To show that Möbius transformation is holomorphic on $\Bbb C_{\infty}$ A function $f: \Bbb C_{\infty}\to \Bbb C_{\infty}$ is biholomorphic if $f$ is bijective, holomorphic and $f^{-1}$ is holomorphic on $\Bbb C_{\infty}$, where $\Bbb C_{\infty}$ is the one point compactification on the complex plane.
Now I want to show that a Möbius transformation $f: \Bbb C_{\infty}\to \Bbb C_{\infty}$ is defined by $$f=\frac{az+b}{cz+d}$$, $a,b,c,d\in \Bbb C$, is holomorphic on $\Bbb C_{\infty}$.
It may have two cases, namely $c=0$ and $c\ne 0$.
If $c=0$, then $f=\frac{a}{d}z+\frac{c}{d}$ is holomorphic on $\Bbb C_{\infty}$ and $f(\infty)=\infty$.
Consider the function $g(z)=f(1/z)$, then $g$ has a pole at $0$. But I have no idea how to use this to show that $f$ is holomorphic at $\infty$.
If $c\neq 0$, then $f$ is holomorphic on $\Bbb C-\{\frac{-d}{c}\}$ and $f(\infty)=\frac{a}{c}$, $f(\frac{-d}{c})=\infty$.
I know that $\lim_{z\to 0}|g(z)|=\frac{a}{c}$, so $g$ is holomorphic at $0$ by Riemann removable singularty theorem, and $f$ is holomorphic at $\infty$.
But I have no idea to prove $f$ is holomorphic at $-\frac{d}{c}$.
In short, when $c=0$, how to prove that $f$ is holomorphic at $\infty$? When $c\ne0$, how to prove that $f$ is holomorphic at $-\frac{d}{c}$ ?
Thanks for your helping.
 A: There are three fundamental types of Möbius transforms: rotation-dilations
$$z\mapsto az,$$
Translations
$$z\mapsto z+b,$$
and inversions
$$z\mapsto\frac{1}{z}.$$
All Möbius transformations are compositions of these three fundamental transformations, so it is enough to show that these three are biholomorphic as functions $\mathbb C_\infty\to\mathbb C_\infty$. And since the inverse of a Möbius transformation is again a Möbius transformation, it is even enough to just show that they are simply holomorphic. Biholomorphicity follows automatically.
So rotation-dilations and translations are obviously holomorphic on $\mathbb C$. We only need to consider their behavior at $\infty$. A function $\varphi:\mathbb C_\infty\to\mathbb C_\infty$ with $\varphi(\infty)=\infty$ is complex differentiable at $\infty$ if $\tilde\varphi(z):=\frac{1}{\varphi(\frac{1}{z})}$ is complex differentiable at $0$. For rotation-dilations and translations, this can be seen by plugging in and looking at the resulting functions.
Now the inversion. On $\mathbb C\backslash\{0\}$, it is obviously holomorphic. So only its behavior at $0$ and $\infty$ needs to be examined further. A function with $\varphi(0)=\infty$, is complex differentiable at $0$ if $\frac{1}{\varphi (z)}$ is complex differentiable at $0$. Similarly, if $\varphi(\infty)=0$, then it is complex differentiable at $\infty$ if $\varphi(\frac{1}{z})$ is complex differentiable at $0$. Both of these are fulfilled (just plug it in), so we're done.
