SOLUTIONS OF HW8
MINGFENG ZHAO
April 15, 2011
1. [Cartan’s book, Problem 9, in Page 77] Let D be a convex open set in the plane, and let f (z) be a holomorphic
function in D. Show that for any pair of points a, b ∈ D, we can choose two points c and d on the line segment joining
a and b such that
f (a) − f (b) = (a − b)[Re f 0 (c) + iIm f 0 (d)].
Proof. For any a, b ∈ D, since D is convex, then for all t ∈ [0, 1], we have ta + (1 − t)b ∈ D. Since f is analytic in D,
then u, v are smooth function on D. And by the Cauchy-Riemann equations, we can get
ux = vy
uy = −vx
f 0 (z) = ux + ivx
Consider the function:
F (t) = f (ta + (1 − t)b),
0 ≤ t ≤ 1.
Figure 1. ta + (1 − t)b ∈ D, 0 ≤ t ≤ 1
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2
MINGFENG ZHAO
Since f is analytic on D, then F is differentiable on [0, 1], and
F 0 (t) = f 0 (ta + (1 − t)b)(a − b)
And also we have
f (a) − f (b)
F (a) − F (b)
=
1
Z
F 0 (t)dt
=
By the fundamental theorem in calculus
0
1
Z
f 0 (ta + (1 − t)b)(a − b)dt
=
0
Z
1
(a − b)
=
(ux (ta + (1 − t)b) + ivx (ta + (1 − t)b))dt
0
Z
(a − b)
=
1
Z
0
For
R1
0
1
ux (ta + (1 − t)b)dt + i
vx (ta + (1 − t)b)dt
0
ux (ta + (1 − t)b)dt, by the mean value theorem in calculus, we know there exists 0 ≤ t1 ≤ 1 such that
1
Z
ux (ta + (1 − t)b)dt = ux (t1 a + (1 − t1 )b)
0
For
R1
0
vx (ta + (1 − t)b)dt, by the mean value theorem in calculus, we know there exists 0 ≤ t2 ≤ 1 such that
Z
1
vx (ta + (1 − t)b)dt = vx (t2 a + (1 − t2 )b)
0
Let c = t1 a + (1 − t1 )b, and d = t2 a + (1 − t2 )b, then c, d ∈ d, and c, d lie on the line segment joining a and b.
Therefore, we get
f (a) − f (b)
=
(a − b)(ux (c) + ivx (d))
=
(a − b)[Re f 0 (c) + iIm f 0 (d)].
SOLUTIONS OF HW8
3
In the following problems, we assume D is a domain in C, and let O(D) denote the holomorphic functions in D.
2. Suppose n ∈ Z+ , and γ is the positively oriented boundary of the unit circle. Compute
Z
ez z −n dz.
γ
Proof. Consider the function f (z) = ez , which is an entire function on C, and also for any n ∈ Z+ , we know
f (n−1) (z) = ez . By Cauchy’s integral formula, we get
1 = f (n−1) (0) =
(n − 1)!
2πi
Z
ez z −n dz
γ
Therefore, we obtain
Z
ez z −n dz =
γ
2πi
(n − 1)!
3. Suppose f ∈ O(BR (0)) for some R > 1. If for |z| < 1, we have
|f (z)| ≤
1
.
1 − |z|
Find the strongest possible upper bound for |f (n) (0)|.
Proof. For any 0 < r < 1, let Cr = {z : |z| = r} with positive orientation. By the Cauchy’s integral formula, we know
for all n ≥ 0, we have
f
(n)
(0)
=
n!
2πi
So we know
|f
(n)
(0)| =
≤
=
Z
n!
2πi
f (z) dz n+1
Cr z
Z f (z) n!
|dz|
2π Cr z n+1 Z
n!
|f (z)|
|dz|
2πi Cr |z|n+1
Z
Cr
f (z)
dz
z n+1
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MINGFENG ZHAO
≤
n!
2π
Z
Cr
1
|dz| By the assumption
(1 − |z|)|z|n+1
=
n!
2πr
·
2π (1 − r)rn+1
=
n!
rn − rn+1
Figure 2. g(r) = rn − rn+1 , 0 < r < 1
Consider the function g(r) = rn − rn+1 , 0 < r < 1, we get g 0 (r) = nrn−1 − (n + 1)rn , let g 0 (r) = 0, we know
r=
n
n+1
< 1. So we know
|g(r)| ≤ g
So when take r =
n
n+1
n
n+1
=
1−
n
n+1
n
n+1
n+1
=
nn+1
(n + 1)n+2
< 1, we get
|f (n) (0)| ≤
n!(n + 1)n+2
nn+1
4. We have discussed simply connected domains. Here we will examine what a non-simply connected domain might
look like-both geometrically and analytically. Please explain your answers. Assume D is a domain in C which is not
simply connected.
a. What is wrong with D? What makes it non-simply connected?
SOLUTIONS OF HW8
5
b. If F = C\D, can F be connected?
c. Suppose a ∈
/ D, and f (z) =
1
z−a .
If γ is a simple closed, piecewise differentiable rectifiable curve in D, is it
possible that
Z
f (z)dz 6= 0.
γ
d. What would that mean?
e. In the previous question, must such a curve γ exist when D is not simply connected?
Proof. a. Since D is not simply connected, then the Cauchy’s integral formula is wrong, that is,
Z
If f (z) is analytic on D, γ is a nice curve in D this does not imply that
f (z)dz = 0
γ
D is not simply connected because there exists some closed curve γ in D can not continuous deformed in D to a
point, that is, D contains holes.
b. Yes. For example D = {z : |z| > 1}, which is not simply connected, but F = C\D = {z : |z| ≤ 1} is connected.
Figure 3. D = {z : |z| > 1}, F = C\D = {z : |z| ≤ 1}
c. Yes. For example D = {z : |z| > 1}, which is not simply connected, a = 0 ∈
/ D = {z : |z| ≥ 1}, consider the closed
curve C2 = {z : |z| = 2} with positive orientation. And
Z
Cr
1
dz = 2πi 6= 0.
z
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MINGFENG ZHAO
Figure 4. D = {z : |z| > 1}
d. No. For example D = {z : |z| > 1, x < 2}, which is not simply connected, a = 3 ∈
/ D = {z : |z| ≥ 1, x ≤ 2}. For
any nice closed curve γ in D, we know a = 3 ∈
/ γ, but we know that
Z
γ
1
dz = 0.
z−3
Figure 5. D = {z : |z| > 1, x < 2}
5. Suppose D ⊂ C is a domain.
SOLUTIONS OF HW8
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a. (Morera’s theorem) Suppose f ∈ C(D), and
Z
f (z)dz = 0
γ
for every nice (i.e. piecewise differentiable, rectifiable) closed curve γ ⊂ D. Then f ∈ O(D). Prove it or give
a counterexample.
b. In Morera’s theorem, does the existence of such a function imply anything about the topology of the domain
D?
Proof. a. Take any point a ∈ D and fix a, now for any z ∈ D, let γz : [0, 1] → D be any nice curve which connects a
and z, then we define
Z
F (z) =
f (w)dw
γz
If we have another nice curve γ̃z : [0, 1] → D which connects a and z. Consider γ̃z ◦ γz−1 : [0, 1] → D is a closed
curve. By assumption, we know
Z
f (w)dw = 0.
γ̃z ◦γz−1
Figure 6. γz , γ̃z
On the other hand, we also know
Z
Z
γ̃z ◦γz−1
Z
f (w)dw −
f (w)dw =
γ̃z
f (w)dw.
γ
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MINGFENG ZHAO
So F (z) does not depend on the choice of γ.
Claim: F (z) is analytic in D, and F 0 (z) = f (z).
In fact, for all z ∈ D and fix z. Since f (z) is continuous on D, then for any > 0, there exists a δ > 0 such that
whenever |z − w| < δ, we have
|f (w) − f (z)| < .
Consider the function h(z) = z, then h0 (z) = 1. Then for any z, w ∈ D, any curve γz,w which connects z and w, we
have
Z
dt = w − z
γz,w
So for any |∆z| < δ, we get
F (z + ∆z) − F (z)
− f (z) =
∆z
=
=
=
R
R
γz+∆z f (w)dw − γz f (w)dw
− f (z)
∆z
R
γz,z+∆z f (w)dw − f (z)∆z ∆z
R
R
γz,z+∆z f (w)dw − f (z) γz,z+∆z dw ∆z
R
(f (w) − f (z))dw
γz,z+∆z
|∆z|
R
≤
γz,z+∆z
|f (w) − f (z)||dw|
|∆z|
If we take γz,z+∆z to be the line segment joining z and z + ∆z, then we have L(γz,z+∆z ) = |z + ∆z − z| = |∆z|,
and all points w ∈ γz,z+∆z , we know |z − w| < δ, which implies that
F (z + ∆z) − F (z)
−
f
(z)
∆z
R
≤
γz,z+∆z
|∆z|
R
≤
|f (w) − f (z)||dw|
γz,z+∆z
|dw|
|∆z|
= ·
L(γz,z+∆z )
|∆z|
= ·
|∆z|
|∆z|
= .
SOLUTIONS OF HW8
9
Figure 7. γz , γz+∆z , γz,z+∆z
Take → 0, we get
lim
∆z→0
F (z + ∆z) − F (z)
= f (z)
∆z
That is, F (z) is analytic in D, and F 0 (z) = f (z). So we get f (z) is holomorphic in D.
b. No. Such as f (z) ≡ 0 for all z ∈ D. Then for any closed nice curve γ in D, we know
Z
f (z)dz = 0.
γ
But D can be any domain in C.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009
E-mail address: mingfeng.zhao@uconn.edu