2025 Fall Qualifying Exam in Complex Analysis

Problem 1.

Let be a holomorphic map from the upper half-plane to itself. Show that for all ,

Proof.

This is the Schwarz-Pick lemma for the upper half-plane.

Let

be the automorphism from to sending to . Similarly, let

Then

is a holomorphic map from to with . By Schwarz's lemma,

Computing the derivative gives

Therefore

Problem 2.

Let be a non-constant entire function. Prove that the image is dense in . Do not use Picard's theorem.

Proof.

Suppose is not dense in . Then there exist and such that

for all .

Then

is an entire function satisfying

on . By Liouville's theorem, is constant. Therefore is constant, contradicting the hypothesis.

Hence is dense in .

Problem 3.

Let be an entire function with . For , let denote the number of zeros of in the closed disk . Let

Prove that for any ,

Proof.

Since , zero is not a zero of . Jensen's formula says that for ,

Since , this becomes

Therefore

Taking , we get

Since the integrand is nonnegative,

Problem 4.

Prove that for every entire function ,

Proof.

Suppose, toward a contradiction, that

for every .

Put . On , we also have , and the inequality becomes

Multiplying by gives

on .

Let

Then is holomorphic on the unit disk and . On ,

By Rouche's theorem, and the constant function have the same number of zeros in . The function has no zeros, while . This contradiction proves

Problem 5.

Let with . Determine all meromorphic functions on whose only singular points are simple poles at for , with

Repeat the problem with the residue condition replaced by

Proof.

First suppose

The series

converges normally on compact subsets of , since for fixed compact sets the terms are . It has exactly the required simple poles and residues. Therefore all solutions are

where is an arbitrary entire function.

Now suppose

The naive series does not converge, but the modified series

does converge normally on compact subsets away from the poles, because

on compact sets. Therefore all solutions are

where is an arbitrary entire function.

Problem 6.

Prove that the sequence of entire functions

converges uniformly on , but does not converge normally, that is, uniformly on compact subsets, in the disk for any .

Proof.

For real ,

The maximum occurs at , and its value is

Hence

Thus uniformly on .

Now fix any and choose . Then lies in the disk , and

This tends to infinity in modulus as . Hence the sequence cannot converge uniformly on any compact subset containing . In particular, it does not converge normally in for any .

Problem 7.

Suppose is entire and there exists such that

for all . Show that

for some polynomial of degree at most .

Proof.

Define

Since

we get

Thus is an entire function of polynomial growth of degree at most .

By Cauchy's estimates, all Taylor coefficients of of degree greater than vanish. Therefore is a polynomial of degree at most . Hence

Problem 8.

Show that for all ,

Discuss the convergence of the series.

Proof.

First note that

For in a compact subset of ,

uniformly as . Therefore the series converges absolutely and uniformly on compact subsets of .

Define

Then is meromorphic with simple poles at the integers, and each pole has residue . The function also has simple poles at the integers, each with residue . Hence

has removable singularities at every integer, so is entire.

The symmetric partial sums

show that is periodic with period in the limit. Therefore is also periodic with period .

On a vertical fundamental strip, both and are bounded away from small neighborhoods of the integers, and the singularities at the integers have already been removed in . Hence is bounded on a fundamental strip. Since is periodic, it is bounded on all of . By Liouville's theorem, is constant.

Finally, both and are odd functions, so is odd. The only constant odd function is . Therefore

which proves

Problem 9.

Let be holomorphic on the disk . Suppose there exists such that for all ,

Show that for some universal constant , independent of and ,

Proof.

Fix . Choose

Since , we have , and the circle lies inside .

By Cauchy's estimate,

For such ,

By the maximum modulus principle and the hypothesis,

Therefore

Since , this implies

Taking the maximum over , the result follows with the universal constant .