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I ran into a bit of a hurdle as I was reading a proof in Newman Bak's Complex Analysis regarding the form of automorphisms of the unit disk.

The proof begins by showing that $g(z)=\frac{z-\alpha}{1-\bar{\alpha}z}$ is an automorphism of the unit disk. This is the part I'm having a bit of trouble with.

By previous work, we have proved that $|g(z)|=1$ for $|z|=1.$ "Since $g(\alpha)=0$, it follows that g is indeed an automorphism of the unit disk." (Newman Bak, 183).

Here is my attempt to fill in the gaps:

If $f(z)=\frac{az+b}{cz+d}$, where $ad-bc\not=0$, then f is a bilinear transformation and is therefore conformal and globally 1-to-1.

In order to show that g is bilinear, we need to show that $1-\alpha\bar\alpha\not=0.$ Suppose by contradiction that $1-\alpha\bar\alpha=0.$ Then $|\alpha| = 1$. However, since we know that $|g(z)|=1$ for $|z|=1,\ g(\alpha)=1$. This is a contradiction since $g(\alpha)=0.$

Likewise, since $|\alpha|\not=1$, $C(\alpha; 1)$ does not pass through the origin. Thus, we know that our bilinear map g sends circles to circles (i.e., it does not send circles to lines). We conclude that g is an automorphism of the unit disk.

Does this seem correct?

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    $\begingroup$ $g(z) = \dfrac{z-\alpha}{1-\overline{\alpha}z}$ is an automorphism of the unit disk if and only if $\lvert \alpha\rvert < 1$. For $\lvert\alpha\rvert = 1$, it is a constant map, not a Möbius transformation, and for $\lvert \alpha\rvert > 1$ it swaps the interior and the exterior of the unit disk. $\endgroup$ – Daniel Fischer Jul 20 '15 at 7:40
  • $\begingroup$ Ahh, I think I misread the beginning of the proof, which was a bit ambiguous. We are to assume that $|\alpha|<1.$ From this, it cannot be the case that $1-\alpha\bar\alpha=0$ and $g(z)$ is bilinear. $\endgroup$ – Greyson Jul 20 '15 at 7:54
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    $\begingroup$ You know that as a Möbius transformation, $g$ is an automorphism of the Riemann sphere. And you know that $g$ maps the unit circle to itself. So the only remaining question is to which part of the complement [in the sphere] of the unit circle it maps the unit disk, the interior or the exterior. Then seeing that $g(\alpha) = 0$ tells you it maps one point of the unit disk to the interior, hence it maps the unit disk to the interior. $\endgroup$ – Daniel Fischer Jul 20 '15 at 7:55
  • $\begingroup$ And if, say, $\alpha$ were a pole of $g$, then we would know that $g$ maps the unit disk to the exterior? $\endgroup$ – Greyson Jul 20 '15 at 8:08
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    $\begingroup$ Right. With the form of $g$ we have, $1/\overline{\alpha}$ is the pole, and that lies in the unit disk if and only if $\lvert\alpha\rvert > 1$, so this method shows the last point in my first comment. $\endgroup$ – Daniel Fischer Jul 20 '15 at 8:11

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