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Let $f$ be a meromorphic function. Is it true that $f(\bar{z})=\overline{f(z)}$ for all complex numbers $z$?

A meromorphic function can be written as a ratio of two holomorphic functions. Moreover, at any point $s\in\mathbb{C}$, the function $f$ can be expressed as a Laurent series in a neighborhood of $s$. But still, the Laurent series is different at $s$ than at $\bar{s}$. How can we show the result?

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It's certainly true if $s$, the center of the series, is real, and the coefficients are real as well. –  user1337 Dec 14 '13 at 20:09
    
It's true if $f(z)$ is real when $z$ is real (Schwarz reflection principle) –  Robert Israel Dec 14 '13 at 20:13
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2 Answers

up vote 4 down vote accepted

The function $f(z) = i$ is meromorphic, but does not satisfy your property.

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According to the Schwarz reflection principle, this is true if $f$ takes real values along the real axis. It is not true in general.

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