2021 Spring Qualifying Exam in Complex Analysis

Problem 1.

Show that for any integer ,

The source PDF appears to omit the factor on the right-hand side; without it the formula is false already for .

Proof.

Use the sector contour for

with angles and . The only pole inside the sector is

Its residue is

Let

The integral along the ray contributes

The circular arcs vanish in the limit. Hence

Using

we obtain

Problem 2.

Let be holomorphic in and continuous on . If is constant on the boundary of , then prove that is either constant or has finitely many zeros in .

Proof.

Let

on .

If , then on the boundary. By the maximum modulus principle, , so is constant.

Assume . Then has no zeros on the boundary. Since is continuous on , there is a collar neighborhood of the boundary in which has no zeros. Thus all zeros of in lie in a compact subset of the disk.

Zeros of a nonzero holomorphic function are isolated. Therefore, if is not identically zero, it can have only finitely many zeros in that compact subset. Hence is either constant or has finitely many zeros in .

Problem 3.

Define the Bernoulli numbers by

(i) Prove that and

(ii) Find , and .

Proof.

Since

we have

Comparing constant terms gives

For , comparing the coefficient of gives

Using this recurrence:

For ,

Thus

Problem 4.

Let

Prove:

(i) is holomorphic in and continuous on .

(ii) Every point of is a singular point for .

Proof.

(i) For ,

Thus the series converges uniformly on , so is continuous there and holomorphic in .

(ii) This is a lacunary series with exponents

and

Its radius of convergence is , since

By the Hadamard gap theorem, the unit circle is a natural boundary. Hence every point of is a singular point for .

Problem 5.

Let be a bounded domain in with boundary and . Prove or disprove: there is a sequence of polynomials such that

uniformly on .

Proof.

The statement is false.

Suppose such polynomials existed. Since is also a polynomial,

for every .

Uniform convergence on gives

uniformly on . Therefore

But , so by the residue theorem,

a contradiction.

Thus no such sequence of polynomials exists.

Problem 6.

Let be harmonic in and suppose

Prove that is harmonic in .

Proof.

An isolated singularity of a harmonic function has an expansion of the form

where is harmonic near , unless the singularity is removable.

The condition

forces the logarithmic coefficient to be zero. It also rules out any principal part terms , since those grow faster than along suitable directions.

Therefore the singular part vanishes, and has a removable singularity at . Hence extends harmonically to all of .

Problem 7.

Prove that

for any .

Proof.

For the upper bound, write

If , then

For the lower bound,

Thus

Since

we get

Problem 8.

Suppose is holomorphic in and

for all positive integers . Prove that

does not exist.

Proof.

Suppose the limit existed. Since

the limit would have to be . Thus as , so would be a removable singularity of and the extension would satisfy .

Then

would be holomorphic near . But

The values do not converge as , contradicting the continuity of at .

Therefore does not exist.

Problem 9.

Prove that all zeros of

lie in the disk , where

Proof.

Let

Suppose . Then . We estimate

Since

we get

Therefore

Thus every zero satisfies

In particular, all zeros lie in up to the usual harmless open/closed boundary convention.