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Given a meromorphic function on $\mathbb{C}$, is the radius of convergence in a regular point exactly the distance to the closest pole?

As Robert Israel points out in his answer, that this is of course an upper bound by the Cauchy-Hadamard principle.

Theo Buehler in the comments gives a refernce for the non obvious direction: Remmert, Theory of complex functions, Chapter 7, §3, p.210ff (p. 164ff of my old German edition). Look for Cauchy-Taylor.

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    Perfect answer, thank you very much. Why did you not provide it directly, and were teasing me so much?;) If you care to pos the answer, I will accept it immediately.2011-06-24

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Yes, it is (that should be "pole", not "pol"). If $r$ is the distance from $z_0$ to the closest pole, the function is analytic in $\{z: |z - z_0| < r\}$, so the radius of convergence is at least $r$, but it can't be more than $r$ because $|f(z)| \to \infty$ as $z$ approaches that pole.

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    @Libertron That is indeed what it means. The radius of convergence $\ge r$ because the function is analytic in \{z: |z - z_0| < r\}, and $\le r$ because $|f(z)| \to \infty$ as $z$ approaches the pole, so you conclude it is exactly $r$.2014-07-21