1. ## Question about mobius transformations

Prove that, every Mobius transformation mapping $D(0;1)=\{z \in \mathbb{C} : |z|<1\}$ onto D(0;1) is of the form $e^{i\lambda}\phi_{\alpha}$ where $\phi_{\alpha}(z)=\frac{z-\alpha}{\overline{\alpha}z-1}$, $\lambda$ is a real constant and $\alpha \in D(0;1)$.

Firstly I take a Mobius transformation $f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb{C}$. I was then thinking of splitting everything into real and imaginiary parts, but the algebra would be horrific. Can anyone help?

2. ## Re: Question about mobius transformations

Originally Posted by Speed1991
Prove that, every Mobius transformation mapping $D(0;1)=\{z \in \mathbb{C} : |z|<1\}$ onto D(0;1) is of the form $e^{i\lambda}\phi_{\alpha}$ where $\phi_{\alpha}(z)=\frac{z-\alpha}{\overline{\alpha}z-1}$, $\lambda$ is a real constant and $\alpha \in D(0;1)$.

Firstly I take a Mobius transformation $f(z)=\frac{az+b}{cz+d}$, $a,b,c,d \in \mathbb{C}$. I was then thinking of splitting everything into real and imaginiary parts, but the algebra would be horrific. Can anyone help?
Step 1. The map $\phi_\alpha$ is a Möbius transformation from D(0;1) to itself.
Proof: $|z-\alpha|^2 = (z-\alpha)(z-\overline{\alpha}) = |z|^2 + |\alpha|^2 - \text{Re}(2z\overline{\alpha})$, and similarly $|\overline{\alpha}z-1|^2 = |\alpha|^2|z|^2 + 1 - \text{Re}(2z\overline{\alpha})$. The inequality $|z-\alpha|^2 < |\overline{\alpha}z-1|^2$ then follows from the fact that $(1-|z|^2)(1-|\alpha|^2) > 0.$ Thus $|\phi_\alpha(z)|<1.$

Step 2. $\phi_\alpha$ maps D(0;1) onto itself. This follows from the fact that $\phi_\alpha$ is its own inverse. If $w = \frac{z-\alpha}{\overline{\alpha}z-1}$ then you can check that $z = \frac{w-\alpha}{\overline{\alpha}w-1}.$

Step 3. If f is a Möbius transformation from D(0;1) onto itself, let $\alpha = f^{-1}(0).$ Then the composition $\phi_\alpha f^{-1}$ is a Möbius transformation from D(0;1) onto itself that fixes the point 0. With luck, you will have come across a theorem which says that this implies that $\phi_\alpha f^{-1}$ is multiplication by a constant of absolute value 1. That will be enough to tell you that $f = e^{i\lambda}\phi_\alpha$ for a real constant $\lambda.$