Contact during exam: Pavel Gumenyuk

Mobile: (+47) 469 50 522

Faglig kontankt under eksamen: Pavel Gumenyuk

Mobil: (+47) 469 50 522

EXAM TMA4175

Complex Analysis, May 31, 2011

9:00 – 13:00

SOLUTIONS

Problem 1. What is the image of the half-strip D := {z : 0 < Re z < 2, Im z > 0} under the

linear-fractional mapping

w = f(z) :=1 + z

1− z?

Solution. The boundary of D is composed of a segment lying on the real axis L1 and two rays

lying on straight lines L2 := {z : Re z = 0} and L3 := {z : Re z = 2}. As usual, for convenience,

we include ∞ to all straight lines we consider.

The lines L1, L2, and L3 divide the extended complex plane C into six domains, with D

being one of these domains. Then, according to properties of linear-fractional mappings, f(D)

is one of the domains into which C is divided by f(L1), f(L2), and f(L3). Let us determine

these lines.

Again by properties of linear-fractional mappings, f(Lj), j = 1, 2, 3, is either a circle, or a

straight line, with the latter being the case if and only if Lj contains the pole of f . In our case,

the pole of f is p = 1, and hence f(L1) is a straight line, while f(L2) and f(L3) are circles.

Now we use the fact that symmetry is preserved under linear-factional mappings. The point

symmetric to p w.r.t. L2 is p∗ = −1. Since f(p) = ∞, we have that f(p∗) = 0 is the center

of f(L2). Furthermore, 0 ∈ L2 implies that f(0) = 1 ∈ f(L2). Thus f(L2) = {w : |w| = 1}. In

a similar way, we establish that f(L3) = {w : |w + 2| = 1}.To find f(L1) we consider points i and −i, which are symmetric w.r.t. L1. Their images

f(i) = i and f(−i) = −i are symmetric w.r.t. the straight line f(L1). Hence f(L1) is the real

axis.

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Now it remains to take one point in D, say z0 := 1 + i. We have w0 := f(z0) = −1 + 2i.

Among the domains onto which the lines f(L1), f(L2), and f(L3) divide the extended complex

plane, the one containing w0 is U := {w : Imw > 0, |w| > 1, |w + 2| > 1} (half-plane minus

two half-disks). This is the image of D we are looking for: f(D) = U . �

Problem 2. Let n 6= 0 be an integer. Find the value of the following integral:∫ 2π

0

sin2(nθ) cos(2nθ) dθ.

Hint: Express the integrand in terms of z = eiθ, 1/z = e−iθ.

Solution. Following the hint and using the identities eiϕ + e−iϕ = 2 cosϕ, eiϕ− e−iϕ = 2i sinϕ,

dθ = deiθ/(ieiθ), we can rewrite the integral to compute as

I :=

∫γ

[1

2i

(zn − 1

zn

)]2 [1

2

(z2n +

1

z2n

)]dz

iz=

∫γ

i

8

(z4n−1 +

1

z4n+1+

2

z− 2z2n−1 − 2

z2n+1

)dz,

where γ is the unit circle oriented counterclockwise, i.e. γ(θ) := eiθ, θ ∈ [0, 2π].

Note that for any integer m, ∫γ

zmdz =

{2πi, if m = −1,

0, if m 6= −1.

Thus

I =i

8· 2 · 2πi = −π

2.

�

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Problem 3. How many zeros, taking into account multiplicities (orders), does the function

f(z) := z5 + 3z2 + ez−2 have in the domains {z : |z| < 1} and {z : 1 < |z| < 2}?

Solution. Consider first the domain D1 := {z : |z| < 1}. Denote f0(z) := 3z2. For all z ∈ ∂D1

we have |f0(z)| = 3 > |f(z)− f0(z)|, because

|f(z)− f0(z)| = |z2 + ez−2| 6 |z5|+ |ez−2| = 1 + eRe z−2 6 1 + 1/e

when z ∈ ∂D1. Therefore, by the Rouche Theorem, the number of zeros of f in D1 is the same

as that of f0, provided the multiplicities are taken into account. The function f0 has the unique

zero at the origin of order 2. Hence, f has two zeros in D1, taking into account multiplicities.

Note also that |f(z)| = |f0(z) + f(z)− f0(z)| > |f0(z)| − |f(z)− f0(z)| > 0 for all z ∈ ∂D1.

Therefore, f has no zeros on the boundary of D1.

In a similar way, considering D2 := {z : |z| < 2}, we may conclude that f has the same

number of zeros in D2 as the function f1(z) := z5, i. e. f has five zeros in D2, taking into

account multiplicities.

Thus the function f has 2 zeros in {z : |z| < 1} and 3 zeros in {z : 1 < |z| < 2}, taking into

account multiplicities. �

Problem 4. Assume that f is holomorphic in {z : 1 < |z| < +∞} and that f(n) = 2n for

each n = 2, 3, . . . What kind of singularity does function f have at ∞?

Solution. By definition, f has an isolated singularity at ∞. An isolated singularity of a

holomorphic function can be of one of the following three types: a pole, a removable, or

essential singularity. It immediately follows from the hypothesis that f is unbounded in any

punctured neighbourhood of ∞. Therefore, ∞ cannot be a removable singular point for f .

Assume for a moment that ∞ is a pole of f . Let ν be its order. Then limz→∞ f(z)/zν exists

and differs from 0 and ∞. However, this contradicts the hypothesis that f(n) = 2n for all

n = 2, 3, . . .. Thus we may conclude that ∞ is an essential singularity of the function f . �

Problem 5. Denote D := {z : |z| < 1}. Fix M > 1. Let SM be the class of all univalent

holomorphic functions f : D→ C such that f(0) = 1, f ′(0) = 1, and |f(z)| 6M for all z ∈ D.

Prove that any sequence fn ∈ SM , n = 1, 2, . . ., has a subsequence fnkthat converges locally

uniformly in D to a function from the class SM .

Hint: follow the argument in the proof of the Riemann Mapping Theorem.

Solution. Let (fn) be any sequence from SM . The functions (fn) are uniformly bounded in D.

Hence the sequence (fn) forms a normal family in D. Therefore there exists a subsequence (fnk)

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that converges locally uniformly in D to some holomorphic function f . By the Hurwitz Theorem,

f is either constant, or univalent in D. By the Weierstraß Theorem, f ′nkconverges to f ′ locally

uniformly in D. So we have 1 = fnk(0) → f(0) and 1 = f ′nk

(0) → f ′(0) as k → +∞, which

means that f ′(0) = f(0) = 1. This excludes the possibility that f is constant in D.

Finally, passing to the limit in the inequality |fnk(z)| 6M we conclude that |f(z)| 6M for

all z ∈ D. Thus f ∈ SM . �

Problem 6. Let f be a holomorphic function in D := {z : |z| < 1}. Assume that |f(z)| < M

for all z ∈ D. Using the Cauchy integral formulas for derivatives, show that

|f (n)(z)| 6 n!

(1− |z|)nM

for any z ∈ D and any n = 1, 2, . . .

Solution. Fix an arbitrary z ∈ D. Let r ∈(0, 1 − |z|

). Then the circle γr(t) = z + reit,

t ∈ [0, 2π], lies in D. Therefore we may express f (n)(z) by means of the Cauchy integral

formula for derivatives, i.e.

f (n)(z) =n!

2πi

∫γr

f(ζ)

(ζ − z)n+1dζ.

Using now the ML-estimate, we get

|f (n)(z)| 6 n!

2π· 2πr · M

rn+1=n!

rnM.

Finally, letting r → 1− |z| − 0, we establish the desired inequality. �

Problem 7. Let M be a positive number, f a holomorphic function in a domain D ⊂ C,

and z0 ∈ D. Assume that |f(z)| 6M for all z ∈ D and that |f(z0)| = M . Explain why in this

case f must be constant in D.

Solution. Apply the Maximum Principle in the same way as it is used in the proof of the

Schwarz Lemma to treat the equality case. If |f(z)| 6 M for all z ∈ D and |f(z0)| = M ,

then f attains its maximum at the point z0. In particular, z0 is a local maximum of |f | in the

domain D. By the Maximum Principle it is possible only if f is constant in D. �

Problem 8. Prove that any bounded holomorphic function f : C∗ → C, C∗ := C \ {0}, is

constant.

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Solution. Let f : C∗ → C be holomorphic and bounded. We have to show that f is constant.

First method. Define g := f ◦exp. Then the function g is entire and bounded. By the Liouville

Theorem, g must be constant. Therefore, f is constant in exp(C) = C∗.

Second method. Here we give a direct proof. Denote by M any upper bound for |f | in C∗. Fix

an arbitrary R > 0. Write the Laurent representation of f :

f(z) =+∞∑

n=−∞

anzn for all z ∈ C∗,

where

an :=1

2πi

∫CR

f(ζ)ζ−n−1dζ for all n ∈ Z, (∗)

and CR is the circle |ζ| = R oriented counterclockwise.

We have to prove that an = 0 for any n 6= 0. To this end we apply the ML-estimate to the

integral in (∗):0 6 |an| 6

1

2π· 2πR ·MR−n−1 = MR−n. (!)

Now passing in (!) to the limit as R → +∞, we conclude that |an| = 0 for any n = 1, 2, 3 . . .

Similarly, letting R→ +0 we see that an = 0 also for any n = −1,−2,−3, . . . �