2013 Spring Qualifying Exam in Complex Analysis

Problem 1.

Find the largest set where the power series

converges.

Proof.

Consider the th term

Its absolute value is

Taking the th root gives

If , this tends to , so the series converges absolutely. If , this tends to infinity, so the terms do not even tend to .

If , then

and

converges. Hence the series converges absolutely on the unit circle.

Therefore the largest set of convergence is

Problem 2.

Let be a sequence of holomorphic functions from to . Prove that if

converges, then

converges absolutely and uniformly on compact sets in .

Proof.

Since , the function

is positive and harmonic on .

By Harnack's inequality, if , then

For compact subsets of , choose with the compact set contained in . Then

Thus, for all in the compact set,

for some . Equivalently,

Hence

Since and ,

Therefore

on the compact set. The series converges, so the Weierstrass -test proves absolute and uniform convergence on compact subsets of .

Problem 3.

Suppose is holomorphic on the upper half-plane

with

and

for all . Prove that

Proof.

The automorphism

maps conformally onto and sends to .

Since and , the Schwarz lemma on the upper half-plane gives

Taking ,

Problem 4.

Suppose

is a holomorphic function. Show that there exist and such that

Proof.

Since

is holomorphic, the Cauchy-Riemann equations give

But depends only on , so . Hence

so depends only on .

Now depends only on , while depends only on . Since

both must be equal to a real constant . Therefore

for real constants .

Thus

where

Problem 5.

Determine the number of roots, counted with multiplicity, of

inside the annulus

Proof.

Let

First count zeros in . On ,

while

Thus, by Rouche's theorem, and have the same number of zeros in . Therefore has zeros in .

Now count zeros in . On ,

while

Again by Rouche's theorem, and have the same number of zeros in . Thus has zeros in .

The inequalities are strict on both circles, so there are no zeros on or . Hence the number of zeros in the annulus is

Therefore the answer is

Problem 6.

Suppose is analytic in an annulus

and there exists a sequence of polynomials converging to uniformly on every compact subset of the annulus. Show that is analytic on the disk

Proof.

Write the Laurent expansion of in the annulus:

We show that all negative coefficients vanish.

Fix with . For ,

Since uniformly on ,

But a polynomial has no negative powers in its Laurent expansion about , so the integral is for every and every . Hence

for all .

Therefore the Laurent expansion is actually a power series

which converges for . This gives the desired analytic extension to the disk .

Problem 7.

Evaluate, for ,

Proof.

The standard residue formula says that for ,

Taking real parts with gives

Thus

Problem 8.

Find explicitly a conformal mapping of

onto the unit disk

Proof.

First send the initial point of the slit, , to by the disk automorphism

This maps onto itself and maps the slit to the slit .

Thus maps conformally onto

On this slit disk, choose the branch of the square root with argument in :

Then maps conformally onto the upper half unit disk

Now use

This maps the upper half unit disk onto the first quadrant. Squaring maps the first quadrant onto the upper half-plane, and the Cayley map

maps the upper half-plane onto the unit disk.

Therefore an explicit conformal map from onto is

where the square root branch is chosen on with argument in after applying .