UMD Analysis Qualifying Exam/Jan09 Complex

Problem 2 Solution 2 Problem 4 Suppose that are analytic on with on . Prove that for all implies Solution 4 Define new function h(z) Define . h is continuous on the closure of D 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 This follows from above Case 1: h(z) non-constant on D 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 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 Solution 6