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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 $-frac{d}{c}$ ?

Thanks for your helping.

Mathematics Asked on January 3, 2021

1 Answers

One Answer

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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