# if $g(z)=\sum_{n=0}^{\infty}c_nz^{-n}$ is meromorphic at $z=0$ then $g$ must have finitely many terms?

well,if $g(z)=\sum_{n=0}^{\infty}c_nz^{-n}$ is meromorphic at $z=0$ then $g$ must have finitely many terms? why? please help if I am missing any trivial arguement. Thank you. Oh! if it has infinitely many terms then it will have infinitely many poles also, but for meromorphic function pole must be finitely many, it is the argument?

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Basically you have the argument. But if you want to be double sure, you should include in the question what you learned as the definition of meromorphicity. – Willie Wong Jul 24 '12 at 17:53
Your title does not match the body of your question. – Chris Eagle Jul 24 '12 at 17:55
:( :( :( :( :( :( – La Belle Noiseuse Jul 24 '12 at 17:56

I don't remember seeing the notion of "meromorphic at a point" anywhere. A function can be meromorphic on an open set. A more precise way to phrase your question would be: if $0$ is a pole, is the negative part of the Laurent series finite? And the answer is yes. If the pole has order $k$, multiply the function by $z^k$ to make the product bounded near $0$, and apply the removable singularity theorem to the product. The conclusion follows.