2013 Fall Qualifying Exam in Complex Analysis

Problem 1.

Describe all entire functions satisfying:

(a) for all ;

(b) for all .

Proof.

(a) The function has the required values:

If is another entire function with the same values, then

has zeros at and for all . These zeros accumulate at , which lies in the domain of analyticity. By the identity theorem,

Thus the only function is

(b) There is no such entire function.

Indeed, the condition

for all implies, by the identity theorem, that

for all . Hence is even, so there is an entire function such that

The value condition gives

Since , we get . Also

so . Therefore

But with ,

and

contradicting . Thus no entire function satisfies part (b).

Problem 2.

Suppose is an entire function such that

Prove that is constant.

Proof.

The hypothesis says as . Let

be the Taylor expansion at .

Fix . For sufficiently large ,

whenever . By Cauchy's estimate,

For , this gives , so after letting .

For , letting gives

so . Hence all coefficients except possibly vanish, and is constant.

Problem 3.

Describe explicitly the automorphism group

Proof.

Every automorphism of extends to a Möbius transformation of the Riemann sphere that permutes the three punctures

Conversely, every Möbius transformation that permutes these three points restricts to an automorphism of .

Thus the automorphism group is isomorphic to and consists of the six maps

Problem 4.

Evaluate

Proof.

For ,

Taking gives

Therefore

Problem 5.

Prove that

is harmonic in and find its harmonic conjugate.

Proof.

Observe that

Its imaginary part is

This is exactly

Since is entire, its real and imaginary parts are harmonic. Therefore is harmonic on .

If , then . To get an analytic function whose real part is , take

Thus a harmonic conjugate of is

Now

Hence

where is a real constant.

Problem 6.

How many solutions does

have in the closed upper half unit disk?

Proof.

Let

On , compare

with

For ,

while

The difference is

Thus

on .

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

This function has exactly two zeros in , both at counting multiplicity. Hence has exactly two zeros in the unit disk.

The polynomial has real coefficients. Also it has no real zeros in , since

on that interval. Therefore the two zeros in the unit disk occur as a conjugate pair, one in the upper half-plane and one in the lower half-plane.

Thus the closed upper half unit disk contains exactly

solution.

Problem 7.

Assume is an essential singularity of a holomorphic function . Show that it is also an essential singularity of , and in fact of for every .

Proof.

Fix . Suppose, for contradiction, that does not have an essential singularity at . Then the isolated singularity of is either removable or a pole.

If has a removable singularity at , then is bounded near . Hence is bounded near , because

Thus would have a removable singularity at , contradicting the assumption that is essential.

If has a pole at , then

Therefore would have a pole at , again a contradiction.

Thus can be neither removable nor a pole at . Hence is an essential singularity of for every .

Problem 8.

Suppose is holomorphic on and

for all . Suppose further that is real on the unit circle. Show that is real for all real .

Proof.

Since is holomorphic on , it has a Laurent expansion

valid on .

The identity gives

for every integer . Therefore

On the unit circle, , so

Since is real for all , the Fourier coefficients must be real. Hence each is real.

Now let . Since all are real and

we have

Thus is real for every real .