UMD Analysis Qualifying Exam/Jan09 Complex

Problem 2Edit

Solution 2Edit

Problem 4Edit

Suppose that   are analytic on   with   on  . Prove that   for all   implies  

Solution 4Edit

Define new function h(z)Edit

Define  .

h is continuous on the closure of DEdit

Since   on  , then by the Maximum Modulus Principle,   is not zero in  .

Hence, since   and   are analytic on   and   on  , then   is analytic on   which implies   is continuous on  

h is analytic on DEdit

This follows from above

Case 1: h(z) non-constant on DEdit

If   is not constant on  , then by Maximum Modulus Principle,   achieves its maximum value on the boundary of  .

But since   on   (by the hypothesis), then

  on  .

In particular  , or equivalently


Case 2: h(z) constant on DEdit

Suppose that   is constant. Then


Then from hypothesis we have for all  ,


which implies


Hence, by maximum modulus principle, for all  




Since  , we also have


Problem 6Edit

Solution 6Edit