2025 Spring Qualifying Exam in Complex Analysis

Proof.

Since is holomorphic on , it has a Laurent expansion

on .

Having a pole of order at means that the negative part stops at and that :

Having a pole of order at means that has a pole of order at . Equivalently, the positive part stops at and :

Therefore the functions are exactly

Proof.

Since the only singularities of are poles at the zeros of , the partial fraction decomposition gives

where is the multiplicity of the zero .

The coefficient is the residue of at :

Let be so large that all zeros of lie inside . By the residue theorem,

If , then as ,

Therefore

as . Hence

so

Proof.

Let

The Cauchy-Riemann equations imply that is holomorphic, and the condition says that has no zeros on .

Now

Therefore

Since is holomorphic on , its real part is harmonic. Hence the given function is harmonic.

A harmonic conjugate is the imaginary part of :

up to an additive real constant. This conjugate exists globally because is a globally defined holomorphic function on .

Proof.

Consider

Although this expression appears to have a pole at , the factor removes all negative powers because has degree . Thus is holomorphic in the unit disk and continuous on its closure.

On , write . Then

Hence

on . By the maximum modulus principle,

for .

Now choose such that

This is equivalent to

so

Then

and therefore

Thus

Proof.

Let

Integrate

over the sector contour with angles and , radius , and a small circle around the origin. The only pole inside the sector is

because

The residue is

The circular arc contribution tends to as . The integral along the positive real axis tends to . Along the ray , oriented inward,

and

Thus the ray contributes

By the residue theorem,

Using

we obtain

Proof.

Let be a disk automorphism sending to , and define

Then is holomorphic and

Also, since is another fixed point of , the point

is a nonzero fixed point of :

By Schwarz's lemma,

for all . Since equality holds at the nonzero point , Schwarz's lemma implies that

for some real . But and , so . Thus

Conjugating back by , we get

for all .

Proof.

Consider

The two removed rays and are mapped to the negative real axis. Hence maps conformally onto

Taking the principal square root maps this slit plane onto the right half-plane:

where the square root is chosen with positive real part.

Finally, the map

maps the right half-plane conformally onto the unit disk. Therefore an explicit conformal map is

Proof.

For ,

Thus is bounded on . The same estimate shows that the series converges uniformly on , by the Weierstrass -test. Hence extends continuously to .

It remains to show that no boundary point admits holomorphic continuation. We first show that every root of unity is a singular boundary point. Let be a root of unity of order . For all sufficiently large , divides , so

Along the radius ,

For all sufficiently large ,

Therefore

becomes unbounded as . If had a holomorphic extension to a neighborhood of , then would be bounded near , a contradiction.

Thus every root of unity is a singular boundary point. Since the roots of unity are dense on , no point of can have a neighborhood across which extends holomorphically. Indeed, any such neighborhood would contain a root of unity on the boundary, contradicting what we just proved.

Proof.

For an example without the growth condition, take

Then and are distinct entire functions, and for every ,

Now assume the growth condition. Let

Then is entire,

for every , and

If is not identically zero, Jensen's formula gives, for large ,

after removing a possible zero at .

The right-hand side is at most for some constant , because

On the other hand, the left-hand side grows like

This contradicts the bound for large .

Therefore , so . Hence no distinct pair can satisfy both the interpolation condition and the growth condition.