2006 Spring Qualifying Exam in Complex Analysis

Problem 1.

Prove or disprove that there exists an analytic function in such that

for all .

Proof.

No such function exists.

The equalities

imply, by the identity theorem, that

on the unit disk. Hence is even, so

for some holomorphic function near .

Then

Let . Then and

Thus and

so . Hence near . But

contradicting . Therefore no such analytic function exists.

Problem 2.

(a) State Liouville's theorem.

(b) Prove Liouville's theorem by calculating

and taking .

Proof.

(a) Liouville's theorem says that every bounded entire function is constant.

(b) Let be bounded and entire, say . For large enough that and lie inside , the residue theorem gives

On the other hand, on ,

Multiplying by the length , the integral tends to as .

Therefore

so for arbitrary . Hence is constant.

Problem 3.

The Bernoulli polynomials are defined by

Prove:

(i) .

(ii)

Proof.

(i) Using the generating function,

But

Comparing the coefficient of gives

(ii) Apply part (i) with in place of :

Sum this from to :

From the generating function, for the relevant normalization in this identity. Hence

Problem 4.

Let be analytic and satisfy

in the strip

Prove that

is constant for .

Proof.

Let . Integrate around the rectangle with vertical sides and and horizontal sides . Since is analytic, the contour integral is .

The horizontal integrals tend to as , because

uniformly for .

Letting leaves

Thus , so is constant.

Problem 5.

Evaluate

Proof.

The standard identity

with gives

Problem 6.

Prove or disprove: there exists a sequence of analytic polynomials such that

uniformly for .

Proof.

No such sequence exists.

On ,

If uniformly on , then

But each is a polynomial, so

This contradiction proves that no such sequence exists.

Problem 7.

Let be analytic in and continuous on . Assume

for all . Find all such .

Proof.

Set

Then is analytic in the disk, continuous on the closed disk, and

on .

Thus is an inner function belonging to the disk algebra. Such functions are precisely finite Blaschke products multiplied by a unimodular constant. Therefore

where and .

Hence all solutions are

The empty product is allowed.

Problem 8.

Let be entire and suppose

and

for all . Prove that is constant.

Proof.

Since has period , define its Fourier coefficients by

By Cauchy's theorem and periodicity, these coefficients are independent of .

For , let . Since

while

the product tends to because . Hence for .

For , let . The same estimate gives for .

Thus all nonzero Fourier coefficients vanish, so is equal to its constant coefficient. Therefore is constant.