I know that an analytic function on C with a nonessential singularity at ∞ is...

nidantasnu

nidantasnu

Answered

2022-07-09

I know that an analytic function on C with a nonessential singularity at is necessarily a polynomial.
Now consider a meromorphic function f on the extended complex plane C ^ . I know that f has only finitely many poles, say z 1 , , z n in C . Suppose also that f has a nonessential singularity at .
Then if z i have orders n i , it follows that ( z z i ) n i f ( z ) is analytic on C , and has a nonessential singularity at , and is thus a polynomial, so f is a rational function.
But I'm curious, what if f doesn't have a singularity at , or in fact has an essential singularity at instead? Is f still a rational function?

Answer & Explanation

Bruno Dixon

Bruno Dixon

Expert

2022-07-10Added 14 answers

The exponential function has an essential singularity at , but is meromorphic everywhere else -- and is not rational.
On the other hand, if the singularity at is removable, then the proof you sketch still works, and shows that f must be a rational function.

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