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?

Question

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&#039;m curious, what if f doesn&#039;t have a singularity at   ∞, or in fact has an essential singularity at   ∞ instead? Is   f still a rational function?

Bruno Dixon · Accepted Answer

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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