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I'm having some trouble with definitions, so I've been stuck with this simples problem:
Let $f$ have a pole at $z_0\in\mathbb{C}$ and $g$ be holomorphic at infinity. Prove that $g\circ f$ is holomorphic at $z_0$. I tried proving by showing the derivative exists, but the that requires calculating $f$ at $z_0$, so we have an infinity at the limit, so I'm not sure what to do...

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    $\begingroup$ Hint. "$f$ has a pole at $z_0$" implies that $1/f(z)$ has a removable singularity at $z_0$, and "$g$ is holomorphic at infinity" means (by definition) that $g(1/z)$ has a removable singularity at $0$. $\endgroup$ – Henning Makholm Sep 25 '18 at 15:00
  • $\begingroup$ I gert that, just don't understand how to use it to prove it's holomorphic at that point... wouldn't I need to show that the derivative exists? $\endgroup$ – MathNewbie Sep 25 '18 at 15:14
  • $\begingroup$ x @MathNewbie: Since $g\circ f$ is strictly speaking not defined at $z_0$, the best you can hope for is to show that it has a removable singularity, so you'll need to interpret your task to be that instead of "is (already) holomorphic". $\endgroup$ – Henning Makholm Sep 25 '18 at 15:45
  • $\begingroup$ If it is holomorphic at infinity, shouldn't it be defined at infinity? I'm considering functions in the extended complex plane $\endgroup$ – MathNewbie Sep 25 '18 at 19:50
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    $\begingroup$ x @MathNewbie: Ah, sorry, that's not the setting I tend to think of by default. But in that case you can say $g\circ f=G\circ F$ with $G(z)=g(1/z)$ and $F(z)=1/f(z)$. Then $F$ is holomorphic at $z_0$, $F(z_0)=0$ and $G$ is holomorphic at $0$. $\endgroup$ – Henning Makholm Sep 25 '18 at 19:54
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Since $g$ is holomorphic at $\infty,$ it is bounded in some $\{|z|>R\}.$ Because $f$ has a pole at $z_0,$ $\lim_{z\to z_0} |f(z)| =\infty.$ This implies that for some $r>0,$ $|f|>R$ in $\{0<|z-z_0|<r\}.$ It follows that $g\circ f$ is bounded in $\{0<|z-z_0|<r\}.$ Therefore $g\circ f$ has a removable singularity at $z_0,$ which is the desired conclusion.

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