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I'm trying to solve the following problem:

Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be a meromorphic function such that $f({1\over z})$ is analytic at $z=0$ and $\displaystyle{\lim_{z\rightarrow\infty}f(z)}=0$. Show that there exists $R>0$ such that $\displaystyle{\int_{|z|=R}{f'(z)\over f(z)}dz}=0$.

If I find a $R>0$ such that $f(z)$ has the same finite number of zeros and poles in $B(0,R)$, then by the Argument Theorem I would have that integral is zero, but how to know that such $R$ exists?

I would appreciate any hint.

Fernando Mauricio Rivera Vega
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    Did you think about using the criterion that $f$ is analytic at $\infty$? Maybe you should make a change of variables and think about things outside $|z|=R$ instead of inside. – Ted Shifrin Dec 09 '20 at 19:56
  • Think rational. – copper.hat Dec 09 '20 at 20:00
  • Show: finitely many poles (using the limit at infinity), hence $f=g/Q$ with $Q$ polynomial and $g$ entire and then show $g$ polynomial etc – Conrad Dec 09 '20 at 20:12
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    Is the claim true? Consider $f(z)=1/z$. Then for any $R>0$, $$\int_{|z|=R}\frac{f'(z)}{f(z)} , \mathrm{d}z=-\int_{|z|=R}\frac{1}{z} , \mathrm{d}z=-2\pi i.$$ I guess you need $\lim_{z\to \infty}f(z) \neq 0$. – Sangchul Lee Dec 09 '20 at 21:04
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    Anyway, assuming that $\lim_{z\to \infty}f(z) \neq 0$, then you may implement the idea in Ted Shifrin's comment by setting $g(z)=f(1/z)$ and noting $$\int_{|z|=R}\frac{f'(z)}{f(z)} , \mathrm{d}z\stackrel{(w=1/z)}=-\int_{|w|=R^{-1}}\frac{g'(w)}{g(w)} , \mathrm{d}w.$$ – Sangchul Lee Dec 09 '20 at 22:15

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