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?