HomeW2

Prévia do material em texto

SOLUTIONS TO HW2
Stein and Shakarchi, Chapter 1
Exercise 13. Let f be a holomorphic function on an open set Ω, and write f(z) = u(x, y) +
iv(x, y) as usual, with u and v differentiable.
(a) Assume <(f) = u(x, y) is constant, so that
∂u
∂x
= 0 and
∂u
∂y
= 0.
Since f is holomorphic, it satisfies the Cauchy-Riemann equations, and so we also get
∂v
∂y
= 0 and − ∂v
∂x
= 0,
which implies that v(x, y) is constant, and hence f is constant.
(b) The proof is the same as the one in (a).
(c) Assume |f | is constant, so that |f(z)|2 = u(x, y)2 + v(x, y)2 is also constant. We
differentiate this relation with respect to x and y and use the chain rule and the Cauchy-
Riemann equations to get the following equations:
2
(
u(x, y)
∂u
∂x
− v(x, y)∂u
∂y
)
CR eqs.
= 2
(
u(x, y)
∂u
∂x
+ v(x, y)
∂v
∂x
)
= 0
2
(
u(x, y)
∂u
∂y
+ v(x, y)
∂u
∂x
)
CR eqs.
= 2
(
u(x, y)
∂u
∂y
+ v(x, y)
∂v
∂y
)
= 0
Multiplying the first equation by u(x,y)
2
and the second equation by v(x,y)
2
and adding, we
obtain
(u(x, y)2 + v(x, y)2)
∂u
∂x
= 0.
If u(x, y)2+v(x, y)2 = 0, then each of u(x, y) and v(x, y) are 0, so the claim follows. Otherwise,
u(x, y)2 + v(x, y)2 6= 0, and we see that ∂u
∂x
= 0. A similar argument shows that ∂u
∂y
= 0, so
that u(x, y) is constant, and we conclude using part (a).
Remark. Part (c) can be done more elegantly as follows. The two equations above are
equivalent to the matrix equation(
u(x, y) −v(x, y)
v(x, y) u(x, y)
)(
∂u
∂x
∂u
∂y
)
=
(
0
0
)
.
The matrix on the left has determinant u(x, y)2+v(x, y)2. Therefore, if u(x, y)2+v(x, y)2 = 0,
we proceed as before, and if u(x, y)2+v(x, y)2 6= 0, then we may invert the matrix to conclude
that ∂u
∂x
= ∂u
∂y
= 0.
1
2 SOLUTIONS TO HW2
Exercise 14. Let {an}Nn=1 and {bn}Nn=1 be two sequences of complex number, and set Bk =∑k
n=1 bn. Note that Bn−Bn−1 = bn. We then have (by reindexing the sum on the third line)
N∑
n=M
anbn =
N∑
n=M
an(Bn −Bn−1)
=
N∑
n=M
anBn −
N∑
n=M
anBn−1
= aNBN − aMBM−1 +
N−1∑
n=M
anBn −
N∑
n=M+1
anBn−1
= aNBN − aMBM−1 +
N−1∑
n=M
(an − an+1)Bn.
Exercise 15. Roughly, the idea is to apply the previous exercise with an = r
n and bn = an.
Let {an}∞n=1 be a sequence such that
∑∞
n=1 an converges, and set Ak =
∑k
n=1 an for k ≥ 1,
A0 = 0, and A = limn→∞An =
∑∞
n=1 an. Then we have, for r < 1,
N∑
n=1
rnan = r
NAN − rA0 −
N−1∑
n=1
(rn+1 − rn)An
= rNAN + (1− r)
N−1∑
n=1
rnAn
= rNAN + (1− r)
N−1∑
n=0
rnAn,
where the last line holds since A0 = 0. Since we are interested in the limit as r approaches 1
from the left, we may assume that 0 < r < 1. Under this assumption, taking the limit as N
goes to infinity in the equation above, we obtain
∞∑
n=1
rnan = (1− r)
∞∑
n=0
rnAn.
SOLUTIONS TO HW2 3
Now let ε > 0 be given, and choose N0 so that |A− An| ≤ ε2 for n > N0. Then we have∣∣∣∣∣
∞∑
n=1
rnan − A
∣∣∣∣∣ =
∣∣∣∣∣(1− r)
∞∑
n=0
rnAn − A
∣∣∣∣∣
?
=
∣∣∣∣∣(1− r)
∞∑
n=0
rn(An − A)
∣∣∣∣∣
≤ (1− r)
(
N0∑
n=0
rn|An − A|+ ε
2
∞∑
N0+1
rn
)
≤ (1− r)
(
N0∑
n=0
rn|An − A|+ ε
2
rN0+1
1− r
)
= (1− r)
N0∑
n=0
rn|An − A|+ ε
2
rN0+1
≤ (1− r)
N0∑
n=0
rn|An − A|+ ε
2
where the equality (?) follows from the identity
∑∞
n=0 r
n = (1− r)−1 when |r| < 1. Now, the
expression (1− r)∑N0n=0 rn|An−A| is a polynomial in r with real coefficients, which takes on
the value 0 when r = 1. Therefore, by continuity of polynomials, there exists a δ > 0 such
that, if 1− r < δ holds, we have
(1− r)
N0∑
n=0
rn|An − A| ≤ ε
2
.
Therefore, taking r in the range 1− δ < r < 1, we obtain∣∣∣∣∣
∞∑
n=1
rnan − A
∣∣∣∣∣ ≤ (1− r)
N0∑
n=0
rn|An − A|+ ε
2
≤ ε,
which shows that limr→1−
∑∞
n=1 r
nan = A.
Exercise 16. (a) When n ≥ 3, we have the inequalities
1 ≤ (log(n))2 ≤ n2,
which implies
1 = lim sup
n→∞
1 ≤ lim sup
n→∞
(log(n))2/n ≤ lim sup
n→∞
(n1/n)2 = 1,
and therefore the radius of convergence is 1.
(b) We proceed by using the Ratio Test. We have
lim
n→∞
∣∣∣∣an+1an
∣∣∣∣ = limn→∞n+ 1 =∞,
and therefore the radius of convergence is 0.
(c) When n ≥ 1, we have
4n ≤ 4n + 3n ≤ 2 · 4n,
4 SOLUTIONS TO HW2
which implies
n2
2 · 4n ≤
n2
4n + 3n
≤ n
2
4n
.
This gives
1
4
= lim sup
n→∞
(n1/n)2
21/n · 4 ≤ lim supn→∞
(
n2
4n + 3n
)1/n
≤ lim sup
n→∞
(n1/n)2
4
=
1
4
,
which implies that the radius of convergence if 4.
(d) A more precise statement of Stirling’s formula says that there exist two constants c and
d such that 0 < c < d and, for all n ≥ 1, we have
cnn+1/2e−n ≤ n! ≤ dnn+1/2e−n.
Therefore, we have the inequalities
c3n3n+3/2e−3n ≤ (n!)3 ≤ d3n3n+3/2e−3n
and
c33n+1/2n3n+1/2e−3n ≤ (3n)! ≤ d33n+1/2n3n+1/2e−3n,
or, equivalently,
d−13−3n−1/2n−3n−1/2e3n ≤ 1
(3n)!
≤ c−13−3n−1/2n−3n−1/2e3n.
We deduce
c3
d
n
33n
√
3
≤ (n!)
3
(3n)!
≤ d
3
c
n
33n
√
3
,
which gives
1
27
= lim sup
n→∞
(
c3
d
)1/n
n1/n
33
√
3
1/n
≤ lim sup
n→∞
(
(n!)3
(3n)!
)1/n
≤ lim sup
n→∞
(
d3
c
)1/n
n1/n
33
√
3
1/n
=
1
27
.
Therefore the radius of convergence is 27.
(e) Note first that if α or β is a negative integer, then the series has only finitely many
nonzero terms, and thus converges on all of C. Assume now that α, β 6∈ {0,−1,−2, . . .}. We
again use the ratio test. Let
an =
α(α + 1) · · · (α + n− 1)β(β + 1) · · · (β + n− 1)
n!γ(γ + 1) · · · (γ + n− 1) ;
we then have
an+1
an
=
(α + n)(β + n)
(n+ 1)(γ + n)
.
Now choose n ≥ |γ|; the triangle equality and reverse triangle inequality give us
(n− |α|)(n− |β|)
(n+ 1)(n+ |γ|) ≤
∣∣∣∣(α + n)(β + n)(n+ 1)(γ + n)
∣∣∣∣ ≤ (n+ |α|)(n+ |β|)(n+ 1)(n− |γ|) .
Therefore, we get
1 = lim sup
n→∞
(n− |α|)(n− |β|)
(n+ 1)(n+ |γ|) ≤ lim supn→∞
∣∣∣∣(α + n)(β + n)(n+ 1)(γ + n)
∣∣∣∣ ≤ lim sup
n→∞
(n+ |α|)(n+ |β|)
(n+ 1)(n− |γ|) = 1,
and the radius of convergence is 1.
SOLUTIONS TO HW2 5
(f) If we let an =
(−1)n
n!(n+r)!4n
, we get
an+1
an
= − 1
(n+ 1)(n+ r + 1)4
;
thus,
lim sup
n→∞
∣∣∣∣− 1(n+ 1)(n+ r + 1)4
∣∣∣∣ = 0,
and the radius of convergence is ∞ (that is, the function converges on the entire complex
plane).
Exercise 23. A function is infinitely differentiable if it has derivatives of all orders. Let f(x)
be defined by
f(x) =
{
0 if x ≤ 0,
e−1/x
2
if x > 0.
It is clear that if x 6= 0, then f has derivatives of all orders at x, and is continuous in a
neighborhood of x; it therefore remains to examine the situation at x = 0. Note first that we
may compute the derivatives away from 0 by using the chain rule, and we obtain
f (n)(x) =
{
0 if x < 0,
e−1/x
2
Pn(x) if x > 0,
where Pn(x) is a polynomial in
1
x
(that is, Pn(x) =
∑N
n=1
an
xn
). By making the change of
variables t = 1/h, we obtain
lim
h→0+
e−1/x
2
Pn(x) = lim
t→∞
∑N
n=1 ant
n
et2
.
The expression inside the limit is a linear combination of functions of the form t
k
et2
; applying
L’Hoˆpital’s rule several times, we obtain
lim
t→∞
tk
et2
= lim
t→∞
k!
et2Qk(t)
= 0,
where Qk(t) is some polynomial in t. Therefore, we conclude that
lim
h→0+
e−1/x
2
Pn(x) = 0.
We must now compute the derivative at 0. We use the definition of derivative from the left
and right:
lim
h→0−
f(h)− f(0)
h
= lim
h→0−
0− 0
h
= 0,
lim
h→0+
f(h)− f(0)
h
= lim
h→0+
e−1/h
2
h
= lim
t→∞
t
et2
?
= lim
t→∞
1
2tet2
= 0,
6 SOLUTIONS TO HW2
where we have used L’Hoˆpital’s rule in (?). Therefore, f ′(0)= 0, f is differentiable, and has
a continuous first derivative.
Assume now by induction that f (n−1)(0) = 0. We have an expression for f (n) away from 0,
and at 0 we again use the definitions:
lim
h→0−
f (n−1)(h)− f (n−1)(0)
h
= lim
h→0−
0− 0
h
= 0,
lim
h→0+
f (n−1)(h)− f (n−1)(0)
h
= lim
h→0+
e−1/h
2
Pn−1(h)
h
= lim
t→∞
tPn−1(t−1)
et2
= 0,
where the last line follows from a computation similar to the one above. Therefore, we
conclude that f (n)(0) = 0 for all n ≥ 1, and therefore f is infintely differentiable. Moreover,
f does not have a converging power series expansion at 0, for if it did, we would necessarily
have
f(x) =
∞∑
n=0
f (n)(0)
n!
xn = 0,
which is absurd.
Exercise 25. (a) We may evaluate this integral directly. Let γ be parametrized by z(t) =
Reit, t ∈ [0, 2pi]. If n 6= −1, we have∫
γ
zndz =
∫ 2pi
0
RneintiReit dt
= iRn+1
∫ 2pi
0
ei(n+1)t dt
= iRn+1
[
1
i(n+ 1)
ei(n+1)t
]2pi
0
= 0;
if n = −1, we have ∫
γ
z−1dz =
∫ 2pi
0
R−1e−itiReit dt
= i
∫ 2pi
0
dt
= 2pii.
(If you’re still feeling shaky about these computations, break this up into real and imaginary
parts).
(b) We can use Corollary 3.3: if γ does not enclose the origin, then f(z) = zn has a
primitive defined on the interior of γ. For n 6= −1, then 1
n+1
zn+1 is a primitive, defined inside
γ (note that this also works for negative exponents). If n = −1, then F (reiθ) = log(r) + iθ
SOLUTIONS TO HW2 7
is a primitive for f(z) = z−1 (provided we restrict the domain of θ). For example, if γ lies
entirely in the right half plane, we take θ ∈ [−pi, pi]. Corollary 3.3 now implies∫
γ
zn dz = 0.
(c) Using partial fractions, we have
1
(z − a)(z − b) =
1
a− b
1
z − a +
1
b− a
1
z − b.
Therefore, if we use part (b) and a slight generalization of part (a), we get∫
γ
1
(z − a)(z − b) dz =
1
a− b
∫
γ
1
z − a dz +
1
b− a
∫
γ
1
z − b dz =
2pii
a− b.
Chapter 2
Exercise 1. Let f(z) = e−z
2
, and let γ = γ1 ∪ γ2 ∪ γ3, where γ1 is the straight line from 0
to R, γ2 is the arc z(t) = Re
it, t ∈ [0, pi
4
], and γ3 is the straight line from R
(
1√
2
+ i 1√
2
)
to 0.
Note first that, since f(z) is holomorphic, Cauchy’s theorem gives∫
γ1
e−z
2
dz +
∫
γ2
e−z
2
dz +
∫
γ3
e−z
2
dz =
∫
γ
e−z
2
dz = 0.
Consider first the integral over γ2. We must show rigorously that this integral goes to 0 as
R goes to ∞. We split the path into an upper and lower part: fix a positive real number h,
and let γL denote the lower part of γ2 with y coordinate less than
R√
2
− h, and let γU denote
the upper part of γ2 with y coordinate greater than
R√
2
− h (draw a picture to see what this
looks like). We will make use of the formula of Proposition 3.1(iii) which states∣∣∣∣∫
γ
f(z) dz
∣∣∣∣ ≤ sup
z∈γ
(|f(z)|) · length(γ).
We have |f(z)| = |e−z2| = e−<(z2) = e−x2+y2 , and on γ2, we have x ≥ R√2 , and therefore
|e−z2| ≤ e−R2/2ey2 on γ. Now, on γL, the function |e−z2| is bounded by e−R2/2e(R/
√
2−h)2 =
e−Rh
√
2+h2 , and the length is bounded by Rpi
4
(the full length of γ2). On γU , |e−z2| is bounded
by e−R
2/2eR
2/2 = 1, and the length is bounded by pi
2
h (notice that the length of γU is bounded
by the arc length of the quarter circle of radius h).
Combining everything, we obtain∣∣∣∣∫
γ2
e−z
2
dz
∣∣∣∣ ≤ ∣∣∣∣∫
γL
e−z
2
dz
∣∣∣∣+ ∣∣∣∣∫
γU
e−z
2
dz
∣∣∣∣
≤ pi
4
Re−Rh
√
2+h2 +
pi
2
h.
If we let h = 1√
R
, this gives∣∣∣∣∫
γ2
e−z
2
dz
∣∣∣∣ ≤ pi4Re−√2R+1/R + pi2√R,
which tends to 0 as R goes to infinity.
8 SOLUTIONS TO HW2
We now examine the integral over γ3. The inverse path −γ3 can be parametrized by
z(t) = t
(
1√
2
+ i 1√
2
)
, t ∈ [0, R], and therefore
−
∫
γ3
e−z
2
dz =
∫
−γ3
e−z
2
dz
=
∫ R
0
e−it
2
(
1√
2
+ i
1√
2
)
dt
=
(
1√
2
+ i
1√
2
)(∫ R
0
cos(t2)− i sin(t2) dt
)
.
To complete the computation, we let R tend to infinity, which shows∫ ∞
0
e−x
2
dx = −
∫
γ3
e−z
2
dz =
(
1√
2
+ i
1√
2
)(∫ ∞
0
cos(t2)− i sin(t2) dt
)
.
This implies∫ ∞
0
cos(t2)− i sin(t2) dt =
(
1√
2
− i 1√
2
)∫ ∞
0
e−x
2
dx =
√
pi
2
(
1√
2
− i 1√
2
)
;
comparing real and imaginary parts shows∫ ∞
0
cos(t2) dt =
∫ ∞
0
sin(t2) dt =
√
2pi
4
.
Exercise 2. Consider the function f(z) = 1
2i
eiz−1
z
, and note that, since f(z) = 1
2
∑∞
n=1
(iz)n−1
n!
,
this function is holomorphic on all of C. We let γε denote the contour for which −γε is
parametrized by z(t) = εeit, t ∈ [0, pi], and let γR denote the contour z(t) = Reit, t ∈ [0, pi].
Then Cauchy’s theorem gives∫
[−R,−ε]
1
2i
eiz − 1
z
dz +
∫
γε
1
2i
eiz − 1
z
dz +
∫
[ε,R]
1
2i
eiz − 1
z
dz +
∫
γR
1
2i
eiz − 1
z
dz = 0.
We evaluate each of these integrals separately. Consider first the integral over γε. Since
f(z) is holomorphic, the quantity |f(z)| is bounded by some M in some small disc of radius
ε0. Taking ε < ε0, we obtain∣∣∣∣∫
γε
1
2i
eiz − 1
z
dz
∣∣∣∣ ≤M · length(γε) = Mεpi.
Note that as long as ε < ε0, the bound on supz∈γε(|f(z)|) is independent of ε. Therefore,
letting ε tend to 0, we obtain ∫
γε
1
2i
eiz − 1
z
dz −→ 0.
We now consider the integral over γR, again by splitting γR into a lower and upper part.
We again fix a positive real number h, and let γL denote the part of γR with y coordinate
less than h, and let γU denote the part of γR with y coordinate greater than h. We have
|eiz| = e<(iz) = e−y, and therefore on γL the maximum of this function is 1, and on γU the
maximum is e−h. The length of γL is bounded by pih (since the length of the two small arc
SOLUTIONS TO HW2 9
is certainly bounded above by the length of half-circle of radius h), and the length of γU is
bounded by Rpi. Therefore, we obtain∣∣∣∣∫
γR
1
2i
eiz
z
dz
∣∣∣∣ = ∣∣∣∣∫
γL
1
2i
eiz
z
dz +
∫
γU
1
2i
eiz
z
dz
∣∣∣∣
≤
∣∣∣∣∫
γL
1
2i
eiz
z
dz
∣∣∣∣+ ∣∣∣∣∫
γU
1
2i
eiz
z
dz
∣∣∣∣
≤ 1
2R
pih+
1
2
e−h
R
Rpi
=
pih
2R
+
pie−h
2
.
Therefore, taking h =
√
R, we get∣∣∣∣∫
γR
1
2i
eiz
z
dz
∣∣∣∣ ≤ pi2√R + pie−
√
R
2
,
which tends to 0 as R goes to infinity. On the other hand, using the parametrization z(t) =
Reit, we get ∫
γR
− 1
2iz
dz = −1
2
∫ pi
0
dt
= −pi
2
.
Therefore, as R goes to infinity, we obtain∫
γR
1
2i
eiz − 1
z
dz −→ −pi
2
.
Upon combining everything and letting R tend to infinity and ε tend to 0, we get∫ ∞
−∞
1
2i
eix − 1
x
dx =
pi
2
;
taking the real part of both sides shows∫ ∞
0
sin(x)
x
dx =
1
2
∫ ∞
−∞
sin(x)
x
dx =
pi
2
.
Remark. Let P (z) and Q(z) be two polynomials such that deg(P (z)) + 1 ≤ deg(Q(z)). Then
the above proof actually shows that we have∫
γR
P (z)
Q(z)
eiz dz −→ 0
as R tends to infinity.
Exercise 3. Let a, b be two positive real numbers (we can always assume b is positive), let
A =
√
a2 + b2, and let 0 ≤ ω ≤ pi
2
be the unique angle for which cos(ω) = a
A
. Note that this
implies sin(ω) = b
A
. Let γ = γ1 ∪ γ2 ∪ γ3, where γ1 is the straight line from 0 to R, γ2 is the
10 SOLUTIONS TO HW2
curve parametrized by z(t) = Reit, t ∈ [0, ω], and γ3 is the straight line from Reiω to 0. By
Cauchy’s theorem, we have∫
γ1
e−Az dz +
∫
γ2
e−Az dz +
∫
γ3
e−Az dz =
∫
γ
e−Az dz = 0.
We first evaluate the integral along γ2. We have∫
γ2
e−Az dz =
[
−e
−Az
A
]z=Reiω
z=R
=
e−AR − e−AReiωA
=
e−AR − e−R(a+ib)
A
;
therefore ∣∣∣∣∫
γ2
e−Az dz
∣∣∣∣ ≤ e−AR + e−RaA ,
which tends to 0 as R tends to infinity.
Thus, letting R tend to infinity, we obtain∫
γ3
e−Az dz =
∫ ∞
0
(cos(ω) + i sin(ω))e−A(cos(ω)+i sin(ω))t dt
=
∫ ∞
0
(
a
A
+ i
b
A
)
e−at−ibt dt
=
∫ ∞
0
e−At dt
=
[
−e
−At
A
]∞
t=0
=
1
A
.
Equating real and imaginary parts (and using the relations cos(ω) = a
A
and sin(ω) = b
A
),
we get the pair of equations
a
A
∫ ∞
0
e−at cos(bt) dt+
b
A
∫ ∞
0
e−at sin(bt) dt =
1
A
b
A
∫ ∞
0
e−at cos(bt) dt− a
A
∫ ∞
0
e−at sin(bt) dt = 0
Writing this in matrix form, we get(
a
A
b
A
b
A
− a
A
)(∫∞
0
e−at cos(bt) dt∫∞
0
e−at sin(bt) dt
)
=
(
1
A
0
)
.
SOLUTIONS TO HW2 11
Inverting this, we get ∫ ∞
0
e−at cos(bt) dt =
a
A2
=
a
a2 + b2
,∫ ∞
0
e−at sin(bt) dt =
b
A2
=
b
a2 + b2
.