UMD Analysis Qualifying Exam/Jan08 Complex

Problem 2 edit

Prove there is an entire function   so that for any branch   of  


for all   in the domain of definition of  

Solution 2 edit

Key steps

  • ratio test

Problem 4 edit

Let   be the domain  . Prove that   is a 1:1 conformal mapping of   onto a domain  . What is  ?

Solution 4 edit

Showing G 1:1 conformal mapping edit

First note that


Also, applying a trigonometric identity, we have for all  ,


Hence if  , then




The latter cannot happen in   since   so




Note that the zeros of   occur at  . Similary the zeros of   occur at  .

Therefore from   and  ,   is a   conformal mapping.

Finding the domain D edit

To find  , we only need to consider the image of the boundaries.

Consider the right hand boundary,  


Since  ,


Now, consider the left hand boundary  .


Since  ,


Now consider the bottom boundary  .


Since  ,


Hence, the boundary of   maps to the real line. Using the test point  , we find


We then conclude  

Problem 6 edit

Suppose that for a sequence   and any  , the series


is convergent. Show that   is analytic on   and has analytic continuation to  

Solution 6 edit

Summation a_n Convergent edit

We want to show that   is convergent. Assume for the sake of contradiction that   is divergent i.e.


Since   is convergent in the upper half plane, choose   as a testing point.


Since   converges in the upper half plane, so does its imaginary part and real part.


The sequence   is increasing ( ) since   and   e.g. the gap between   and   is grows as   grows. Hence,


This contradicts that   is convergent on the upper half plane.

Show that h is analytic edit

In order to prove that   is analytic, let us cite the following theorem

Theorem Let   be a sequence of holomorphic functions on an open set  .   Assume that for each compact subset   of   the sequence converges uniformly on  , and let the limit function be  .   Then   is holomorphic.

Proof See Theorem 1.1 in Chapter V, Complex Analysis Fourth Edition, Serge Lang.

Now, define  .   Let   be a compact set of  .   Since   is continuous