Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
    
:( :( :( :( :( :( –  Bunuelian Trick Jul 24 '12 at 17:56

1 Answer 1

up vote 1 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.