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Suppose $f$ is meromorphic on $\mathbb{C}_{\infty}$.

Is it true that if $f$ has a pole only at $\infty$ $\iff$ $\operatorname{Res}[f,\infty]=0$ ?

My attempt: If $f$ has a pole only at $\infty$ then $g(z)=f(\frac{1}{z})$ has pole at $0$ of order $n$ say, so $z^ng(z)$ is bounded near $0$ and that means, $z^{-n}f(z)$ is bounded near $\infty$ and hence $f$ is a polynomial. So $\operatorname{Res}[f,\infty]=0$.

Am I right?

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It looks fine as far as it goes, but that shows only one direction of the equivalence. The opposite direction is in fact false, as the example $f(z)=z^{-2}$ shows. –  Harald Hanche-Olsen Aug 4 '12 at 12:17
    
thank you........... –  miosaki Aug 4 '12 at 19:52
    
@Harald: You could turn that into an answer. –  joriki Aug 5 '12 at 7:23
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@joriki: Done. I had left it as a comment because I thought it was too brief, maybe someone wanted to flesh it out a bit, but nobody has done so, so now it's an answer too. I guess it's important to have an answer so the question doesn't sit around as unanswered for an eternity. –  Harald Hanche-Olsen Aug 5 '12 at 8:04
    
@Harald: Exactly, thanks. –  joriki Aug 5 '12 at 8:34
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up vote 4 down vote accepted

It looks fine as far as it goes, but that shows only one direction of the equivalence. The opposite direction is in fact false, as the example $f(z)=z^{−2}$ shows.

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