2015 Spring Qualifying Exam in Complex Analysis

Problem 1.

Prove that for each , every solution of

must satisfy

Proof.

Let be a solution. Since does not satisfy the equation, . The equation gives

Hence

Write . Then

Therefore

Since the moduli are equal,

Thus

and so

Problem 2.

Classify all singularities and find the associated residues for

Proof.

The possible singularities are , , and .

At , the factor has an essential singularity, while the denominator is nonzero. Hence is an essential singularity.

At , the denominator has a simple zero and is analytic and nonzero, so is a simple pole. Its residue is

At , the denominator has a double zero, so is a double pole. Write

For a double pole, the residue is . Since

we get

Thus

It remains to find the residue at the essential singularity . Since

there is no term in the Laurent expansion at infinity, so the residue at infinity is . By the residue theorem on the Riemann sphere,

Therefore

Problem 3.

Expand in a series of powers of each branch of defined by

where one branch satisfies and the other satisfies .

Proof.

The equation can be rewritten as

Equivalently,

Thus

Using the binomial expansion

we obtain the branch with :

The branch with is

In compact form,

and

valid for .

Problem 4.

Evaluate

Proof.

Consider

The desired integral is .

Factor the denominator:

Close the contour in the upper half-plane. Since decays there, Jordan's lemma applies. The only pole in the upper half-plane is

Hence

The residue is

Now

and

Therefore

Taking imaginary parts gives

Problem 5.

Suppose is holomorphic and

Give an upper bound for , and characterize the functions for which equality holds.

Proof.

By the Schwarz-Pick lemma,

Since ,

Equality in Schwarz-Pick occurs exactly when is an automorphism of the unit disk. The disk automorphisms mapping to are

For these functions,

Thus the sharp upper bound is

with equality exactly for the automorphisms above.

Problem 6.

Let be meromorphic in a neighborhood of the closed unit disk , and suppose it has only one singular point on the circle , which is a simple pole. Show that

where .

Proof.

Let the residue of at the simple pole be . Then near ,

where is holomorphic in a neighborhood of the closed unit disk, and in fact holomorphic in some larger disk with after removing the pole.

For ,

Thus the coefficient of coming from the pole is

Write the Taylor expansion of at as

Since is holomorphic on for some , Cauchy's estimates give

Therefore

where

Since and with ,

This proves the desired asymptotic formula.

Problem 7.

True or false: there exists a bounded harmonic function on the upper half-plane that cannot be extended to any larger domain. Explain your answer.

Proof.

The statement is true.

Choose a measurable set such that both and its complement have positive measure in every nonempty interval. Let

be the Poisson integral of . Then is harmonic on the upper half-plane and satisfies

so it is bounded.

The nontangential boundary values of are almost everywhere. Suppose extended harmonically across some open interval . Then the extended harmonic function would be continuous in a neighborhood of , so its boundary values on would agree almost everywhere with a continuous function.

But is not equal almost everywhere to any continuous function on any interval, because both and have positive measure in every subinterval. This contradiction shows that cannot be extended harmonically across any boundary interval.

Therefore cannot be extended to any larger domain containing the upper half-plane.

Problem 8.

Suppose is analytic in an annulus

and there exists a sequence of polynomials converging to uniformly on compact subsets of the annulus. Show that is analytic on the disk .

Proof.

Since is analytic in the annulus, it has a Laurent expansion

valid for .

We show that all negative Laurent coefficients vanish. Fix with

For , the Laurent coefficient is

Because uniformly on the circle ,

But each is a polynomial, so it has no negative powers in its Laurent expansion about . Therefore

for every . Hence for all .

Thus the Laurent expansion of has no principal part:

This power series converges for , so it defines an analytic extension of to the whole disk .