2018 Fall Qualifying Exam in Complex Analysis

Problem 1.

Let be analytic on , with there. Let

Prove that has an essential singularity at if and only if has an essential singularity at .

Proof.

An isolated singularity is removable, a pole, or essential.

If has a removable singularity at , then either the removable value is nonzero, in which case is also removable, or the removable value is , in which case has a pole.

If has a pole at , then tends to , so has a removable singularity.

Thus taking reciprocals interchanges poles and removable singularities, except for nonzero removable singularities which remain removable. Therefore the only remaining possibility is essential. Hence is essential at if and only if is essential at .

Problem 2.

Let and be entire functions. Suppose:

(a) for all .

(b) for all .

Prove that there exists with such that

for all .

Proof.

Since has no zeros, the only possible zero of is at , of order . The inequality forces to vanish at to order at least . Therefore

extends to an entire function.

For ,

By removability this also holds at . Hence is a bounded entire function. By Liouville's theorem, is constant:

Thus

Problem 3.

Let be a uniformly bounded sequence of analytic functions on . Suppose

exists for . Prove that there exists an analytic function on such that

uniformly on compact subsets of .

Proof.

The family is uniformly bounded, hence locally bounded. By Montel's theorem, every subsequence has a further subsequence converging uniformly on compact subsets to a holomorphic function.

The set

has an accumulation point at . Since the limits exist for every , any two subsequential limits must agree at every point . By the identity theorem, the two subsequential limits are the same holomorphic function.

Therefore all convergent subsequences have the same limit, and the whole sequence converges uniformly on compact subsets to that holomorphic function.

Problem 4.

Find the number of solutions, counted with multiplicity, of

in the open unit disk.

Proof.

Consider

On ,

By Rouche's theorem, and have the same number of zeros in , counted with multiplicity. The function has exactly zeros there, counting multiplicity.

Thus the equation has

solutions in the open unit disk.

Problem 5.

Evaluate

when:

(a) is the circle of radius around , counterclockwise.

(b) is the circle of radius around , counterclockwise.

Proof.

The poles occur where

that is,

Each is a double pole. Near ,

Therefore

so the residue at every pole is .

For radius , the only enclosed pole is . Hence

For radius , the enclosed poles are

The sum of residues is , so

Problem 6.

Find a surjective holomorphic map from the open unit disk to the punctured disk , with

for every .

Proof.

The map

maps conformally onto the right half-plane. Define

Since , we have

and . The exponential maps the right half-plane, after the negative sign, onto the punctured unit disk, so is surjective onto .

Also,

which never vanishes in . Thus this has the required properties.

Problem 7.

Let be analytic on and suppose

Show that for ,

Proof.

Fix with . Let . Apply the residue theorem to

on the annulus . The boundary consists of positively oriented and negatively oriented. Thus

As , the outer integral tends to because and the denominator is of size on a circle of length .

Therefore

Problem 8.

Suppose is holomorphic and

Give an upper bound for , and characterize the functions for which equality holds.

Proof.

By Schwarz-Pick,

Since ,

Equality holds exactly for disk automorphisms. Thus equality holds precisely when

where

Problem 9.

Let

(a) Find all harmonic conjugates of in an open ball of radius centered at .

(b) Prove that there is no harmonic conjugate of in , where is any open set containing .

Proof.

Since

a harmonic conjugate locally is

(a) The disk centered at of radius does not wind around , so a single-valued branch of exists there. Hence all harmonic conjugates are

where is any chosen branch on that disk.

(b) If a harmonic conjugate existed on , then

would be a holomorphic branch of

there. But no single-valued branch of exists on a punctured neighborhood of : going once around changes by . Therefore no global harmonic conjugate exists on .