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Math 116 - Homework 3 Solutions
February 1, 2017
Problem 1 Solution
1. Apply theorem 4.8 Chapter 2 page 52 to f − g: since f − g is holomorphic and vanishes on the
real axis it must be identically 0.
2. Fix w ∈ R. Then the function ezew is a product of an entire function and a real constant, so it is
entire. ez+w is of the form f(z+w), when f(z) = ez and w a constant. f(z+w) satisfies the C-R
equations, since a translation of a function just translates its derivatives. So both ez+w, ezew are
entire, and we know that these two functions agree on z ∈ R. Therefore, for any w ∈ R we have
ez+w = ezew
Now, fix z ∈ C and consider these two functions as functions of w. They are entire (by the same
reasoning), and since they agree on w ∈ R, they have to agree on the entire complex plane.
Problem 2 Solution Let f(z) =
∑
anz
n and g(z) =
∑
bnz
n. Then f and g are holomorphic
functions around 0: One way to show it would be to use Morera’s theorem: Define fN =
∑N
n=1 anz
n.
fN converge uniformly to f , so
´
T
fN converges to
´
T
f . Since
´
T
fN = 0 (each fN is holomorphic)
so is
´
T
f . Now, by using analytic continuation (Theorem 4.8 in page 52), we see that f agrees with
g where they are defined, since f − g vanishes on xn = 1/n and limxn = 0 is in this domain. We
conclude that an = bn.
Problem 3 Solution sin z would be an example: it vanishes on npi, but it is not the zero function.
sin z is holomorphic by Theorem 2.6, page 16.
Problem 4 Solution Since f ◦g is an entire function which agrees with the identity function on the
real axis, they must agree everywhere by analytic continuation. Therefore f ◦ g(z) = z for all z ∈ C.
In particular this requires g to be 1 − 1. Now choose xn = 1/n, then g(xn) will be a set in C with
an accumulation point by continuity. Thus g ◦ f(g(xn)) = g(xn) , and g ◦ f = id on some set with
accumulation point, which again means g ◦ f is identity by analytic continuation again.
Problem 5 Solution Let x be a real number, consider C be the circle r = 1/2 center at x. If η ≥ 0
|f (n)(x)| ≤ n!||f ||C
rn
≤ n!A · sup
z∈C
(1 + |z|)η2n
≤ 2nn!A · (1 + 1
2
+ |x|)η
≤ 2nn!A
(
3
2
)η
(1 + |x|)η
1
If η < 0 we have
|f (n)(x)| ≤ n!||f ||C
rn
≤ n!A · sup
z∈C
(1 + |z|)η2n
≤ 2nn!A · (1
2
+ |x|)η
≤ 2nn!A · (1
2
+
1
2
|x|)η
≤ 2nn!A
(
1
2
)η
(1 + |x|)η
Problem 6 Solution First, we can assume that z0 = 0 since we can define g(z) on Ω − z0 as
g(z) = ϕ(z + z0)− z0. g(0) = 0 and if g is linear so is ϕ. Assume there exists an, the first coefficient
after n = 1 in the power expansion of ϕ, so that an 6= 0. The power series of ϕ is
ϕ = z + anz
n +O(zn+1)
Note that
ϕ2 = ϕ ◦ ϕ = ϕ+ anϕn +O(φn+1) = z + 2anzn +O(zn+1)
To understand the last equation, one just have to identify the contribution to zk from powers of ϕ.
The only contribution to z is from ϕ’s first term. Only ϕ’s second term and ϕn’s first term contribute
to zn. All other terms of ϕ will be of the form czk for k greater than n.
By repeating this argument we have ϕk(z) = z + kan + O(zn+1). Let C be a circle centered at 0
with radius r. Apply Cauchy inequality to ϕk’s nth term, and we have
|k||an| =
∣∣∣∣ϕ(n)(0)n!
∣∣∣∣ ≤ ||ϕ||Crn
||ϕ||c is bounded since ϕ maps into Ω, which is bounded. Since this inequality holds for any k, an
must be zero - and we have a contradiction. So, all the terms anzz after the first one in the power
series expansion of ϕ must be zero and ϕ(z) = z.
Problem 7 Solution We claim that there exists n and uncountable number of points z ∈ C such
that f (n)(z) = 0. In fact, consider the closed disk centered at 0 with radius R, call this disc D. For
every z0 ∈ D, there exists m such that f (m)(z0) = 0, since the power series expansion at that point
has 0 coefficient for some term. Let Un =
{
z ∈ D : f (n)(z) = 0}. We have:⋃
n∈N
Un = D
then one of the sets Un has to be uncountable, thus infinite. Since D is compact, it means f (n) vanishes
on some converging sequence in D, which means it vanishes on C by analytic continuation.
Problem 8 Solution Assume f(x+ iy) = u(x, y) + iv(x, y). Then for g(z) = f
(
1
z¯
)
, which is defined
on C − D, we have that g(t + is) = f
(
t+is
t2+s2
)
. If g(t, s) = U(t, s) + iV (t, s), we have the following
relations:
U(t, s) = u
(
t
t2 + s2
,
s
t2 + s2
)
V (t, s) = −v
(
t
t2 + s2
,
s
t2 + s2
)
We see that
Ut = ux
∂
∂t
(
t
t2 + s2
)
+ uy
∂
∂t
(
s
t2 + s2
)
= ux
(
−t2 + s2
(t2 + s2)
2
)
+ uy
(
−2st
(t2 + s2)
2
)
Vs = −vx ∂
∂s
(
t
t2 + s2
)
− vy ∂
∂s
(
s
t2 + s2
)
= −vx
(
−2ts
(t2 + s2)
2
)
− vy
(
t2 − s2
(t2 + s2)
2
)
2
Since f satisfies the C-R equations, ux = vy and vx = −uy, therefore Ut = Vs. Similarly Us = −Vt,
so g is holomorphic. Since f is non vanishing, so is g, and in this case 1/g(z) is also holomorphic in
C − D. 1/g and f agree on the unit circle. Using the symmetry principle (together with the mobius
transformation that takes the unit disk to the uppe half plane, τ = z−iz+i ) the function
F (z) =
{
f(z) z ∈ D
1/g(z) z ∈ C− D
is entire. Since f is non vanishing, on the unit disk r < |f | < R for some r,R > 0 by compactness.
Hence |F | < max {R, 1/r} , and we conclude that F is constant.
An Alternative solution would be to apply the maximum modulus principle (that you will learn
later in this course) to f and 1/f .
Problem 9 Solution If f and g both go to 0 at 0, then we can assume that
f(z) = znh1(z)
g(z) = zmh2(z)
where n ≥ 1 and m ≥ 1 are the orders of zeros of f and g at 0, and h1(z) and h2(z) are holomorphic
non vanishing function on some small neighborhood around 0. Taking the derivatives of f and g we
shall have
f ′(z) = nzn−1h1(z) + znh′(z) = zn−1(nh1(z) + z · h′1(z))
g′(z) = mzm−1h2(z) + zmh′(z) = zm−1(mh2(z) + z · h′2(z))
where
(mh2(z) + z · h′2(z))
and
(nh1(z) + z · h′1(z))
are also non vanishing holomorphic function around 0. Now, for the limit
lim
z→0
f/g = lim
z→0
znh1(z)
zmh2(z)
If n 6= m, this limit will go to 0 if n > m or diverge if m > n. We see the same behavior in the limit of
zn−1(nh1(z) + z · h′1(z))
zm−1(mh1(z) + z · h′2(z))
since the factors multiplying zn−1 and zm−1 do not vanish when |z| is small.
If n = m then
lim f/g = lim
h1(z)
h2(z)
= lim
nh1(z) + z · h′1(z)
nh2(z) + z · h′2(z)
We conclude that
lim f/g = lim
f ′
g′
in all cases.
Problem 10 Solution Let Pn → f . Theorem 5.2 page 53 states that such limit of holomorphic
function is holomorphic, and we need to show that the limit is a polynomial of degree at most N .
First Pn(0) has to converge to f(0), which means that the constant coefficient in the polynomials Pn
is a converging sequence. Theorem 5.3 says that under this condition we also have that P ′n converges
to f ′ uniformly on every compact sets. Thus P ′n(0) has to converge to f ′(0), which means that the
coefficient for the linear terms of Pn is also a converging sequence. We can repeat this argument N
times and conclude that the coefficient of those polynomials all converges.
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Consider the closed unit disk and write
Pn =
N∑
k=0
an,kz
k
we know that for any k, an,k → bk for some bk ∈ C. Let g(z) =
∑N
k=0 bkz
k. Then on the closed unit
disk
Pn(z)→ g(z)
and g(z) is a polynomial. Since the limit is unique, we have g(z) agree with f(z) on the unit disk, thus
everywhere on C by analytic continuation.
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