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Given that $f:\mathbb{C}\rightarrow \mathbb{C}$ is meromorphic, analytic at $0$ and satisfies $f(1/n)=\frac{n}{2+n}$ I want to know whether I can say $f(z)=\frac{1}{2+z}$, I have considered $g(z)=f(z)-\frac{n}{2+n}$ and zero set of $g$ is $\{\frac{1}{n}\}$ has limit point $0\in\mathbb{C}$, so can I say $f(z)=\frac{1}{2+z}$ by Identity Theorem?If not where is the problem? I know that $f$ is analytic except finitely many points where it has poles.

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I think you want to write $f(z)=\frac{1}{2z+1}$..am I wrong ? –  users31526 Jun 17 '12 at 13:24
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up vote 1 down vote accepted

Yes, that were correct reasoning if your $f$ would coincide with your sequence in the points you considered (it does not, but you fooled me for a minute ;-). If $\{w:f(w)=g(w)\}$ contains an accumulation point in $\Omega$ where $f, g:\Omega \rightarrow \mathbb{C}$ are holomorphic then $f=g$, provided $\Omega\subset \mathbb{C}$ is a connected domain. This quite obviously applies here if you take $\Omega$ as $\mathbb{C}$ minus the possibly existing poles.

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