2022 Spring Qualifying Exam in Complex Analysis

Problem 1.

Let and let

be a polynomial with complex coefficients such that

for all . Show that for sufficiently small, has simple zeros in the annulus

Proof.

Let

The zeros of are the st roots of unity other than . Thus has simple zeros, all lying on .

Choose pairwise disjoint small disks around these zeros, all contained in the annulus

On the union of the boundary circles , the continuous function has a positive minimum:

Now

On the compact set , this satisfies

for some constant depending only on the chosen disks and .

Choose so small that . Then on every ,

By Rouche's theorem, and have the same number of zeros in each . Since each contains exactly one simple zero of , each contains exactly one zero of .

Thus has zeros in the annulus. Since has degree for , these are all its zeros. Moreover, by choosing the disks small and using the fact that zeros vary continuously under small perturbations of the coefficients, the zeros remain simple for sufficiently small .

Problem 2.

Let

Compute

Proof.

Let

Then is conformal and . Also

For any holomorphic , the function

maps to . By Schwarz-Pick,

Since

and , we get

Therefore

This bound is attained by taking

Then

Hence the supremum is

Problem 3.

Let be holomorphic on

and continuous on

Suppose there exist constants such that

Prove that

for all .

Proof.

Fix . For , integrate

over the positively oriented contour consisting of the interval and the upper semicircle .

By Cauchy's integral formula,

On the semicircle , for large ,

Since the length of is , the arc integral is , hence tends to .

Letting , we obtain

which proves the formula.

Problem 4.

Let be holomorphic on a neighborhood of

Assume that whenever ,

Show that is constant.

Proof.

Write

The hypothesis says that

on . Since is harmonic in a neighborhood of , the maximum principle gives

throughout . Hence

in .

Differentiate to get

The Cauchy-Riemann equations give

Combining these,

Thus

So is constant, and therefore is constant. Hence is constant.

Problem 5.

Find all holomorphic functions such that for all ,

Proof.

Define

Then

and

Thus is entire and doubly periodic with periods and .

An entire doubly periodic function is bounded on a fundamental parallelogram, hence bounded on all of . By Liouville's theorem, is constant.

Therefore all solutions are

Problem 6.

Let be holomorphic on

and assume that

for . Prove that for all ,

Proof.

Since has no zeros in , the function

is positive and harmonic in . Harnack's inequality for positive harmonic functions on the unit disk gives, for ,

Substitute

Since , exponentiating the inequalities gives

This is the desired estimate.

Problem 7.

Let be the set of functions holomorphic on

and continuous on

that satisfy

Show that is a normal family.

Proof.

It is enough to prove local boundedness. Fix . For , Cauchy's integral formula gives

Therefore

Since , Cauchy-Schwarz gives

Thus

for all and all .

Hence is locally bounded. By Montel's theorem, is a normal family.

Problem 8.

Determine all entire functions such that

for all sufficiently large , and

for all .

Proof.

The function

satisfies the conditions.

We prove it is the only solution. Let

Then is entire and

for every .

The growth hypothesis gives, for large ,

Thus has at most exponential growth of order .

If were not identically zero, Jensen's formula would imply that the number of zeros of in is at most for this growth. But the zeros

with all lie in , so there are at least on the order of zeros. This contradiction implies

Therefore