2023 Spring Qualifying Exam in Complex Analysis

Problem 1.

Prove or disprove that there exists a holomorphic function

such that

Solution.

Such a holomorphic function does exist.

Let

Since

we have

Thus

The poles of this rational function are , , and , all of which lie inside the circle .

The residue at is

The sum of the residues of at and is , because the coefficient of in its Laurent expansion at infinity is . Therefore

Hence the sum of all residues inside is

Consequently, for every ,

Every closed curve in the domain has winding number equal to an integer multiple of the winding number of such a circle. Therefore every period of on this domain is zero. By the primitive criterion, has a holomorphic primitive on .

Equivalently, one can see this directly from the Laurent expansion at infinity. For ,

and

The coefficient of in the first series is , while that in the second series is also , so these coefficients cancel in . Therefore the Laurent series of contains no term and can be integrated term by term to produce the desired holomorphic function .

Remark.

An explict expression of is

Problem 2.

Let be a holomorphic function on

such that and whenever . Let . Prove that there exists a holomorphic function on such that

for every .

Proof.

Since has no zeros in , the point is the only zero of . Let be its order. Then

where is holomorphic and nonvanishing on .

Because is simply connected and has no zeros, has a holomorphic logarithm. Thus there exists a holomorphic function on such that

For ,

Define

Then is holomorphic on and

Therefore the required holomorphic function exists.

Problem 3.

Find all entire functions such that

for every , and

for all .

Solution.

The only such entire function is

Suppose, toward a contradiction, that is not identically zero. If vanishes at , let be the order of the zero at and write

where is entire and . If , take and .

For and ,

Let denote the number of zeros of in , counted with multiplicity. Jensen's formula gives

where are the zeros of . Therefore

Every zero satisfying contributes at least , so

Thus

On the other hand, for sufficiently large , every point

has modulus at most . Hence has at least

zeros in . This grows on the order of , contradicting .

Therefore must be identically zero.

Problem 4.

Let be an entire function. Suppose there exist constants and , and a sequence , such that for every ,

Prove that is a polynomial of degree at most .

Proof.

Let be any integer satisfying . By Cauchy's estimate on the circle ,

Using the assumed bound,

Since and ,

Therefore

for every integer .

The Taylor series of about is

All coefficients with vanish, so this series is a polynomial. Its degree is at most , and hence at most .

Problem 5.

Let and let be the family of holomorphic functions on such that for every ,

Prove that is a normal family.

Proof.

By Montel's theorem, it is enough to prove that is locally uniformly bounded.

Fix , and choose such that

For , Cauchy's integral formula gives

Therefore

Since

we obtain

By the Cauchy-Schwarz inequality,

Hence

for every and every .

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

Problem 6.

Let be a real-valued harmonic function on satisfying

and

Prove that

and show that the constant is sharp by providing an example that achieves the bound.

Proof.

Fix . By the Poisson formula on the disk of radius ,

Differentiating at the origin gives

and

Let be a unit vector. Then

Since ,

Taking the supremum over all unit vectors gives

Letting , we obtain

To show sharpness, consider

This maps conformally onto the right half-plane. The principal logarithm is holomorphic there, and

Define

Then is real-valued and harmonic on , , and .

Moreover,

It follows from the Cauchy-Riemann equations that

Hence

Therefore the constant is sharp.

Problem 7.

Use contour integration to prove that for every ,

Proof.

Let

For a positive integer , let be the positively oriented square with vertices

For sufficiently large , the square contains the poles and , as well as the integers .

On , the function is uniformly bounded independently of . Also,

for all sufficiently large and some constant . Since the length of is ,

At an integer , the residue of is , so

At ,

Using

we get

Similarly,

Therefore the sum of these two residues is

By the residue theorem,

Letting , the contour integral tends to zero. Hence

Problem 8.

Find the number of zeros of the polynomial

in the unit disk .

Solution.

We first show that has no zeros on .

Suppose and . Then

Equality holds in the triangle inequality. Since each of the three terms , , and has modulus , they must all have the same argument. Therefore

and

Since

we must have . But

a contradiction. Thus has no zeros on the unit circle.

For , define

On ,

By Rouché's theorem, and have the same number of zeros in . Hence has exactly five zeros in , counted with multiplicity.

Now consider the continuous family

For , the same strict Rouché estimate shows that has no zeros on . For , this was proved above. Thus no zero crosses the unit circle as varies from to . By the argument principle, the number of zeros inside remains constant.

Therefore has exactly zeros in the unit disk, counted with multiplicity.

Problem 9.

For any domain , let denote the class of automorphisms of , that is, invertible conformal maps from to itself. Let

(a) Show that every extends uniquely to a function , and explain why must map the set of punctures

onto itself.

(b) Find explicitly. Justify your answer.

Solution.

(a) Let . Since , the function is bounded near each puncture. Therefore the isolated singularities at and are removable. Hence extends to a holomorphic map

The extension is unique by the identity theorem.

The inverse map is also bounded and therefore extends holomorphically to a map

On ,

and

By the identity theorem, both identities hold on all of . Thus is an automorphism of .

Let

Since maps bijectively onto itself, its extension cannot map a point of into . Otherwise, applying the extended inverse would imply that a point of has a preimage in , contradicting the fact that maps into . Therefore

Since is bijective, we have

(b) Put

An automorphism of preserving the set either fixes both points or interchanges them.

If an automorphism fixes both and , then it is the identity. Indeed, after conjugating by a disk automorphism sending to , it becomes a disk automorphism fixing and another nonzero interior point. An automorphism fixing has the form

and fixing a nonzero point forces .

It remains to find the automorphism interchanging and . For , define

Then is a disk automorphism sending to . Set

Since

the map

is an automorphism of . By construction,

and

A simplification gives

There can be at most one automorphism interchanging and , because the composition of any two such automorphisms fixes both points and is therefore the identity. Consequently,