UMD Analysis Qualifying Exam/Jan09 Complex

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Problem 2[edit]

Solution 2[edit]

Problem 4[edit]

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

Solution 4[edit]

Define new function h(z)[edit]

Define .

h is continuous on the closure of D[edit]

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[edit]

This follows from above

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

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[edit]

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[edit]

Solution 6[edit]