2022 Fall Qualifying Exam in Complex Analysis

Proof.

Let

Then is harmonic on and satisfies

The sharp gradient estimate for bounded harmonic functions on the unit disk gives

Since is holomorphic, the Cauchy-Riemann equations imply

Therefore

In particular,

Proof.

Let be compact. Let

Since is entire, it is bounded on ; say

for .

For every , Cauchy's estimate on the circle gives

Therefore

By Stirling's formula,

so

converges. The Weierstrass -test now gives uniform convergence on .

Since was arbitrary, the series converges uniformly on compact subsets of .

Proof.

For holomorphic functions with values in , normality implies local boundedness. Hence the family

is locally bounded on the annulus .

Choose the compact subannulus

Then there exists such that

for all and all .

The sets

cover all sufficiently large , because the intervals overlap for large . Therefore is bounded outside a sufficiently large disk. Since is entire, it is also bounded on the remaining compact disk. Thus is bounded on all of .

By Liouville's theorem, is constant.

Proof.

(i) Since has no zeros on and , the only zero of in is . Since , this zero is simple.

On , let

Let

Choose such that

Then for and ,

By Rouche's theorem, and have the same number of zeros in , counted with multiplicity. Since has exactly one zero in , and it is simple, has exactly one zero in , counted with multiplicity. Therefore this zero is unique and simple.

(ii) Define

At the zero , we have

The zero is simple, so

By the holomorphic implicit function theorem, depends holomorphically on for .

Proof.

The function

clearly works, because

and

so all the required derivative bounds hold.

We prove it is the only solution. The derivative bounds imply that the Taylor series of at ,

has infinite radius of convergence. Hence it defines an entire function agreeing with near , and therefore agreeing with on the whole upper half-plane by analytic continuation.

Let

Then is entire and

for all .

The derivative bounds at show that has growth at most exponential of type :

Hence also has growth at most exponential of order .

But the zeros have counting function comparable to in the disk . Jensen's formula would then force an entire function of exponential type with these zeros to grow at least like , unless it is identically zero. Therefore

Thus

Proof.

Let

On ,

By Rouche's theorem, and

have the same number of zeros in , counted with multiplicity.

But

and

Thus has a zero of order at least at . Therefore has at least two zeros in , counting multiplicity.

Proof.

Since is bounded and simply connected and contains , let

be a Riemann map with . Define

Then is holomorphic,

and

By Schwarz's lemma,

Since , is not a rotation. Therefore

for every .

Fix . By compactness,

Hence

for . Thus uniformly on compact subsets of .

Conjugating back by , we get

uniformly on compact subsets of .

Proof.

We construct first. Let be the zeros of , with multiplicities . Since and have no common zeros,

for every .

At each zero of , prescribe the Taylor polynomial of up to order so that

Equivalently, should match the Taylor expansion of at through order .

By the Mittag-Leffler interpolation theorem, there exists an entire function satisfying all these prescribed finite jet conditions at the discrete set of zeros of .

Then

vanishes at each zero of to order at least . Therefore

has removable singularities at the zeros of and is entire. Hence