2007 Fall Qualifying Exam in Complex Analysis

Problem 1.

Prove Jordan's lemma. Let be continuous in

and suppose uniformly in as . Then for every ,

where is the upper semicircle .

Proof.

Parametrize the upper semicircle by , . Then

Hence

Using on and symmetry,

Therefore

Since , the integral tends to .

Problem 2.

Let be holomorphic in the closed unit disk. Prove that, for ,

Proof.

Write

Also,

Using orthogonality on the disk,

Therefore

Problem 3.

Let . Find the radius of convergence of

Proof.

Let

Then

By the ratio test, the radius of convergence is

Problem 4.

Show that

is holomorphic in .

Proof.

For each fixed , the function

is entire. On every compact set ,

Thus differentiation under the integral sign is justified on compact sets. Hence

and similarly all higher derivatives exist. Therefore is entire.

Problem 5.

Let be holomorphic. Prove that

Proof.

This is the Schwarz-Pick lemma. For fixed , let

and

Then

maps the unit disk into itself and satisfies . By Schwarz's lemma,

Taking gives the desired inequality.

Problem 6.

Let be a simply connected domain in . Let be holomorphic and suppose fixes two distinct points . Prove that

on .

Proof.

By the Riemann mapping theorem, choose a conformal map

Then

is a holomorphic self-map of the unit disk fixing the two distinct points and .

An automorphism or self-map of the unit disk with two fixed points in the disk must be the identity. Indeed, conjugate one fixed point to and apply Schwarz's lemma; equality at another nonzero fixed point forces a rotation, and the second fixed point forces that rotation to be .

Hence is the identity, so is the identity on .

Problem 7.

Let be real. Evaluate

Proof.

Use

Then

Thus

The standard partial fraction expansion for gives

Therefore

Problem 8.

Let be analytic on and have a simple pole at with residue . Prove that for every ,

Proof.

Write

where is entire. Since

we get at :

Thus it remains to show

Since is entire, Cauchy's estimates on the circle give

Choose . Then

Problem 9.

Suppose is entire and

for some positive constants . Let be the zeros of , listed with multiplicity. Prove that

for all .

Proof.

By Jensen's formula, the number of zeros of in , counted with multiplicity, satisfies

for all sufficiently large . This follows from the growth bound

Thus the zero-counting function grows at most linearly.

Group the zeros into dyadic annuli:

The number of zeros in this annulus is . Therefore

Since ,

converges. Hence