2014 Fall Qualifying Exam in Complex Analysis

Problem 1.

Let be an entire holomorphic function such that

for all , where is the real line in the complex plane. Prove or disprove: is constant.

Proof.

The statement is true.

Since does not meet the real axis and is connected, the imaginary part of has one fixed sign. Thus either

In either case, is a harmonic function on the whole plane that is bounded from one side.

A harmonic function on that is bounded above or bounded below is constant. Therefore is constant. If

and is constant, then the Cauchy-Riemann equations give

Hence is constant too. Therefore is constant.

Problem 2.

Evaluate the real integral

Proof.

For , define

The standard beta-integral formula gives

Differentiating with respect to ,

Thus the desired integral is . Since

we get

Therefore

Problem 3.

Let be entire holomorphic such that

for all . If

find .

Proof.

The line is the line through the origin with angle . Reflection across this line is

Since is entire and real-valued on this line, the Schwarz reflection principle gives

Taking , we have

Therefore

Thus

Problem 4.

Let be a twice differentiable function on such that

Prove that

defines a holomorphic function on which is continuous on .

Proof.

Since is twice differentiable at and

Taylor's formula at gives

Therefore

So there is a constant such that

for all sufficiently large .

For ,

for all large . Since converges, the Weierstrass -test shows that the series converges uniformly on .

Each term is continuous on and holomorphic on . The uniform limit on the closed disk is continuous there, and the locally uniform convergence inside the disk implies that the limit is holomorphic in .

Problem 5.

Let be a bounded domain in with piecewise boundary. Let be holomorphic in , continuous on , and suppose all zeros of are

counting multiplicity. Let be holomorphic in and continuous on . Evaluate

Proof.

Near a zero of of multiplicity , write

where is holomorphic and . Then

Thus

has residue

at .

Since the zeros are listed counting multiplicity, the residue theorem gives

Problem 6.

Let

Construct a conformal holomorphic map from onto the unit disk .

Proof.

The domain is the quarter unit disk in the first quadrant.

First use

This maps conformally onto the upper half of the unit disk:

Next apply

This maps the unit disk to the upper half-plane. On the upper half-disk, it maps the boundary diameter to the positive imaginary axis and the semicircular boundary to the positive real axis. Hence it maps the upper half-disk onto the first quadrant.

Squaring maps the first quadrant onto the upper half-plane:

Finally, the Cayley transform

maps the upper half-plane conformally onto the unit disk.

Therefore one conformal map from onto is

Problem 7.

Let be a simply connected domain in and . Let be the set of all functions

such that:

(i) ;

(ii) ;

(iii) is one-to-one.

Prove that is not empty.

Proof.

As stated, the problem needs the usual assumption that . If , then any bounded entire function is constant by Liouville's theorem, so no one-to-one map exists.

Assume now that is a proper simply connected domain. By the Riemann mapping theorem, there exists a conformal bijection

Let

The disk automorphism

maps to . Hence

is one-to-one, maps into , and satisfies

Now . Choose so that

Then

still maps one-to-one into , satisfies , and has

Thus , so is not empty.

Problem 8.

Let be harmonic in

such that

Prove that can be extended to be harmonic in .

Proof.

An isolated singularity of a harmonic function has the form

near , where is harmonic in a full neighborhood of . The constant is the coefficient of the logarithmic singularity.

Dividing by , we get

Since is harmonic near , it is bounded near . Also

Therefore

The hypothesis says

so necessarily .

Thus the logarithmic singularity is absent, and near . Hence extends harmonically across to all of .