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Math Help - 2 complex analysis questions

  1. #1
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    2 complex analysis questions

    1.) Let G be a region and let f and g be analytic functions on G such that
    f(z)g(z) = 0 on G. Show that either f(z) = 0 or g(z) = 0 on G.

    2.) Let f be an entire function with constants 0<M, R>0 and positive integer n>1 such that |f(z)|<M|z|^n for all |z|<R. Prove that f is a polynomial of degree less than or equal to n.
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  2. #2
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    Hey

    Quote Originally Posted by grad444 View Post
    1.) Let G be a region and let f and g be analytic functions on G such that
    f(z)g(z) = 0 on G. Show that either f(z) = 0 or g(z) = 0 on G.
    I dont know

    Quote Originally Posted by grad444 View Post
    2.) Let f be an entire function with constants 0<M, R>0 and positive integer n>1 such that |f(z)|<M|z|^n for all |z|<R. Prove that f is a polynomial of degree less than or equal to n.
    f is an entire function, a holomorphic function on D_\infty(0) = D_\infty(z_0)

    => f(x) = \sum^\infty_{n=0} a_n(z-z_0)^n = \sum^\infty_{n=0} a_n*z^n

    show that a_m = 0 if m > N

    |a_m| \le max_{\eta \in \partial D_R(0)} \frac{|f(\eta)|}{r^m}

    For | \eta | = r \ge R is  |f(\eta )| \le M * r^N

    That means |a_m| \le M*r^N / r^m = M * \frac{1}{r^{m-N}}

    -> 0, r -> \infty

    => |a_m| = 0

    => f(z) = \sum^N_{n=0} a_n z^n

    => f is a polynomial of degree less than or equal to N.
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  3. #3
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    Quote Originally Posted by grad444 View Post
    1.) Let G be a region and let f and g be analytic functions on G such that
    f(z)g(z) = 0 on G. Show that either f(z) = 0 or g(z) = 0 on G.
    Hint: If \{ z_n \} is a sequence of convergent distinct points in a region with limit point in the region and h(z_n) = 0 where h is analytic on the region then h is identically zero.
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