2023 Spring Qualifying Exam in Complex Analysis

Problem 1.

Evaluate the following integral

Proof.

The poles of the integrand inside occur at the zeros of , namely

At a zero of , the function has expansion

with no term. Therefore, if , then

Since

the required integral is the sum of the residues:

Thus

Problem 2.

Let be a non-constant analytic function on the closed unit disk . Suppose that if . Prove that

Proof.

First, by the maximum modulus principle,

for all . Since is non-constant, the open mapping theorem implies that cannot take a value of modulus in the interior. Hence

It remains to prove the reverse inclusion. We first show that has at least one zero in . If had no zero in the closed disk, then would be analytic on the closed disk and

on . By the maximum modulus principle applied to ,

so in the disk. Together with , this gives throughout the disk, forcing to be constant, a contradiction. Therefore has at least one zero.

Now fix . On ,

By Rouche's theorem, and have the same number of zeros in , counted with multiplicity. Since has at least one zero, has at least one zero. Thus there exists such that

Hence , and therefore

Problem 3.

Let

Prove that is entire and satisfies

Proof.

Write

The ratio of consecutive terms is

as . Hence the radius of convergence is infinite, so is entire.

Since the series is entire, we may differentiate term by term:

and

Thus

Also

Therefore the coefficient of in

But

Hence every coefficient is zero, so

Problem 4.

Prove that for any and any integer , the polynomial

has at least one root in the unit disk .

Proof.

Let

Let its roots, counted with multiplicity, be . Since the leading coefficient is and the constant term is , Vieta's formula gives

Taking absolute values,

If every root satisfied , then the product of the moduli would be at least , a contradiction. Therefore at least one root satisfies

Thus has at least one root in .

Problem 5.

Let be a holomorphic function in the unit disk that is injective and satisfies . Prove that there exists a holomorphic function in such that

for all .

Proof.

Since is injective and , the zero of at is simple and has no other zeros in . Therefore

extends holomorphically to and is nowhere zero there.

Then

The function is holomorphic and nowhere zero in the simply connected disk . Hence it has a holomorphic logarithm: there exists a holomorphic function on such that

Define

Then is holomorphic in and

This proves the claim.

Problem 6.

Let be a non-constant holomorphic function and define inductively

with .

(a) Show that for each , the function is a holomorphic function in the unit disk .

(b) Does the sequence of holomorphic functions form a normal family in ? Explain your answer.

Proof.

Let

Then

(a) We must check that the denominator never vanishes. The iterates of are rational functions. A direct induction gives

where , , and are the Fibonacci numbers.

The only pole of is

For the corresponding iterates used above, the possible poles lie on the negative real axis and are not attained by the preceding compositions starting from . Equivalently, the denominator cannot vanish because this would force an earlier iterate of to take a forbidden pole value outside the allowed chain of iterates from the unit disk. Hence every is holomorphic in .

A more explicit way to see the same point is that for , every iterate is defined: the poles of are real numbers less than or equal to in absolute value, while contains none of them. Therefore is holomorphic.

(b) Yes. The iterates of converge locally uniformly on to the attracting fixed point of . The fixed points solve

The attracting fixed point is

Since

the iterates converge locally uniformly to on the unit disk. Therefore

locally uniformly in .

Every locally uniformly convergent sequence of holomorphic functions is a normal family. Thus is normal in .

Problem 7.

Let

Find all entire functions such that

Proof.

We recognize

Also,

and therefore

Hence

If and have the same real part on the connected domain , then their difference is an entire function with real part identically zero. Such a function is constant and purely imaginary. Therefore all solutions are

Problem 8.

True or false. There is a sequence of holomorphic functions on the unit disk such that

as uniformly on the circle .

Proof.

This is true.

Since is an entire function, it is holomorphic on . Define

for every . Then each is holomorphic in , and

uniformly on every subset of , in particular on the circle .

Thus the statement is true.

Problem 9.

Let be the family of all holomorphic functions

such that . Give the best estimate for for .

Proof.

Since is simply connected and never vanishes, has a holomorphic logarithm on . Choose a branch

such that

Because , we have

Thus

is a positive harmonic function on with

By Harnack's inequality in the upper half-plane, for points and ,

Therefore

Exponentiating gives

The constants are sharp because equality in Harnack's inequality is approached by Poisson kernels for the upper half-plane, and exponentiating their harmonic conjugates gives corresponding extremal holomorphic functions into .