# 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 $$f=frac{az+b}{cz+d}$$, $$a,b,c,din Bbb C$$, is holomorphic on $$Bbb C_{infty}$$.

It may have two cases, namely $$c=0$$ and $$cne 0$$.

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 $$cneq 0$$, then $$f$$ is holomorphic on $$Bbb C-{frac{-d}{c}}$$ and $$f(infty)=frac{a}{c}$$, $$f(frac{-d}{c})=infty$$.

I know that $$lim_{zto 0}|g(z)|=frac{a}{c}$$, 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 $$cne0$$, how to prove that $$f$$ is holomorphic at $$-frac{d}{c}$$ ?

Mathematics Asked on January 3, 2021

There are three fundamental types of Möbius transforms: rotation-dilations

$$zmapsto az,$$

Translations

$$zmapsto z+b,$$

and inversions

$$zmapstofrac{1}{z}.$$

All Möbius transformations are compositions of these three fundamental transformations, so it is enough to show that these three are biholomorphic as functions $$mathbb C_inftytomathbb C_infty$$. 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 $$varphi:mathbb C_inftytomathbb C_infty$$ with $$varphi(infty)=infty$$ is complex differentiable at $$infty$$ if $$tildevarphi(z):=frac{1}{varphi(frac{1}{z})}$$ 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 $$mathbb Cbackslash{0}$$, 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.

Correct answer by Vercassivelaunos on January 3, 2021

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