2024 Fall Qualifying Exam in Complex Analysis

Problem 1.

Find all real-valued harmonic functions on the whole complex plane such that for all ,

Proof.

Observe that

and

Thus

is harmonic on .

Then

is harmonic on and satisfies

for all . An entire harmonic function bounded above is constant. Therefore

for some real constant .

Hence all such functions are

Problem 2.

Let

(a) Prove that is an entire function.

(b) Evaluate

Proof.

(a) Let

Using Stirling's formula , we get

Hence

so the radius of convergence is infinite. Therefore is entire.

(b) By the residue theorem,

Now

and

The coefficient of comes only from the first term in . All powers with have only powers with .

Thus the residue is , and

Problem 3.

Let be a polynomial of degree , and let

Prove that is increasing and that

is decreasing.

Proof.

The function is increasing by the maximum modulus principle. Indeed, if , then the maximum of on is attained on , so

For the second assertion, define the reciprocal polynomial

This is a polynomial of degree at most . For ,

Since is increasing in , if then , so

Equivalently,

Thus is decreasing.

Problem 4.

Prove that

admits a Laurent expansion in the annulus

and compute this expansion.

Proof.

For and ,

Therefore

For and ,

Thus

Combining the two expansions, for ,

Problem 5.

Let and be two complex numbers and assume that is not equal to any integer . Show that the radius of convergence of

is at least and can be infinity in some cases.

Proof.

Let

Since is not a non-positive integer, none of the denominator factors is zero.

If the numerator does not eventually vanish, then

Therefore the radius of convergence is .

If the numerator eventually vanishes, then the series is actually a polynomial. This happens, for example, when

In that case the radius of convergence is infinite.

Thus the radius of convergence is always at least , and it can be infinity.

Problem 6.

Let and let be the family of all power series

with coefficients in such that for all ,

Show that such power series converge on the unit disk and form a normal family on the unit disk. Is the statement still true if ?

Proof.

Assume . Then

Hence every such power series has radius of convergence at least , so it converges in .

To prove normality, fix . For ,

Since , the series on the right converges. Therefore the family is uniformly bounded on every compact subset of . By Montel's theorem, it is a normal family.

If , the statement is false. Take

Then

has radius of convergence , so it does not even converge on the whole unit disk.

Problem 7.

Let be holomorphic on the disk for some . Show that there exists with such that

Proof.

Suppose, toward a contradiction, that

for every . Multiplying by , this is equivalent to

on .

Let

Then is holomorphic in a neighborhood of the closed unit disk and .

On ,

By Rouche's theorem, and the constant function have the same number of zeros in . The constant function has no zeros, so should have no zeros in the unit disk. But , a contradiction.

Therefore there exists such that

Problem 8.

Let be an entire function with , and suppose has at least zeros in the unit disk . Show that for each , there exists a point on the boundary of such that

Proof.

Let

If , then by the maximum modulus principle,

Thus the result holds for .

Now assume . Let be zeros of in , counted with multiplicity. Then . Jensen's formula gives

Since , this implies

Therefore

Hence there exists with such that

Problem 9.

Let be a bounded sequence of holomorphic functions on the unit disk and let be a sequence in that converges in . Assume that

exists for all . Show that converges uniformly on compact subsets of .

Proof.

We use Vitali's theorem. The family is bounded on , so it is locally bounded. By Montel's theorem, every subsequence has a further subsequence converging uniformly on compact subsets of to a holomorphic limit.

The pointwise limits

exist for all , and the sequence has a limit point in . Interpreting the hypothesis in the standard way, the points give a set with an accumulation point in . Therefore any two subsequential holomorphic limits must agree on a set with an accumulation point. By the identity theorem, the two limits are identical.

Thus every subsequence of has a further subsequence converging to the same holomorphic limit. It follows that the whole sequence converges uniformly on compact subsets of .