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well, could any one give me a hint for this one? $f$ is meromorphic and defined on $\mathbb{C}\setminus F$ where $F$ is a finite set. Assume $f$ has no essesntial singularity, then we need to show $f$ is a rational function.

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Sorry, my answer was not correct,I removed it. –  user20266 Jun 24 '12 at 10:12
    
You should add explicitly that $f$ is also meromorphic at infinity, else any entire non polynomial function, like $e^z$, is a counterexample . –  Georges Elencwajg Jun 24 '12 at 10:12
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Dear @ Thomas, please undelete it (click on "undelete" right under your answer or write a new answer) after a little modification: conclude your correct argument at the beginning with a reminder that an entire function meromorphic at infinity is a polynomial, because if its Taylor series were infinite it would yield a singularity at infinity. Just do it, Thomas! –  Georges Elencwajg Jun 24 '12 at 10:26
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up vote 2 down vote accepted

Each point $a\in F$ is either removable singularity or a pole. The latter is removable for $(z-a)^nf(z)$ where $n$ is large enough. In this way we get a polynomial $q$ such that $qf$ is entire. Since there is no essential singularity at $\infty$ either, $qf$ is a polynomial. $\Box$

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