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I was looking this question and answer Take $f$ and entire function such that $f(z) = f(z+1)=f(z+\sqrt{2})$, then it is constant and I wonder what would happen if $f$ is not entire but meromorphic.

Thank you.

Rub
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  • Pole is a very nice type of singularity, allowing you to conclude the same answer as in the link. – Sangchul Lee Feb 16 '19 at 17:54
  • How could I do it? I thought it was enough to say that poles must be isolated and there must be a finite number of them, so taking a neighbourhood of $\sqrt(2)$ it would be right. However, I did not know either if it was okay or how to formalise it. – Rub Feb 16 '19 at 18:41
  • Your argument is just fine. Since the set of all poles of a meromorphic function is discrete, the complement is connected and open. So you can apply the identity theorem to conclude. – Sangchul Lee Feb 16 '19 at 18:51
  • The same logic José Carlos Santos uses in the other thread shows that if $p$ is a pole, then poles must be dense on the line $p + \Bbb R$. – Paul Sinclair Feb 17 '19 at 03:48
  • If you are discussing meromorphic function on $\mathbb{C}$, by definition, the set of poles must be discrete. This restriction is essential because people want to regard meromorphic functions as holomorphic functions as holomorphic function to the Riemann sphere $\hat{\mathbb{C}} = P^1\mathbb{C}$. But if you are willing to restrict the domain of $f$ to a smaller region $\Omega$ of $\mathbb{C}$, then we can make poles 'accumulate' towards the boundary of $\Omega$. – Sangchul Lee Feb 17 '19 at 17:49
  • Still, the conclusion will continue to hold for any domain $\Omega$ which is connected and is invariant under shifts by $m+n\sqrt{2}$ for $m, n \in \mathbb{Z}$, and this can be proved by almost the same argument as in the link. – Sangchul Lee Feb 17 '19 at 17:57

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