Hw6 - Mathematics 185 – Intro to Complex Analysis Fall 2009 – M Christ Solutions to Selecta from Problem Set 6 2.4.3 Let f be an entire

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Unformatted text preview: Mathematics 185 – Intro to Complex Analysis Fall 2009 – M. Christ Solutions to Selecta from Problem Set 6 2.4.3 Let f be an entire function. Suppose there exist a positive integer n and positive real numbers M,N such that | f ( z ) | ≤ M | Z | n whenever | z | ≥ N . Show that f is a polynomial of degree ≤ n . Solution. By a problem from earlier in the semester, it suffices to show that f ( n +1) ( z ) = 0 for every z ∈ C . We follow the same method as in the proof of Liouville’s Theorem. Fix any z . Let R be a large positive parameter, which will eventually be allowed to tend to infinity but which should be regarded as being fixed, for the present. Note that if | w- z | = R then | w | ≥ R- | z | ; if we choose R ≥ N + | z | then | w | ≥ N , so | f ( w ) | ≤ M | w | n . Moreover | w | ≤ R + | z | , so | f ( w ) | ≤ M ( R + | z | ) n . Therefore Cauchy’s inequality says that | f ( n +1) ( z ) | ≤ ( n + 1)! R n +1 M ( R + | z | ) n . This holds for every R ≥ N + | z | . Clearly the right-hand side tends to 0 as R → ∞ . Thus | f ( n +1) ( z ) | is smaller than every positive number, so it vanishes. 2.4.4 Let f be analytic “inside and on” a simple closed curve γ . Suppose that f ( w ) = 0 for all w “in” γ . Show that f ( z ) = 0 for all z “inside” γ . Solution. Assume that γ is piecewise C 1 ; otherwise we cannot attack this problem using what we’ve learned so far. Also assume that f is analytic in some open set which contains γ and the region “inside” γ . (Both the problem statement, and Cauchy’s Theorem 2.2.1 for simple closed curves, are ambiguous on this point.)Theorem 2....
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This note was uploaded on 10/10/2009 for the course MATH 185 taught by Professor Lim during the Spring '07 term at University of California, Berkeley.

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Hw6 - Mathematics 185 – Intro to Complex Analysis Fall 2009 – M Christ Solutions to Selecta from Problem Set 6 2.4.3 Let f be an entire

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