UMD Analysis Qualifying Exam/Jan09 Complex

Problem 2

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Solution 2

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Problem 4

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

Solution 4

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Define new function h(z)

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Define  .

h is continuous on the closure of D

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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 D

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This follows from above

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

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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 D

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Suppose that   is constant. Then


  where  


Then from hypothesis we have for all  ,


 


which implies


 


Hence, by maximum modulus principle, for all  


 


i.e.


 


Since  , we also have


 

Problem 6

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Solution 6

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