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Math Help - A proof about zero of order

  1. #1
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    A proof about zero of order

    Suppose that f(z) is analytic at z_0.
    Prove that a necessary and sufficient condition that z_0 is a zero of order m is that f(z) can be written as

    \begin{align*}f(z)=(z-z_0)^m g(z)\end{align*} ,

    where g(z) is analytic and nonzero at z_0.

    The proof seems intuition, but I don't know where to start. Thank you.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: A proof about zero of order

    Use that (i) z_0 is a zero of order m if and only if f(z_0)=f'(z_0)=\ldots=f^{(m-1)}(z_0)=0 and f^{(m)}(z_0)\neq 0 (ii) f(z)=\sum_{n=0}^{+\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n in a neighborhood of z_0 .
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    Re: A proof about zero of order

    Thank you for you help!
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