2024 Spring Qualifying Exam in Complex Analysis

Proof.

Let . We prove that is not an isolated point of .

Since is harmonic, it is real analytic. In a small disk centered at , write

where each is a homogeneous harmonic polynomial of degree . Since , the constant term is zero.

If all terms vanish, then is identically zero in a neighborhood of , and then is certainly not isolated in the zero set.

Otherwise, let be the first degree for which . A nonzero homogeneous harmonic polynomial in two variables has the form

for some constants and . This function vanishes along rays through . By the implicit structure of real analytic zero sets, the zero set of near consists of arcs tangent to those rays. In particular, there are zeros of arbitrarily close to other than itself.

Thus no point of is isolated.

Proof.

No such entire function exists.

Suppose such an entire function existed. Then for all sufficiently large ,

Hence has no zeros outside a large disk. Therefore has only finitely many zeros in .

Let be the polynomial whose zeros, with multiplicity, are exactly the zeros of . Then

is entire. For sufficiently large ,

for some constants . In particular,

as .

Thus is an entire function bounded near infinity, hence bounded on all of . By Liouville's theorem, is constant. Since , this constant must be . But is not identically zero. This contradiction proves that no such entire function exists.

Proof.

Let

We compare and on a large right half-disk

On the imaginary axis, , we have

and

Thus

On the large semicircle , , we have

For sufficiently large, this ratio is greater than uniformly on the semicircle. Hence, by Rouche's theorem on , the functions and have the same number of zeros in .

The function has exactly zeros in the right half-plane, all at , counted with multiplicity. Therefore has exactly zeros in .

It remains to prove the zeros are simple. Suppose is a multiple zero. Then

and

Using the first equation in the second gives

Thus

Equivalently,

So is purely imaginary. This contradicts the fact that all zeros lie in the open right half-plane. Therefore all zeros are simple.

Proof.

Let

This is holomorphic on , so it has a Laurent expansion

On the unit circle ,

Therefore

The Laurent coefficients are the Fourier coefficients:

Since is real and even as a function of , its sine Fourier coefficients vanish. Hence

Also,

so . Thus

as required.

Proof.

(a) If has a removable singularity at , then extends holomorphically to . Since is entire, also extends holomorphically to . Thus has a removable singularity at .

(b) Suppose has a pole at . Then as .

If is a non-constant polynomial of degree , then

as , with . Hence , and has a pole at .

If is transcendental entire, then has an essential singularity at infinity. Since a pole maps a sufficiently small punctured disk around onto the exterior of a sufficiently large disk, the composition has an essential singularity at .

Thus in case (b), has a pole if is a polynomial, and an essential singularity if is transcendental entire.

(c) Suppose has an essential singularity at . By the Casorati-Weierstrass theorem, the image of every punctured neighborhood of under is dense in .

If had a removable singularity or a pole at , then would tend to a limit in or to as . But since takes values arbitrarily close to every complex number near the essential singularity, this would force the non-constant entire function to have only one limiting value on a dense subset of , which is impossible.

Therefore has an essential singularity at .

Proof.

We prove that is locally bounded. Then Montel's theorem implies normality.

Fix . For , choose a number such that every disk lies in for some . On , we have

Hence

for every .

Since is holomorphic, is subharmonic. By the mean-value inequality, for ,

where depends only on , not on .

Also,

For ,

Thus is uniformly bounded on every compact subset of . By Montel's theorem, is a normal family.

Proof.

First assume is entire. On the circle , we have

Thus the condition becomes

The function

is holomorphic on the punctured disk . Since is holomorphic on and on the circle , the identity theorem gives

on the annulus .

Write

Then

For to extend holomorphically to , all negative powers must vanish. Thus for every , so must be constant. This contradicts the assumption that is non-constant. Therefore no such exists for non-constant entire .

If is only assumed holomorphic on , the answer depends on . For example, if

then on ,

so works.

However, not every such works. For instance, gives the same obstruction as above:

which cannot be the restriction of a holomorphic function on the whole unit disk.

Proof.

Let

By Stirling's formula,

with . Also,

Hence

Taking th roots,

Therefore the radius of convergence is

So the radius of convergence is

Proof.

Suppose, toward a contradiction, that

for every .

Define

Since

we have

which is a polynomial. On ,

By the maximum modulus principle,

for all . But

Thus attains its maximum modulus at an interior point. By the maximum modulus principle, is constant. Since , we get

Therefore

Taking the contrapositive, if , then there exists such that