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I know that if $f$ is meromorphic then $\exists A\subset \Omega$ s.t $f$ is holomorphic on $\Omega \setminus A$ and $A$ is discrete, and $A$ are the poles of $f$. I want to show that $f'$ is meromorphic, but this seems trivial for $\forall x\in \Omega\setminus A$ we know that $f$ is analytic on a region of $x$, so $f$ is analytic there and thus $f'$ is holomorphic in $x$. this implies $f'$ holomorphic on $\Omega\setminus A$,hence meromorphic. Am I missing something? This seems too obvious.

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    $\begingroup$ You have to show that $f'$ has a pole at each $x \in A$. $\endgroup$ – Paul Frost Dec 8 '18 at 10:10
  • $\begingroup$ Why? The definition just requires that the set of poles is discrete. I showed that the set of poles has to be a subset of $A$, thus has to be discrete. $\endgroup$ – Simon Green Dec 8 '18 at 11:58
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    $\begingroup$ A meromorphic function $f$ on $\Omega$ is a a holomorphic function on $\Omega \setminus A$, where $A$ is discrete, such that $f$ has a pole at each $x \in A$. It is not allowed to have a removable or an essential singularity at $x$. You correctly state that $f'$ is holomorphic on $\Omega \setminus A$, but you do not analyze the singularities at points of $A$. You have to show that if $f$ has a pole at $x$, then so has $f'$. $\endgroup$ – Paul Frost Dec 8 '18 at 13:23
  • $\begingroup$ Very satisfaying answer, thank you very much. $\endgroup$ – Simon Green Dec 8 '18 at 14:02
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Your argument is incomplete. You have to show that $f'$ doesn't have an essential singularity at points of $A$. Let $c \in A$. We can write $(z-c)^{n}f(z)=g(z)$ for some non-negative integer $n$ and some holomorphic function $g$ in a neighborhood of $c$. We then have $(z-c)^{n}f'(z)+n(z-c)^{n-1}f(z)=g'(z)$ from which it is easy to see that $c$ is indeed a pole of $f'$.

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Hint
Expand the function as Laurent series at the poles. You know the rest part is holomorphic, and expanding it as Laurent series gives the meromorphicity of the poles in the original function.

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    $\begingroup$ The word should be meromorphicity. $\endgroup$ – Szeto Dec 8 '18 at 10:43

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