UMD Analysis Qualifying Exam/Jan09 Complex
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Problem 2[edit | edit source]
Solution 2[edit | edit source]
Problem 4[edit | edit source]
Suppose that are analytic on with on . Prove that for all implies |
Solution 4[edit | edit source]
Define new function h(z)[edit | edit source]
Define .
h is continuous on the closure of D[edit | edit source]
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 | edit source]
This follows from above
Case 1: h(z) non-constant on D[edit | edit source]
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 | edit source]
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