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    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?
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    Re: Question about mobius transformations

    Quote Originally Posted by Speed1991 View Post
    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.
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