A proof about zero of order

**fareastmovement** · December 8th 2011, 12:35 AM

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.

**FernandoRevilla** · December 8th 2011, 08:16 AM

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 $z_0$ .

**fareastmovement** · December 8th 2011, 10:40 AM

Thank you for you help!

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