iff
∞∑
n=1
[
log
(
1 − z
ak
)
+ Pn(z)
]
converges with respect to a suitable
branch of log. Let R > 0 be given. Choose n0 ∈ N such that
|an| > 2R, ∀n ≥ n0. If |z| ≤ R and n ≥ n0, then log
(
1 − z
an
)
is analytic and its Taylor series is given by
log
(
1 − z
an
)
= −
∞∑
k=1
1
k
(
z
an
)k
= −
n∑
k=1
1
k
(
z
an
)k
−
∞∑
k=n+1
1
k
(
z
an
)k
.
Therefore, for n ≥ n0, if Pn(z) =
n∑
k=1
1
k
(
z
an
)k
, ∀z ∈ B(0, R) then
∣∣∣∣log
(
1 − z
an
)
+ Pn(z)
∣∣∣∣ ≤
∞∑
k=n+1
∣∣∣∣∣1k
(
z
an
)k
∣∣∣∣∣
≤ 1
n + 1
∞∑
k=n+1
(
R
|an|
)k
�
�
“book” — 2014/6/4 — 21:09 — page 273 — #51
�
�
�
�
�
�
Series Developments and Infinite Products 273
≤
(
R
|an|
)n+1 ∞∑
k=0
(
R
|an|
)k (
since
1
n + 1
≤ 1
)
=
(
R
|an|
)n+1
⎛
⎜⎜⎝ 1
1 − R
|an|
⎞
⎟⎟⎠
≤ M
(
1
2
)n+1
for some M > 0.
Since
⎛
⎜⎜⎝ 1
1 − R
|an|
⎞
⎟⎟⎠ converges to 1, such an M > 0 exists, by
Corollary 1.3.4. Hence, by comparison test,
∞∑
n=1
log
(
1 − z
an
)
+
Pn(z) converges absolutely on B(0, R). Therefore, a most general
entire function that has the zeroes {an : n ∈ N}, is of the form
f (z) = zm exp(g(z))
∞∏
n=1
(
1 − z
ak
)
exp(Pn(z)), ∀z ∈ C,
for some entire function g. �
COROLLARY 5.4.15 Every meromorphic function on C is a quotient of two
entire functions at all points of C except at the zeroes of f .
Proof: Let f be an arbitrary meromorphic function. From Corollary 5.2.12,
we observe that f has at most countable number of poles. If there are
countably infinite number of poles, then they must tend to ∞, otherwise,
it should have a finite limit point, which is a contradiction to the fact that the
poles are isolated. Hence, we can apply the previous theorem, and we can find
an entire function G such that its zeroes are the poles of f . Hence, f · G has
removable singularity at every pole of f, and it is analytic at every other points
of C. If F is an entire function such that ( f · G)(z) = F(z), for all point of C
except at the poles of f , then clearly, f = F
G
at all points of C except at the
poles of f . �
THEOREM 5.4.16 (Euler’s product for sin)
sin(πz) = πz
∞∏
n=1
(
1 − z2
n2
)
.
�
�
“book” — 2014/6/4 — 21:09 — page 274 — #52
�
�
�
�
�
�
274 Infinite Product
Proof: As sin(πn) = 0, for every n ∈ Z, from Theorem 5.4.14, we write
sin(πz) = z exp(g(z))
∞∏
n=−∞
n�0
(
1 − z
n
)
exp
( z
n
)
,∀z ∈ C,
for some entire function g. Applying logarithmic derivative on both sides, we
get
π cos(πz)
sin(πz)
= 1
z
+ g′(z) +
∞∑
n=−∞
n�0
(
1
z − n
+ 1
n
)
.
From Example 5.3.3, we have
π cot(πz) = 1
z
+
∞∑
n=−∞
n�0
(
1
z − n
+ 1
n
)
.
Hence, it follows that g′(z) = 0, ∀z ∈ C ⇒ g(z) = c, ∀z ∈ C for some
constant c ∈ C. Therefore,
sin(πz) = z exp(c)
∞∏
n=−∞
n�0
(
1 − z
n
)
exp
( z
n
)
,∀z ∈ C.
Using the easy fact that
sin(πz)
πz
→ 1 as z → 0, we observe that
1 = lim
z→0
sin(πz)
πz
= lim
z→0
z exp(c)
∞∏
n=−∞
n�0
(
1 − z
n
)
exp
( z
n
)
πz
= 1
⇒ exp(c)
π
= 1
⇒ exp(c) = π
⇒ sin(πz) = πz
∞∏
n=−∞
n�0
(
1 − z
n
)
exp
( z
n
)
⇒ sin(πz) = πz
∞∏
n=1
(
1 − z
n
) (
1 + z
n
)
exp
( z
n
)
exp
(
− z
n
)
⇒ sin(πz) = πz
∞∏
n=1
(
1 − z2
n2
)
.
�
�
�
“book” — 2014/6/4 — 21:09 — page 275 — #53
�
�
�
�
�
�
Series Developments and Infinite Products 275
COROLLARY 5.4.17 cos(πz) =
∞∏
n=1
(
1 − 4z2
(2n − 1)2
)
Proof: Using sin(2πz) = 2 sin(πz) cos(πz), we have
cos(πz) = sin(2πz)
2 sin(πz)
=
2πz
∞∏
n=1
(
1 − 4z2
n2
)
2πz
∞∏
n=1
(
1 − z2
n2
)
=
∞∏
n=1
(
1 − 4z2
n2
)
∞∏
n=1
(
1 − z2
n2
) . (5.9)
Applying the following observation
∞∏
n=1
(
1 − 4z2
n2
)
=
∞∏
k=1
(
1 − 4z2
(2k − 1)2
) ∞∏
k=1
(
1 − 4z2
(2k)2
)
=
∞∏
k=1
(
1 − 4z2
(2k − 1)2
) ∞∏
k=1
(
1 − z2
k2
)
in equation (5.9), we get
cos(πz) =
∞∏
n=1
(
1 − 4z2
(2n − 1)2
)
.
�
COROLLARY 5.4.18 (Wallis’ formulae)π
2
=
∞∏
n=1
4n2
4n2 − 1
.
√
π = lim
n→∞
1√
n
n∏
j=1
2j
2j − 1
.
�
�
“book” — 2014/6/4 — 21:09 — page 276 — #54
�
�
�
�
�
�
276 Infinite Product
Proof: Substituting z = 1
2
in the Euler’s product for sin, we get
sin
(π
2
)
= π
2
∞∏
n=1
(
1 − 1
4n2
)
⇒ 1 = π
2
∞∏
n=1
(
4n2 − 1
4n2
)
,
which implies the first formula.
We rewrite the first formula as
π
2
=
∞∏
n=1
4n2
4n2 − 1
= lim
n→∞
n∏
j=1
(2j)2
(2j − 1)2
2j − 1
2j + 1
= lim
n→∞
⎡
⎣ n∏
j=1
(2j)2
(2j − 1)2
×
(
1
3
· 3
5
· · · 2n − 1
2n + 1
)⎤⎦
= lim
n→∞
⎡
⎣ n∏
j=1
(2j)2
(2j − 1)2
× 1
2n + 1
⎤
⎦
= lim
n→∞
1
2n
n∏
j=1
(2j)2
(2j − 1)2
(
as lim
n→∞
1
1 + 1
2n
= 1
)
.
This implies that
√
π = lim
n→∞
1√
n
n∏
j=1
2j
2j − 1
. �
Exercise 5.4.19
1. Prove the following:
(a) sinh(πz) = πz
∞∏
n=1
(
1 + z2
n2
)
.
(b) cosh(πz) =
∞∏
n=1
(
1 + z2
(2n − 1)2
)
.
2. Prove that
√
2 =
∞∏
n=1
(
1 + (−1)n+1
2n − 1
)
. Hint: Substitute z = π
4
in the
expansion of cos(πz).
3. Prove that π
2 =
∞∏
n=1
(2n)2
(2n − 1)(2n + 1)
. Hint: Substitute z = 1
2
in the
expansion of
sin(πz)
πz
.
�
�
“book” — 2014/6/4 — 21:09 — page 277 — #55
�
�
�
�
�
�
Series Developments and Infinite Products 277
RESULT 5.4.20
∞∑
n=1
1
n2
= π2
6
and
∞∑
n=1
1
n4
= π4
90
.
Proof: 4Recall that the Euler’s product for sin is given by
sin(πx)
πx
=
∞∏
n=1
(
1 − x2
n2
)
, ∀x ∈ [0, 1).
If g(x) =
∞∑
n=1
log
(
1 − x2
n2
)
, then from the proof of Lemma 5.4.4, we see
that
sin(πx)
πx
= exp(g(x)). Therefore,
log
(
sin(πx)
πx
)
= g(x) =
∞∑
n=1
log
(
1 − x2
n2
)
=
∞∑
n=1
∞∑
k=1
(
x2
n2
)k
k
, ∀x ∈ [0, 1).
Since
∞∑
n=1
∞∑
k=1
∣∣∣∣∣∣∣
(
x2
n2
)k
k
∣∣∣∣∣∣∣ =
∞∑
n=1
log
(
1 − |x|2
n2
)
= g(|x|) < +∞,
we can apply Fubini’s theorem5 to get
log
(
sin(πx)
πx
)
=
∞∑
k=1
( ∞∑
n=1
1
n2k
)
x2k
k
=
∞∑
k=1
( ∞∑
n=1
1
n2k
)
x2k
k
, ∀x ∈ [0, 1).
(5.10)
On the other hand, we have
log
(
sin(πx)
πz
)
= log
(
1 −
[
(πx)2
3!
− (πx)4
5!
+ · · ·
])
=
[
(πx)2
3!
− (πx)4
5!
+ · · ·
]
+ 1
2
[
(πx)2
3!
− (πx)4
5!
+ · · ·
]2
+ · · ·
= 1
3!
(πx)2 +
(
− 1
5!
+ 1
2 × (3!)2
)
(πx)4
4There are plenty of proofs for these identities. We present here the Euler’s proof.
5Fubini’s theorem is a well-known big theorem on integral over a product measure space. As
it is beyond the scope of this book, we have not even stated this theorem. Interested readers can
refer to any book on measure and integration to see this theorem.
�
�
“book” — 2014/6/4 — 21:09 — page 278 — #56
�
�
�
�
�
�
278 Gamma Function and Its Properties
+
(
1
7!
− 1
3! × 5!
+ 1
3 × (3!)3
)
(πx)6 + · · · . (5.11)
Comparing the coefficients of x2 and x4 from equations (5.10) and (5.11), we
get
π2
6
=
∞∑
n=1
1
n2
and π4
(
− 1
120
+ 1
72
)
= 1
2
∞∑
n=1
1
n4
.
Thus, we get
π2
6
=
∞∑
n=1
1
n2
and
π4
90
=
∞∑
n=1
1
n4
. �
Exercise 5.4.21 Prove that
π6
945
. Hint: Compare the coefficients of x6 from
equations (5.10) and (5.11).
5.5 GAMMA FUNCTION AND ITS PROPERTIES
Definition 5.5.1 The gamma function � is defined by
�(z) = exp(−γ z)
z
∞∏
n=1
(
1 + z
n
)−1
exp
( z
n
)
, ∀z ∈ C\{−n : n ∈ N},
where γ = lim
n→∞
(
1 + 1
2
+ 1
3
+ · · · + 1
n
− ln(n)
)
, which is called the
Euler’s constant.
First, we note that the infinite product
∞∏
n=1
(
1 + z
n
)
exp
(
− z
n
)
converges
uniformly on the compact subsets of C. If |z| ≤ r and n > 2r, then
∣∣∣log
(
1 + z
n
)
− z
n
∣∣∣ =
∣∣∣∣
(
z
n
− z2
2n2
+ z3
3n3
− · · ·
)
− z
n
∣∣∣∣
≤
∞∑
k=2
rk
knk
<
∞∑
k=2
rk
nk
= r2
n2
∞∑
k=0
rk
nk
�
�
“book” — 2014/6/4 — 21:09 — page 279 — #57
�
�
�
�
�
�
Series Developments and Infinite Products 279
= r2
n2
(
1 − r
n
)−1
(as 0 <
r
n
<
1
2
< 1)
= 2
r2
n2
(as
(
1 − r
n
)−1
< 2).
Hence,
∞∑
n=1
log
(
1 + 1
n
)
− z
n converges uniformly on compact subsets
of C. Equivalently, the infinite product
∞∏
n=1
(
1 + z
n
)
exp
(
− z
n
)
converges
uniformly on the compact subsets of C. If G(z) =
∞∏
n=1
(
1 + z
n
)
exp
(
− z
n
)
,
then �(z) = 1
z exp(γ z)G(z) , and hence, � is well defined on C \ {−n : n ∈ N}.
THEOREM 5.5.2 (Properties of �)
The � function satisfies the following properties:
1. �(z + 1) = z�(z), ∀z ∈ C,
2. �(n) = (n − 1)!, ∀n ∈ N,
3. �
(
1
2
)
= √
π ,
4. Legendre’s duplication formula
√
π�(2z) = 22z−1�(z)�
(
z + 1
2
)
, ∀z ∈ C.
Proof:
1. Let G(z) =
∞∏
n=1
(
1 + z
n
)
exp
(
− z
n
)
, ∀z ∈ C. Hence, the zeroes of
G are −1,−2,−3, . . . . This implies that the zeroes of G(z − 1) are
0,−1,−2,−3, . . . . Therefore, from Theorem 5.4.14, we have
G(z − 1) = z exp(φ(z))G(z), ∀z ∈ C,
for some entire function φ. That is, for each z ∈ C, we have
∞∏
n=1
(
1+ z − 1
n
)
exp
(
− (z − 1)
n
)
= z exp(φ(z))
∞∏
n=1
(
1+ z
n
)
exp
(
− z
n
)
.
�
�
“book” — 2014/6/4 — 21:09 — page 280 — #58
�
�
�
�
�
�
280 Gamma Function and Its Properties
Finding the logarithmic derivative of the above equation, we get
∞∑
n=1
(
1
z − 1 + n
− 1
n
)
= 1
z
+ φ′(z) +
∞∑
n=1
(
1
z + n
− 1
n
)
. (5.12)
Now,
∞∑
n=1
(
1
z − 1 + n
− 1
n
)
= 1
z
− 1 +
∞∑
n=2
(
1
z − 1 + n
− 1
n
)
= 1
z
− 1 +
∞∑
m=1
(
1
z + m
− 1
m + 1
)
= 1
z
− 1 +
∞∑
n=1
(
1
z + n
− 1
n
)
+
∞∑
n=1
(
1
n
− 1
n + 1
)
= 1
z
− 1 +
∞∑
n=1
(
1
z + n
− 1
n
)
+ lim
n→∞
(
1 − 1
n + 1
)
= 1
z
+
∞∑
n=1
(
1
z + n
− 1
n
)
.
Applying the last observation in equation (5.12), we get φ′(z) =
0, ∀z ∈ C, and hence, it is a constant, say, C. To find the value of
C, substitute z = 1 in G(z − 1) = z exp(φ(z))G(z) to get
G(0) = exp(C)G(1) ⇒ 1 = exp(C)
∞∏
n=1
(
1 + 1
n
)
exp
(
−1
n
)
,
which implies that
−C =
∞∑
n=1
(
ln
(
1 + 1
n
)
− 1
n
)
= lim
N→∞
N∑
n=1
(
ln
(
1 + 1
n
)
− 1
n
)
�
�
“book” — 2014/6/4 — 21:09 — page 281 — #59
�
�
�
�
�
�
Series Developments and Infinite Products 281
= lim
N→∞
(
ln (N + 1)−
(
1 + 1
2
+ 1
3
+ · · · + 1
N
))
= lim
N→∞
(
ln (N)+ ln
(
N + 1
N
)
−
(
1 + 1
2
+ 1
3
+ · · · + 1
N
))
= lim
N→∞
(
ln (N)−
(
1 + 1
2
+ 1
3
+ · · · + 1
N
))
,
as lim
N→∞
ln
(
N+1
N
)
= ln(1) = 0. Thus, C = γ , the Euler’s constant.
Thus, we have
G(z − 1) = z exp(γ )G(z), ∀z ∈ C.
From the definition of � and G, we can easily write that
�(z) = 1
z exp(γ z)G(z)
�(z + 1) = 1
(z + 1) exp(γ (z + 1))G(z + 1)
= 1
(z + 1) exp(γ (z + 1))G(z)(z + 1)−1 exp(−γ )
= 1
exp(γ z)G(z)
= z�(z).
2. We prove this identity by induction on n.
�(1) = 1
exp(γ )G(1)
= 1 = 0!, as 1 = G(0) = exp(γ )G(1).
Assume that this identity holds for some n ∈ N. Now, by using (1) and
induction hypothesis, we get �(n + 1) = n�(n) = n(n − 1)! = n! thus,
�(n) = n!, ∀n ∈ N.
3. From the definition of G and Example 5.3.3, we have zG(z)G(−z) =
z
∞∏
n=1
(
1 − z2
n2
)
= sin(πz)
π
. Therefore, for every z ∈ C,
�(1 − z)�(z) = (−z)�(−z)�(z)
= (−z)
1
(−z) exp(−γ z)G(−z)
1
z exp(γ z)G(z)
= 1
zG(−z)G(z)
= π
sin(πz)
.
�
�
“book” — 2014/6/4 — 21:09 — page 282 — #60
�
�
�
�
�
�
282 Gamma Function and Its Properties
Substituting z = 1
2
, we get
�
(
1 − 1
2
)
�
(
1
2
)
= π
1
⇒ �
(
1
2
)
= √
π .
4. Finding the logarithmic derivative on both sides of
�(z) = exp(−γ z)
z
∞∏
n=1
(
1 + z
n
)−1
exp
( z
n
)
,
we get
�′(z)
�(z)
= −1
z
− γ +
∞∑
n=1
( −1
z + n
+ 1
n
)
.
Finding the derivative on both sides of the last equation, we get
d
dz
(
�′(z)
�(z)
)
= 1
z2
+
∞∑
n=1
1
(z + n)2
=
∞∑
n=0
1
(z + n)2
.
Therefore,
d
dz
(
�′(z)
�(z)
)
+ d
dz
⎛
⎜⎜⎝
�′
(
z + 1
2
)
�
(
z + 1
2
)
⎞
⎟⎟⎠
=
∞∑
n=0
1
(z + n)2
+
∞∑
n=0
1(
z + 1
2
+ n
)2
= 4
( ∞∑
n=0
1
(2z + 2n)2
+
∞∑
n=0
1
(2z + 2n + 1)2
)
= 4
∞∑
m=0
1
(2z + m)2
= 4 × 1
2
d
dz
(
�′(2z)
�(2z)
)
= 2
d
dz
(
�′(2z)
�(2z)
)
.
�
�
“book” — 2014/6/4 — 21:09 — page 283 — #61
�
�
�
�
�
�
Series Developments and Infinite Products 283
Integrating on both sides, we get
�′(z)
�(z)
+
�′
(
z + 1
2
)
�
(
z + 12
) = 2
�′(2z)
�(2z)
+ a,
for some constant a ∈ C Again integrating on both sides, we get
log (�(z))+ log
(
�
(
z + 1
2
))
= log (�(2z))+ az + b,
which implies the following equation.
�(z) × �
(
z + 1
2
)
= exp(az + b)�(2z). (5.13)
Substituting z = 1
2
and z = 1 in equation (5.13), we get
�
(
1
2
)
�(1) = exp
(a
2
+ b
)
�(1) ⇒ a
2
+ b = log(
√
π ), (5.14)
�(1)�
(
3
2
)
= exp(a + b)�(2) ⇒ a + b = log
( √
π
2
)
. (5.15)
Equation (5.15)−equation (5.14) ⇒ a
2
= log
( √
π
2
)
− log(
√
π )
= − log(2) ⇒ a = −2 log 2.
Using this value in equation (5.15), we get b = log(2) + 1
2
log(π).
Therefore, from equation (5.13), we get
�(z)�
(
z + 1
2
)
= exp
(
(−2 log(2))z + log(2) + 1
2
log(π)
)
�(2z)
= 2−2z+1 √π�(2z).
Thus, we get
√
π�(2z) = 22z−1�(z)�
(
z + 1
2
)
.
�
�
�
“book” — 2014/6/4 — 21:09 — page 284 — #62
�
�
�
�
�
�
284 Gamma Function and Its Properties
Example 5.5.3 Prove that �(z) = lim
N→∞
NzN!
z(z + 1) · · · (z + N)
, ∀z ∈
C\{−1,−2,−3, . . .}.
Solution:
For z ∈ C\{0,−1,−2,−3, . . .}, we have
�(z) = exp(−γ z)
z
∞∏
n=1
(
1 + z
n
)−1
exp
( z
n
)
= lim
N→∞
exp
(
−
[
N∑
n=1
1
n − ln(N)
]
z
)
z
· lim
N→∞
N∏
n=1
(
1 + z
n
)−1
exp
( z
n
)
= lim
N→∞
⎛
⎜⎜⎜⎝
exp
(
−
[
N∑
n=1
1
n − ln(N)
]
z
)
z
·
N∏
n=1
(
z + n
n
)−1
exp
( z
n
)
⎞
⎟⎟⎟⎠
= lim
N→∞
N∏
n=1
exp
(
− z
n
) exp (z ln(N))
z
·
N∏
n=1
(
n
z + n
)
exp
( z
n
)
= lim
N→∞
Nz
z
·
N∏
n=1
(
n
z + n
)
= lim
N→∞
NzN!
z(z + 1) · · · (z + N)
.
In the following example, we justify that � has some other integral rep-
resentation. However, this justification depends on dominated convergence
theorem. Hence, the following example is for those who are familiar with the
dominated convergence theorem.
Example 5.5.4 Prove that �(z) =
∞∫
0
tz−1 exp(−t) dt, ∀z = (x, y) ∈ C with
x > 0.
Solution:
First, we claim that the above integral exists for all z = (x, y) ∈ C with x > 0.
Since exp(−t) ≤ 1,∀t ∈ [0,∞], we have∣∣∣tz−1 exp(−t)
∣∣∣ < |exp((z − 1) log t)|
�
�
“book” — 2014/6/4 — 21:09 — page 285 — #63
�
�
�
�
�
�
Series Developments and Infinite Products 285
= |exp((x − 1) log t) exp(iy log t)|
= |exp((x − 1) log t)| .
Since for x > 0,
∞∫
0
tx−1 exp(−t) dt =
1∫
0
tx−1 exp(−t) dt +
∞∫
1
tx−1 exp(−t) dt
= lim
ε→0
1∫
ε
tx−1 exp(−t) dt +
∞∫
1
tx−1 exp(−t) dt
= lim
ε→0
1 − εx
x
+
∞∫
1
tx−1 exp(−t) dt
<
1
x
+
∞∫
1
exp(−t/2) dt < +∞,
our claim holds. Next, we claim that if G(z) =
∞∫
0
tz−1 exp(−t) dt, ∀z ∈ {z ∈
C : Re z > 0}, then G is an analytic function. Fix z0 ∈ C such that Re z > 0.
Now, for z0 ∈ C such that Re z > 0, by applying dominated convergence
theorem, we get
lim
z→z0
∣∣∣∣∣∣
G(z) − G(z0)
z − z0
−
∞∫
0
tz−1 log(t) exp(−t) dt
∣∣∣∣∣∣
= lim
z→z0
∞∫
0
(
tz−1 − tz0−1
z − z0
− tz0−1 log(t)
)
exp(−t) dt
=
∞∫
0
lim
z→z0
(
tz−1 − tz0−1
z − z0
− tz0−1 log(t)
)
exp(−t) dt
= 0.
Hence, G is analytic on {z ∈ C : Re z > 0}.
Using exp(−t) = lim
n→∞
(
1 − 1
n
)n
, ∀t > 0, we get G(z) = lim
n→∞ Gn(z),
∀z ∈ {z ∈ C : Re z > 0}, where Gn(z) =
n∫
0
tz−1
(
1 − t
n
)n
dt, ∀z ∈ C,
with Re z > 0, ∀n ∈ N. We note that
�
�
“book” — 2014/6/4 — 21:09 — page 286 — #64
�
�
�
�
�
�
286 Gamma Function and Its Properties
Gn(z) =
n∫
0
tz−1
(
1 − t
n
)n
dt
= nz
1∫
0
sz−1(1 − s)n ds
(by using the change of variable t = ns)
=
⎧⎨
⎩
1
z
(
n
n−1
)z+1
Gn+1(z + 1), n ≥ 2
1
z(z+1) , n = 1
.
(by applying integration by parts)
By iterating n − 1 times, we get
Gn(z) = nzn!
z(z + 1)(z + 2) · · · (z + n)
, ∀z ∈ C with Re z > 0, ∀n ≥ 1,
and hence,
G(z) = lim
n→∞
nzn!
z(z + 1)(z + 2) · · · (z + n)
, ∀z ∈ C with Re z > 0.
Thus, by using the previous example, we get
G(z) = �(z), z ∈ C with Re z > 0.
Remark 5.5.5: The integral
∞∫
0
tz−1 exp(−t) dt does not exist for z ∈ C with
Re z < 0.
Indeed, if z = x + iy with x < 0 and ε > 0, then
∞∫
0
|tz−1 exp(−t)| dt ≥
1∫
ε
tx−1 exp(−t) dt = exp(−1)
1 − εx
x
→ ∞ as ε → 0.
�
�
“book” — 2014/6/4 — 21:16 — page 287 — #1
�
�
�
�
�
�
6
Residue Calculus
6.1 RESIDUE
Definition 6.1.1 Let f be an analytic function except at a singularity ‘a’.
The residue of f at ‘a’ is defined to be a complex number R, which makes
f (z) − R
z − a
as the derivative of an analytic function on B(a, δ)\{a} for some
δ > 0.
Residue of f at a is unique, and hence, it is denoted by (Res f )(a).
LEMMA 6.1.2 If f is an analytic function on � except at an isolated
singularity a, then residue of f at a exists and is unique.
Proof: Since a is an isolated singularity, then there exists δ > 0 such that f is
analytic on (Cl B(a, δ))\{a}. For 0 < r < δ, let C denote the circle |z−a| = r
and R = 1
i2π
∫
C
f (z) dz. We claim that
∫
γ
(
f (z) − R
z − a
)
dz = 0 for every
closed curve γ in B(a, δ)\{a}.
Case 1: γ is homologous to 0 in B(a, δ)\{a}.
If
F(z) = f (z) − R
z − a
, ∀z ∈ Cl B(a, δ)\{a}.
then F is analytic on Cl B(a, δ)\{a}, and by general version of
Cauchy’s theorem, we get
∫
γ
F(z) dz = 0.
Case 2: γ is not homologous to 0 in B(a, δ)\{a}.
Then, WN(γ , a) = m for some m ∈ Z\{0}. We claim that γ−mC ∼ 0
287
�
�
“book” — 2014/6/4 — 21:16 — page 288 — #2
�
�
�
�
�
�
288 Residue
in B(a, δ)\{a}. If z0 � B(a, δ)\{a}, then either z0 = a or z0 � B(a, δ).
If z0 = a, then
WN(γ − mC, z0) = WN(γ , a) − mWN(C, a) = m − m × 1 = 0.
If z0 � B(a, δ), then
WN(γ , z0) = 0 and WN(C, z0) = 0 and WN(γ − mC, z0) = 0.
Hence, by general version of Cauchy’s theorem, we get
0 =
∫
γ−mC
(
f (z) − R
z − a
)
dz
=
∫
γ−mC
(
f (z) − R
z − a
)
dz
=
∫
γ
(
f (z) − R
z − a
)
dz − m
∫
C
(
f (z) − R
z − a
)
dz,
which implies
∫
γ
(
f (z) − R
z − a
)
dz = m
⎛
⎝∫
C
f (z) dz − R
∫
C
dz
z − a
⎞
⎠
= m
⎛
⎝i2πR − R
∫
C
dz
z − a
⎞
⎠
= m (i2πR − Ri2π) = 0.
Hence, by Result 4.1.30, F is the derivative of an analytic function.
Thus, residue of f at a exists, and (Res f )(a) = R = 1
i2π
∫
C
f (z) dz.
If there are R, S ∈ C such that f (z) − R
z − a
and f (z) − S
z − a
are derivatives of the analytic functions � and � on B(a, δ)\{a}
and B(a, ε)\{a}, respectively, then obviously, R − S ∈ C and
(� − �)′(z) = S − R
z − a
in B(a, min{δ, ε}) \ {a}. Hence, by Corollary
4.1.29, we get∫
C
S − R
z − a
dz = 0 ⇒ (R − S)i2π = 0 ⇒ R = S.
Thus, the lemma follows. �
�
�
“book” — 2014/6/4 — 21:16 — page 289 — #3
�
�
�
�
�
�
Residue Calculus 289
LEMMA 6.1.3 If ‘a’ is an isolated singularity of f and if the Laurent series
expansion of f in B(a, r)\{a} ( for some r > 0) is f (z) =
∞∑
n=−∞
cn(z − a)n,
then (Res f )(a) = c−1.
Proof: We know that (z − a)n is the derivative of
(z − a)n+1
n + 1
inside
B(a, r)\{a}. Using the fact that
lim sup
n→∞
∣∣∣∣ cn
n + 1
∣∣∣∣
1
n
= lim sup
n→∞
|cn| 1
n and lim sup
n→∞
∣∣∣∣ c−n
−n + 1
∣∣∣∣
1
n
= lim sup
n→∞
|c−n| 1
n ,
we conclude that
∞∑
n=−∞
n�−1
cn
n + 1
(z − a)n converges in the neighbourhood of a
(except at a) on which
∞∑
n=−∞
n�−1
cn(z− a)n+1 converges (cf. the proof of Theorem
2.3.2). Hence,
f (z) − c1
z − a
=
∞∑
n=−∞
n�−1
cn(z − a)n = d
dz
⎛
⎜⎝ ∞∑
n=−∞
n�−1
cn
n + 1
(z − a)n+1
⎞
⎟⎠ ,
for every 0 < |z − a| < r. Thus, (Res f )(a) = c−1. �
RESULT 6.1.4 If f is a meromorphic function with a pole ‘a’ of order m, then
(Res f ) (a) = 1
(m − 1)!
lim
z→a
dm−1
dzm−1
((z − a)mf (z)).
Proof: If f has a pole of order m at a, then by Result 5.2.14, we have
f (z) =
∞∑
n=−m
cn(z − a)n, ∀z ∈ B(a, r)\{a},
for some r > 0. From the previous lemma, we get
1
(m − 1)!
lim
z→a
dm−1
dzm−1
(
(z − a)mf (z)
)
= 1
(m − 1)!
lim
z→a
dm−1
dzm−1
(
(z − a)m
∞∑
n=−m
cn(z − a)n
)
�
�
“book” — 2014/6/4 — 21:16 — page 290 — #4
�
�
�
�
�
�
290 Residue
= 1
(m−1)!
lim
z→a
dm−1
dzm−1
( ∞∑
n=0
cn−m(z − a)n
)
which is equal to c−1 if m = 1.
= 1
(m − 1)!
lim
z→a
∞∑
n=m−1
cn−m
(
m−2∏
ν=0
(n − ν)
)
(z − a)n−m+1if m ≥ 2
= 1
(m − 1)!
(m − 1)!c−1
= c−1
= (Res f )(a).
Thus, the result follows. �
Example 6.1.5 Find the residue of f at all poles.
(1) f (z) = z2 + 3z − 1
; (2) f (z) = z
(z + 1)(z − 2)
;
(3) f (z) = 2z + 3
(z + 2)2(z − 3)
; (4) f (z) = z2 + 2
(z + 2)3
.
Solution:
1. If f (z) = z2 + 3
z − 1
, then 1 is the only simple pole of f . Then,
(Res f )(1) = lim
z→1
(z − 1)f (z) = lim
z→1
(z2 + 3) = 4.
2. If f (z) = z
(z + 1)(z − 2)
, then −1 and 2 are the simple poles of f . Then,
(Res f )(−1) = lim
z→−1
(z + 1)f (z) = lim
z→−1
z
z − 2
= 1
3
and (Res f )(2) = lim
z→2
(z − 2)f (z) = lim
z→2
z
z + 1
= 2
3
.
3. If f (z) = 2z + 3
(z + 2)2(z − 3)
, then −2 is a double pole of f , and 3 is a
simple pole of f . Then,
(Res f )(−2) = lim
z→−2
d
dz
[(z + 2)2f (z)] = lim
z→−2
d
dz
(
2z + 3
z − 3
)
= lim
z→−2
−9
(z − 3)2
= −9
25
and (Res f )(3) = lim
z→3
(z − 3)f (z) = lim
z→3
2z + 3
(z + 2)2
= 9
25
.
�
�
“book” — 2014/6/4 — 21:16 — page 291 — #5
�
�
�
�
�
�
Residue Calculus 291
4. If f (z) = z2 + 2
(z + 2)3
, then −2 is the only pole of f of order 3. Then,
(Res f )(−2) = 1
2
lim
z→−2
d2
dz2
[(z + 2)3f (z)]
= 1
2
lim
z→−2
d2
dz2
(
z2 + 2
)
= 1
2
× 2 = 1.
Example 6.1.6 Evaluate the residues of the following functions at their
poles:
(1)
z2 + 1
z2 + 5z + 6
; (2)
z + 1
(z2 + 1)2
; (3)
exp(z)
(z + 3)(z − 2)2
.
Solution:
1. Let f (z) = z2 + 1
z2 + 5z + 6
, ∀z ∈ C. The poles of f are the zeroes of the
polynomial z2 + 5z + 6. Therefore, z = −2 and z = −3 are the simple
poles of f . Hence,
(Res f ) (−2) = lim
z→−2
(z + 2)f (z)
= lim
z→−2
(z + 2)
z2 + 1
z2 + 5z + 6
= lim
z→−2
z2 + 1
z + 3
= 5.
(Res f ) (−3) = lim
z→−3
(z + 3)f (z)
= lim
z→−3
(z + 3)
z2 + 1
z2 + 5z + 6
= lim
z→−3
z2 + 1
z + 2
= −10.
2. Let f (z) = z + 1
(z2 + 1)2
, ∀z ∈ C. z = i and z = −i are the poles of the
order 2 for f .
(Res f ) (i) = 1
(2 − 1)!
lim
z→i
d
dz
[
(z − i)2 z + 1
(z2 + 1)2
]
�
�
“book” — 2014/6/4 — 21:16 — page 292 — #6
�
�
�
�
�
�
292 Residue
= lim
z→i
d
dz
[
z + 1
(z + i)2
]
= lim
z→i
[
(z + i)2 − (z + 1)2(z + i)
(z + i)4
]
= (i2)2 − (i + 1)2(i2)
(i2)4
= −4 − i4 + 4
16
= − i
4
.
(Res f ) (−i) = 1
(2 − 1)!
lim
z→−i
d
dz
[
(z + i)2 z + 1
(z2 + 1)2
]
= lim
z→−i
d
dz
[
z + 1
(z − i)2
]
= lim
z→−i
[
(z − i)2 − (z + 1)2(z − i)
(z − i)4
]
= i
4
.
3. Let f (z) = exp(z)
(z + 3)(z − 2)2
, ∀z ∈ C. z = −3 is a simple pole, and
z = 2 is a pole of the order 3 for f .
(Res f ) (−3) = lim
z→−3
[
(z + 3)
exp(z)
(z + 3)(z − 2)2
]
= lim
z→−3
exp(z)
(z + 3)(z − 2)2
= exp(−3)
25
.
(Res f ) (2) = 1
(3 − 1)!
lim
z→2
d2
dz2
[
(z − 2)3 exp(z)
(z + 3)(z − 2)2
]
= 1
2
lim
z→2
d2
dz2
[
exp(z)
z + 3
]
= 1
2
lim
z→2
d
dz
[
(z + 2) exp(z)
(z + 3)2
]
�
�
“book” — 2014/6/4 — 21:16 — page 293 — #7
�
�
�
�
�
�
Residue Calculus 293
= 1
2
lim
z→2
exp(z)[(z + 3)2 − 2(z + 2)]
(z + 3)3
= 17
250
exp(2).
Example 6.1.7 Find the residues of f at the given singularities.
(1)
sin(3z)
z
at 0; (2)
cos(z)
2z
at 0; (3) exp
(
2
3z
)
at 0; (4)
1
cos(z)
at
π
2
.
Solution:
1. Let f (z) = sin(3z)
z
, ∀z � 0. Since sin(3z) =
∞∑
k=0
(−1)k(3z)2k−1
(2k − 1)!
, we
have lim
z→0
sin(z)
z
= lim
z→0
(
3 +
∞∑
k=1
(−1)k(3z)2k
3(2k + 1)!
)
= 3 in C. Hence, 0 is
a removable singularity for f , and hence, (Res f )(0) = 0.
2. Let f (z) = cos(z)
2z
, ∀z � 0. By a similar argument we get
cos 3z
z
= 1
2z
+
∞∑
k=1
(−1)k(z)2k−1
2(2k)!
, which is the Larurent series
expansion of f in 0 < |z| <∞. Hence, (Res f )(0) = 1
2
(the coefficient
of
1
z
).
3. If f (z) = exp
(
2
3z
)
, then f (z) =
∞∑
k=0
(
2
3z
)k
k!
. Hence, 0 is an essential
singularity of f , and hence, (Res f )(0) = 2
3
.
4. Let f (z) = 1
cos(z)
, ∀z � (2k − 1)
π
2
, k ∈ Z. Clearly, z = π
2
is a simple pole. Hence, (Res f )(0) = lim
z→
π
2
(
z − π
2
) 1
cos(z)
=
lim
z→
π
2
(
z − π
2
) −1
sin(z − π
2
)
= −1.
�
�
“book” — 2014/6/4 — 21:16 — page 294 — #8
�
�
�
�
�
�
294 Cauchy’s Residue Theorem
Exercise 6.1.8 Evaluate the residues of f at its singularities.
(1)
sin(z)
z2
at 0; (2)
1
sin2(z)
at π ; (3)
exp(z) − 1
z4
at z = 0.
Answers: (1) 1; (2) 0; (3)
1
6
.
RESULT 6.1.9 If f and g are analytic functions on a region � such that
f (z0) � 0 and g has a simple zero at z0, then
(
Res
f
g
)
(z0) = f (z0)
g′(z0)
.
Proof: By assumption, it is clear that
f
g
has a simple pole at z0. Hence, by
Lemma 6.1.4, (
Res
f
g
)
(z0) = lim
z→z0
(z − z0)
f (z)
g(z)
= lim
z→z0
f (z)
g(z)
z−z0
= lim
z→z0
f (z)
g(z)−g(z0)
z−z0
=
lim
z→z0
f (z)
lim
z→z0
g(z)−g(z0)
z−z0
= f (z0)
g′(z0)
.
Hence, the result follows. �
6.2 CAUCHY’S RESIDUE THEOREM
THEOREM 6.2.1 (Cauchy’s residue theorem)
Let f be an analytic function on a region � except at a finite number of
singularities. Then,
∫
γ
f (z) dz = i2π
n∑
j=1
WN(γ , aj)(Res f )(aj),
for every closed curve γ in � \ {a1, a2, . . . , an} such that γ ∼ 0 in �.
�
�
“book” — 2014/6/4 — 21:16 — page 295 — #9
�
�
�
�
�
�
Residue Calculus 295
Proof: Let a1, a2, a3, . . . , an be the isolated singularities of f such that aj �
ak if j � k. Choose δj > 0 such that
1. Cl B(aj, δj) ⊆ � \ γ and
2. Cl B(aj, δj) ∩ Cl B(ak , δk) = ∅ for every j, k ∈ {1, 2, . . . , n} with j � k.
In view of proof of Lemma 6.1.2, if Cj is the circle |z − aj| = δj, then we
have (Res f )(aj) = 1
i2π
∫
Cj
f (z) dz. We claim that γ −
n∑
j=1
WN(γ , aj)Cj ∼ 0
in �∗ = �\{a1, a2, a3, . . . , an}. If z0 � �∗, then either z0 � � or z0 = aj
for some j ∈ {1, 2, 3, . . . , n}. If z0 � �, then WN(γ , z0) = WN(Cj, z0) = 0
for all j ∈ {1, 2, 3, . . . , n}, and hence, WN
(
γ −
n∑
j=1
WN(γ , aj)Cj
)
= 0. If
z0 = ak , then WN(Cj, z0) =
{
1 if j = k
0 if j � k
for all j ∈ {1, 2, 3, . . . , n}, and
hence, WN
(
γ −
n∑
j=1
WN(γ , aj)Cj
)
= WN(γ , ak) − WN(γ , ak) = 0.
Hence, our claim holds. Therefore, by general version of Cauchy’s theorem,
we get
0 =
∫
γ−
n∑
j=1
WN(γ ,aj)Cj
f (z) dz =
∫
γ
f (z) dz −
n∑
j=1
WN(γ , aj)
∫
Cj
f (z) dz,
which implies that
∫
γ
f (z) dz = i2π
n∑
j=1
WN(γ , aj)(Res f )(aj). �
Example 6.2.2 Evaluate the following integrals using Cauchy’s residue the-
orem:
1.
∫
|z|=3
z + 1
z2 + 7z + 10
dz.
2.
∫
|z|=4
3z2 − 2z + 1
z2 − 2z
dz.
3.
∫
|z|=3
z + 3
(z2 + 6z + 8)(z2 + 1)
dz.
�
�
“book” — 2014/6/4 — 21:16 — page 296 — #10
�
�
�
�
�
�
296 Cauchy’s Residue Theorem
4.
∫
|z−1|=3
z2 + 2
(z − 2)2(z + 4)
dz.
5.
∫
|z+1−i|=6
z
(z2 − 2z − 15)(z2 + 1)
dz.
6.
∫
γ
sin(z)
(z2 + 1)2
dz, where γ is the rectangle with vertices (−1, 0), (1, 0),
(1, 2), and (−1, 2).
7.
∫
|z|=1
4z + 1
(z2 + 1)(z2 + 2)
dz.
8.
∫
|z|=4
z3 + 1
z(z + 1)(z + 2)
dz.
9.
∫
|z−1|=5
exp(z) − 1
sin(z)
dz.
10.
∫
|z|=1
exp(z)
sin(z)
dz.
Solution:
1. Let f (z) = z + 1
z2 + 7z + 10
, ∀z ∈ C\{−2,−5}. Clearly, −2 and −5 are
the simple poles of f . We note that −2 is enclosed by |z| = 3 and −5
is not enclosed by |z| = 3. Therefore, by Cauchy’s residue theorem, we
get ∫
|z|=3
z + 1
z2 + 7z + 10
dz = i2π (Res f )(−2)
= i2π lim
z→−2
(z + 2)f (z)
= i2π lim
z→−2
z + 1
z + 5
= −i2π
3
.
2. Let f (z) = 3z2 − 2z + 1
z2 − 2z
dz, ∀z ∈ C\{0, 2}. Clearly, 0 and 2 are the
simple poles of f and both are enclosed by |z| = 4. Therefore, by
Cauchy’s residue theorem, we get
�
�
“book” — 2014/6/4 — 21:16 — page 297 — #11
�
�
�
�
�
�
Residue Calculus 297
∫
|z|=4
3z2 − 2z + 1
z2 − 2z
dz = i2π [(Res f )(0) + (Res f )(2)] .
Since
(Res f )(0) = lim
z→0
zf (z)
= lim
z→0
3z2 − 2z + 1
z − 2
= −1
2
(Res f )(2) = lim
z→2
(z − 2)f (z)
= lim
z→2
3z2 − 2z + 1
z
= 9
2
,
we get ∫
|z|=4
3z2 − 2z + 1
z2 − 2z
dz = i2π
[−1
2
+ 9
2
]
= i8π .
3. Let f (z) = z + 3
(z2 + 6z + 8)(z2 + 1)
,∀z ∈ C\{−2,−4, i,−i}. Clearly,
−2,−4, i, and −i are simple poles of f , and −2, i, and −i are enclosed
by |z| = 3, but −4 is not enclosed by |z| = 3. Therefore, by Cauchy’s
residue theorem,∫
|z|=3
z + 3
(z2 + 6z + 8)(z2 + 1)
dz
= i2π [(Res f )(−2) + (Res f )(−i) + (Res f )(i)] .
Since
(Res f )(−2) = lim
z→−2
(z + 2)f (z) = lim
z→−2
z + 3
(z + 4)(z2 + 1)
= −1