2020 Fall Qualifying Exam in Complex Analysis

Problem 1.

Prove that there exists no holomorphic function such that

where is real-valued.

Proof.

If such an existed, then

Thus the imaginary part would be

The Cauchy-Riemann equations give

and

The first equation says

while the second says

Then

which is impossible for a function. Therefore no such holomorphic function exists.

Problem 2.

Let be entire and suppose

for all , where

Prove that

Proof.

First, has no zeros in . Indeed, if had a zero of order at some point in the disk, and in particular if , the lower bound near that zero would be impossible. At , for example, a zero of order would give

but , so this contradicts

for small .

Thus is holomorphic in . On ,

By the maximum modulus principle,

Therefore

Problem 3.

Prove that for any and , the polynomial

has at least one root in the disk

Proof.

If , then is a root. Assume .

Let the roots be . Suppose, toward a contradiction, that

for every . Put

Then .

The reciprocal polynomial is

After dividing by , its roots are , and it has the form

Thus the elementary symmetric functions satisfy

and

Newton's identities then imply

But

for , a contradiction. For , the condition

is already impossible with both .

Therefore at least one root satisfies .

Problem 4.

Evaluate

Proof.

Use the partial fraction decomposition

The apparent singularities are removable in the original integrand. We use

and, in the principal-value sense,

Therefore

Since , this becomes

Problem 5.

Suppose the radius of convergence of

is equal to . Find the radius of convergence of

Proof.

Since the original radius of convergence is ,

It follows that

Let

Then

The new series is

As a power series in , it has radius

Therefore it converges when

The radius of convergence in is

Problem 6.

Find explicitly a conformal mapping of

onto the unit disk.

Proof.

One explicit construction is as follows. First square:

This maps the right half of the unit disk with the slit removed onto

Next use the disk automorphism

which sends the slit to a slit on the negative real axis. Choose the branch of the square root on this slit disk with positive imaginary part on the upper side of the slit:

This opens the slit and maps the domain conformally to a half-disk. A final Mobius transformation maps that half-disk to the unit disk. Thus one explicit map is

with the square-root branch chosen as above. This is a composition of conformal maps and hence is conformal onto the unit disk.

Problem 7.

Let , , be families of holomorphic functions on :

and

Which of are normal?

Proof.

The family is normal. Indeed, if

with , then for ,

Thus is locally bounded, hence normal by Montel's theorem.

The family is not normal. For example, the constant polynomials

belong to , but they are not locally bounded, so the family is not normal as a family of holomorphic functions into .

The family is normal because , and every subfamily of a normal family is normal.

Thus

Problem 8.

Let be simply connected and open. Suppose is holomorphic and not identically equal to . Prove or disprove:

(1) There exists at most one fixed point of in .

(2) There exists exactly one fixed point of in .

Proof.

Since is simply connected and proper, by the Riemann mapping theorem there is a conformal map

Then

is a holomorphic self-map of and is not the identity.

(1) This is true. If had two distinct fixed points in , then by the Schwarz-Pick lemma would have to be the identity. Hence would be the identity, contrary to the assumption. Therefore has at most one fixed point.

(2) This is false. For example, on take

This maps into , but its only fixed point is , which lies on the boundary and not in . Thus it has no fixed point in .

Conjugating this example by a Riemann map gives a counterexample on any simply connected proper domain .