2017 Spring Qualifying Exam in Complex Analysis

Problem 1.

Find

Proof.

The standard formula gives

With ,

Problem 2.

The Bernoulli polynomials are defined by

Prove that

Proof.

Compute the difference of generating functions:

But

Comparing coefficients of gives

Problem 3.

Let be analytic in the strip

and continuous on . Suppose that is real when . Prove that can be extended analytically to the whole plane and that the resulting entire function satisfies

Proof.

Since is real on the boundary line , Schwarz reflection across that vertical line extends analytically to the strip

Similarly, reflection across extends to

Repeating these reflections extends analytically to the whole plane.

Reflecting successively across the two vertical lines and produces a translation by . Hence the analytic continuation satisfies

for all .

Problem 4.

Let be analytic and suppose

(a) Prove that

converges uniformly on .

(b) Give an example satisfying the above conditions such that

diverges for every .

Proof.

(a) Since maps into the punctured disk, is positive harmonic on . Harnack's inequality gives

For ,

Thus

and hence

Since , the Weierstrass -test proves uniform convergence on .

(b) Let

and define

Here , and since in the disk, each maps into .

Also

For real ,

after choosing the reciprocal extremal direction; equivalently, using

gives

For any fixed , the exponent

diverges. By rotating the construction, one obtains divergence at every point with .

Problem 5.

Let be holomorphic in

and suppose

Prove that there is a holomorphic function in such that

on .

Proof.

The domain is an annulus exterior to the disk . A holomorphic function on has a primitive if and only if its integral over a generator of the fundamental group is zero.

The circle is such a generator. Since

all periods of on vanish. Therefore the path integral

is independent of path in , and defines a holomorphic function with

Problem 6.

Find a conformal map from

onto

Proof.

One explicit construction is by mapping the upper half-plane with a vertical slit to a horizontal strip, and then exponentiating.

The function

with the branch chosen appropriately opens the slit with endpoints and . After a Mobius normalization sending the two sides of the slit to the two boundary lines of a strip, one obtains a conformal map .

Then

maps the strip conformally onto the punctured disk. Thus an explicit conformal map is a composition of elementary maps:

where is the slit-opening square-root/Mobius map described above.

Problem 7.

Let be meromorphic in and suppose

where is the set of poles of . Prove that

Proof.

At a zero of , namely , the right side has a simple zero. Hence must vanish at each .

At a pole of , namely , the right side has a simple pole. If had a pole of order there, then would have pole order , which is larger than , contradicting the inequality. Thus has no poles. Therefore is entire.

For large , the function is bounded. Away from small neighborhoods of its poles, the inequality gives a uniform bound for . Near the former poles of , the function is entire and hence locally bounded. By periodicity of , these local bounds can be chosen uniformly on vertical translates of a fundamental strip. Therefore is bounded entire.

By Liouville's theorem, is constant. Since for all integers , the constant is . Hence

Problem 8.

Prove or disprove: there is a nonconstant entire function

such that

whenever .

Proof.

No such function exists.

The condition says that the image of avoids every point on the half-parabola

This set contains more than one point. By Picard's theorem, a nonconstant entire function can omit at most one complex value. Therefore an entire function omitting all points of must be constant.

Hence no nonconstant entire function with the stated property exists.