2015 Fall Qualifying Exam in Complex Analysis

Proof.

For ,

Since converges, the Weierstrass -test shows that

converges uniformly on the closed unit disk .

The term-by-term derivative is

This series does not converge uniformly on the closed unit disk, because at it becomes the harmonic series

which diverges. Thus uniform convergence on is impossible.

On every smaller disk , however,

and converges. Therefore the derivative series converges uniformly on compact subsets of the open unit disk, but not uniformly on the whole closed unit disk.

Proof.

There is a small hypothesis issue in the statement as written: if the bounded sequence is eventually constant, the conclusion need not follow. The intended argument requires the sequence to have a real accumulation point with infinitely many distinct terms.

Under that intended interpretation, since is bounded, it has a convergent subsequence. Passing to a subsequence of distinct terms, write

Define

Because is entire, the function is also entire. Hence is entire.

For each , is real, so

because is real. Thus the zeros of the entire function have an accumulation point in the complex plane. By the identity theorem,

Therefore, for every real ,

So , which means is real for all real .

Proof.

Let

Differentiate with respect to :

The standard residue computation gives, for ,

Taking real parts and using evenness,

Therefore

Also . Hence

Thus

Proof.

Write

We count zeros in the right half-plane by Rouche's theorem on the large right half-disk

On the imaginary-axis part of the boundary, , so

because .

On the semicircular part , , we have

while . For ,

Thus, on the boundary of ,

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

The function has exactly one zero, namely , and since , this zero lies in the right half-plane. Therefore

has exactly one root in the right half-plane.

Proof.

The boundary data have Fourier expansion

For the disk of radius , the harmonic function with boundary term is

Therefore

This function is continuous on , harmonic for , real-valued, and has the required boundary values.

Equivalently, since ,

Proof.

First prove the upper bound. Suppose . Then , and

Since , we have

Thus

So the leading term cannot be cancelled by the remaining terms, and . Hence every root satisfies

For the lower bound, apply the upper-bound result to the reciprocal polynomial

The leading coefficient of is , and the maximum of the absolute values of its lower coefficients is

Therefore every zero of satisfies

If is a zero of , then and is a zero of . Hence

so

Combining the two inequalities gives the stated annulus.

Proof.

The statement is true.

Fix . For ,

The numerical series

converges for every . Therefore the family is uniformly bounded on each compact disk .

By Montel's theorem, every locally bounded family of holomorphic functions on the unit disk is normal. Hence this family is normal.

Proof.

(a) The map sends to and preserves the angle , so it is a one-to-one continuous map from onto . Its inverse is

which is also continuous. Thus is a homeomorphism.

It is not conformal. In polar coordinates, a conformal radial map preserving angle has the form

with radial and angular scale factors equal:

Here , so , while

These are not equal for . Hence the map does not preserve infinitesimal angles and is not conformal.

(b) No. The annulus has finite conformal modulus

The punctured unit disk has infinite modulus; it is conformally equivalent to a punctured disk with the inner radius tending to .

Conformal modulus is invariant under conformal maps between doubly connected domains. Since the moduli are different, no conformal mapping exists.