# Why is the derevative of meromorphic function is meromorphic?

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.

• You have to show that $f'$ has a pole at each $x \in A$. – Paul Frost Dec 8 '18 at 10:10
• 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. – Simon Green Dec 8 '18 at 11:58
• 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 $f'$. – Paul Frost Dec 8 '18 at 13:23
• Very satisfaying answer, thank you very much. – Simon Green Dec 8 '18 at 14:02

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 $$f'$$.