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I came up with the following idea but I don't have a proof for it.

If $f $ is a meromorphic function on $\mathbb{C}$ with finite poles and zeros such that $\lim_{z\to 0} f(1/z) $ exists on $\mathbb{C}$, $\oint_{| z|=R} \frac{f'}{f}\;dz =0$ for a big enough $R>0 $.

Even though I don't have a proof for this fact, I have good ideas to think it is true. From a fact that I'm trying to prove, the meromorphic functions $ f$ such that $\lim_{z\to 0} f(1/z) $ exists, are only the reational functions which are actually meromorphic functions on $\overline{\mathbb{C}} $. So the previous limit can be thought as $f(\infty) $.

The argument principle told us that $\oint_{| z|=R} \frac{f'}{f}\;dz =N-P$ where $P $ is the number of poles and $ N$ is the number of zeros of $ f$ (with multiplicity), for a big enough $R>0 $. Given that $ f$ will turn out to be a quotient of polynomials, $ N$ will be the degree of the polynomial in the numerator and $P $ the degree of the denominator. Then the only way for $\lim_{z\to 0} f(1/z) $ to be complex is $N=P$ .

This is kind of a proof but I don't want to use the fact of $f $ being rational since that is what I want to prove. So what I'm asking you guys is to prove it with out using that fact, perhaps Laurent series will be useful.

Thank you all in advance

Natalio
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    Just to clarify: is the question to prove that the limit implies that $f$ is rational or how to do this following your suggested approach (as the title suggest)? Related question: https://math.stackexchange.com/questions/273491/must-a-meromorphic-function-on-a-compact-set-have-same-number-of-zeros-and-poles – Winther May 05 '19 at 20:39
  • To prove $f$ is rational you can define $g(z) = f(z) \prod (z-z_i)$ killing off the (finite number) of poles of $f$ to get a holomorphic function that grows as $\lesssim |z|^n$ as $z\to\infty$. Liouville's theorem gives you that $g$ is a polynomial and therefore $f$ is rational. – Winther May 05 '19 at 20:41
  • @Winther I want to prove that $ f$ is rational but I would like to use my idea since it is good to try you own way even if it is a harder one. So never mind I want to prove $f$ is rational. All I want to prove is the satement in the sqare without ever using the word rational. – Natalio May 06 '19 at 00:55
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    Yes it's good to try, but it's also good to realize when an idea is not the best way to go. You have a simple one-liner for proving that $f$ is rational and then the statement here is then proven by another one-liner. I don't know for sure, but I bet any way of showing your statement will effectively involve showing that $f$ is rational in one way or another first. – Winther May 06 '19 at 15:54

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