UMD Analysis Qualifying Exam/Jan08 Complex

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Problem 2Edit

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


 


for all   in the domain of definition of  

Solution 2Edit

Key steps

  •  
  •  
  • ratio test

Problem 4Edit

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

Solution 4Edit

Showing G 1:1 conformal mappingEdit

First note that


 


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


 


Hence if  , then


 


or


 


The latter cannot happen in   since   so


 


i.e.


 


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


Therefore from   and  ,   is a   conformal mapping.

Finding the domain DEdit

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 6Edit

Suppose that for a sequence   and any  , the series


 


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


Solution 6Edit

Summation a_n ConvergentEdit

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 analyticEdit

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