2012 Fall Qualifying Exam in Complex Analysis

Proof.

Consider

For , close the contour in the upper half-plane. The only pole inside is the double pole at .

Since

the residue at is

Now

At ,

Thus the residue is

Therefore

Taking real parts and using evenness,

Hence

Proof.

Write the Laurent expansion of in the annulus:

We prove that all negative coefficients vanish.

Fix with . For ,

Since uniformly on ,

Each is a polynomial, so this integral is for every . Hence for all .

Thus the Laurent expansion has no principal part and is a power series

with radius of convergence at least . Therefore extends analytically to .

Proof.

The intended statement is for nonconstant polynomials. A nonzero constant polynomial would have .

Assume is nonconstant. Since has no zeros in the right half-plane, every zero of lies in the closed left half-plane

By the Gauss-Lucas theorem, every zero of lies in the convex hull of the zeros of .

The closed left half-plane is convex, so the convex hull of the zeros of is still contained in

Therefore has no zeros in the right half-plane. Hence

whenever .

Proof.

Let

On ,

while

By Rouche's theorem, has the same number of zeros in as , namely .

On ,

while

Again by Rouche's theorem, has the same number of zeros in as , namely .

The inequalities are strict, so there are no zeros on the boundary circles. Therefore the number of zeros in is

Thus the answer is

Proof.

Use the conformal map

which maps the upper half-plane to the unit disk and sends to . Its derivative is

so

Let

Then is holomorphic and . By the Schwarz lemma,

Since

we get

Equality holds exactly when

Thus the extremal functions are

Proof.

Since ranges over , the hypothesis is equivalent to

for all .

By continuity,

Thus

has a removable singularity at and extends to an entire function. For ,

By Liouville's theorem, is constant. Therefore

for some constant with .

Proof.

The isolated singularity at is either removable, a pole, or essential. We rule out the last two possibilities.

If were a pole of order , then near ,

for some . Hence

In polar coordinates,

would have the same convergence behavior as

This diverges for every integer .

An essential singularity is also impossible by the standard isolated-singularity theorem: if a holomorphic function on a punctured disk is in near the puncture, then the pole order must be strictly less than . Here , so any possible pole order would have to be less than , forcing order .

Therefore the singularity is removable.

Proof.

Split the series into its positive and negative tails.

For , the terms are

Taking the th root of the absolute value gives

This tends to if , and it fails to tend to if . Hence the positive tail converges only for

For , the terms are

Taking the th root of the absolute value gives

This tends to if , and it fails to tend to if .

Thus the positive tail requires , while the negative tail requires . No complex number satisfies both conditions. Therefore the Laurent series converges nowhere:

Proof.

The only possible obstruction to having a global harmonic conjugate on is the period around a loop enclosing the origin.

Let

The function

is harmonic on . The choice of makes the period of the conjugate differential of around the unit circle equal to . Since every closed curve in has winding number an integer multiple of the unit circle, all periods vanish.

Therefore has a single-valued harmonic conjugate on . Hence

is holomorphic on and

Thus

as required.