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Thread: A proof about zero of order

  1. #1
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    A proof about zero of order

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

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

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

    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) $\displaystyle z_0$ is a zero of order $\displaystyle m$ if and only if $\displaystyle f(z_0)=f'(z_0)=\ldots=f^{(m-1)}(z_0)=0$ and $\displaystyle f^{(m)}(z_0)\neq 0$ (ii) $\displaystyle f(z)=\sum_{n=0}^{+\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$ in a neighborhood of $\displaystyle z_0$ .
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  3. #3
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    Re: A proof about zero of order

    Thank you for you help!
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