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Let f be an analytic function on $\mathbb{C}$ except for isolated singularities and assume that there exists an $R>0$ and $M>0$ such that for $|z|>R \,\,\,|f(z)|\leq \frac{M}{|z|^{\alpha}}$ where $\alpha >1$. I want to show that f has finitely many singularities and all of them inside $|z|\leq R$ so f is meromorphic. Any help would be great.

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  • $\begingroup$ The only other possibility is an essential singularity, for which points near the singularity can have any complex value except for one number $w$, and you cannot write the bound on $|f(z)|$ as you have above. $\endgroup$ Nov 12, 2014 at 3:00
  • $\begingroup$ I know the properties about essential singularities you wrote but how that means f has only finitely many singularities, thanks. $\endgroup$
    – user135582
    Nov 12, 2014 at 3:18

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Since $f$ is bounded on $|z|> R$, any singularity in $\{|z|> R\}$ is removable. All other singularities are in $\{|z|\le R\}$. Since the singularities are isolated and $\{|z|\le R\}$ is compact, there are only a finite number of them. This does not imply that $f$ is meromorphic, since some of the singularities could be essential. Consider for example $f(z)=e^{1/z^2}-1$. If $|z|>1$ then $|f(z)|\le M/|z|^2$, and $z=0$ is an essential singularity.

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