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Let $f$ be analytic at $z=z_0$ and have a zero of $n$th order at $z=z_0$. Then $1/f(z)$ has a pole of $n$th order at $z=z_0$.

I want to prove this, and for this I expand $f(z)$ as a power series, \begin{align*} f(z) = \sum_{k=0}^\infty c_k (z-z_0)^k \end{align*} Since we know that $(z-z_0)$ is zero at $z=z_0$ all the way up to order $n$, i.e. $(z-z_0)^k = 0$ all the way up to $k=n$ (since it can be rewritten as $(z-z_0$ and by the definition of a zero $a_0=0$ such that $f(z_0) = 0$, it follows that \begin{align*} \frac{1}{f(z)} = \frac{1}{\sum_{k=0}^\infty c_k (z-z_0)^k}, \end{align*} such that $1/f \rightarrow \infty$ about the same point. Is this proof complete enough or am I missing something?

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  • $\begingroup$ Well... You said nothing about the order of that pole, it seems. $\endgroup$ – user228113 Dec 2 '15 at 18:24
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Yeah, you're probably going to kick yourself. That expansion doesn't prove anything, for a simple reason:

You're starting these from $k = 0$ but they should start from $k = n$.

Done.

Furthermore an easier way to write this is $f(z) = (z - z_0)^n g(z)$ for some analytic $g$ with $g(z_0) \ne 0,$ and then you'll want to say that $1/f(z) = (z - z_0)^{-n} h(z)$ for $h(z) = 1/g(z)$ also analytic in this neighborhood with $h(z_0) \ne 0$, so the lowest term in the Laurent series is $h(z_0) / (z - z_0)^n,$ and the pole exists and is of order $n$.

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  • $\begingroup$ +1 ... But, if $c_k=0$ for $k<n$, then there was nothing wrong with the OP's series. One needs only recognize that $c_k=0$ for $k<n$ and $c_n\ne 0$. $\endgroup$ – Mark Viola Dec 2 '15 at 18:44

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