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Suppose $f(z)$ is analytic on $ {z : 0<|z-\alpha|<r}$, where $r>0$. If $m$ is a positive integer and $\lim_{z \rightarrow \alpha} (z-\alpha)^m f(z) =A,$ for $A $ not equal to $0$ or $\infty$, prove that $f$ has a pole of order $m$ at $\alpha$.

I think this is pretty straightforward from the definition of pole. When it is a pole of order m then expanding to the Laurent series we get the principal part has m terms with the lowest power being $-m$. Then multiplying by $(z-\alpha)^m $ and taking limit as $z\rightarrow \alpha.$ we see the non zero guy $a_{-m}$.

I am just confused what is there to be rigorous. Isn't this just a definition? Any suggestion will be appreciated.

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what is your definition of a pole? –  Alex R. Dec 28 '12 at 3:25

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As $\,f\,$ is analytic in $\,1<|z-\alpha|<r\,$ , we can develop $\,f\,$ as a Laurent series

$$f(z)=\sum_{n=-\infty}^\infty a_n(z-\alpha)^n$$

That $\,(z-\alpha)^mf(z)\xrightarrow[z\to\alpha]{} A\neq 0\,$ means that $\,a_n=0\,\,\,\forall n<-m\,$ , and thus $\,\alpha\,$ is a pole of order m

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