# Holomorphic at infinity

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...

• 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 $0$. – Henning Makholm Sep 25 '18 at 15:00
• 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? – MathNewbie Sep 25 '18 at 15:14
• 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". – Henning Makholm Sep 25 '18 at 15:45
• If it is holomorphic at infinity, shouldn't it be defined at infinity? I'm considering functions in the extended complex plane – MathNewbie Sep 25 '18 at 19:50
• 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 $0$. – Henning Makholm Sep 25 '18 at 19:54

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 $$\{0<|z-z_0| It follows that $$g\circ f$$ is bounded in $$\{0<|z-z_0| Therefore $$g\circ f$$ has a removable singularity at $$z_0,$$ which is the desired conclusion.