2006 Fall Qualifying Exam in Complex Analysis

Problem 1.

Show that

is a meromorphic function on .

Proof.

The possible poles are at

On every compact set avoiding these points, for all sufficiently large ,

Hence

and the series converges uniformly by the Weierstrass -test. Thus the sum is holomorphic away from the points . At each such point, only one summand has a simple pole. Therefore the sum is meromorphic on .

Problem 2.

Show that for ,

Proof.

Consider

For , close the contour in the upper half-plane. There is a double pole at . Since

the residue is

Therefore

Taking real parts and using evenness gives

Problem 3.

Let be a polynomial. Assume

for . Show that

for .

Proof.

For a nonconstant polynomial, all zeros of lie in the closed left half-plane. By the Gauss-Lucas theorem, the zeros of lie in the convex hull of the zeros of . Since the closed left half-plane is convex, that convex hull is still contained in

Thus has no zeros in the right half-plane.

If is constant, the statement needs to be interpreted separately, since then .

Problem 4.

Let be distinct points in , let be analytic in , and let

Prove that

is a polynomial of degree at most with

Proof.

For fixed ,

is a polynomial in of degree at most , because is divisible by . Hence is a polynomial of degree at most .

If , then , so

by Cauchy's integral formula.

Problem 5.

Let be analytic in and suppose

on . Show that

Proof.

Let

By the maximum modulus principle,

in the unit disk. The Schwarz-Pick derivative estimate for gives

At ,

Since

we get

Problem 6.

Let be real. Prove that

has a single solution in , and that this solution is real and positive.

Proof.

Rewrite the equation as

On ,

By Rouche's theorem, and have the same number of zeros in , namely one.

For , the function

is strictly increasing because

and is positive on . Also

Thus there is a unique with . Since the disk zero is unique, it is this real positive solution.

Problem 7.

Let be a bounded domain in , and let be analytic on with

Prove that is a normal family in .

Proof.

Fix a compact set . Choose such that for every . Since is subharmonic,

Thus the family is uniformly bounded on . Since was arbitrary, the family is locally uniformly bounded. By Montel's theorem, it is normal.

Problem 8.

Let be bounded and analytic. Let be the nonzero zeros of in , counted with multiplicity. Prove that

Proof.

This is the Blaschke condition. If has no zeros, there is nothing to prove. Otherwise apply Jensen's formula to :

Since is bounded, the right-hand side is bounded above uniformly in . Letting gives

For ,

Therefore