Verifying Möbius Transformations using Hyperbolic Geometry Verify that every transformation from $$H = \left\{Tz = e^{i\theta} \frac{z-z_0}{1-z_0 z} \right\}$$ can be written as $Tz = \frac{az-b}{\bar{b}z+\bar{a}}$ with $|a|^2 - |b|^2 = 1$.
The book gives the hint that we need to use algebraic manipulation to convert the first equation to the second using $$a = \frac{e^{\frac{1}{2}i\theta}}{\sqrt{1-|z_0|^2}}.$$ But I've spent several hours using this manipulation with no success. Any ideas? 
 A: Well, let us try your book's hint and some educated guess:
$$|b|^2=|a|^2-1=\frac1{1-|z_0|^2}-1=\frac{|z_0|^2}{1-|z_0|^2}$$
So we could try (the educated guess kicks in)
$$b:=\frac{z_0e^{i\theta/2}}{\sqrt{1-|z_0|^2}}$$
and then
$$\frac{az-b}{\overline bz+\overline a}=\frac{e^{i\theta/2}z-z_0e^{i\theta/2}}{\overline{z_0}ze^{-i\theta/2}+e^{-i\theta/2}}=e^{i\theta}\frac{z-z_0}{1+\overline {z_0}z}$$
A: I'll assume that the original question had a couple of typos, and that the task is to match
$$e^{i\theta}\frac{z-z_0}{1-\overline{z_0}z} \qquad \text{with}\qquad \frac{\phantom{-}a z-b}{-\overline{b}z+\overline{a}}$$
where the second fraction's nature as linear fractional transformation meshes better with the condition $1 = |a|^2-|b|^2 = a \overline{a}-(-b)\overline{(-b)}$. Even so, the strategy I describe would also be suitable to prove the version that @DonAntonio demonstrated.
We'll ignore the hint and dive right into the search for an $a$ and $b$ that make the second fraction match the first.

Let's write that second fraction's denominator so that it matches the first's better:
$$\frac{a z-b}{\overline{a}-\overline{b} z}$$
Now, consider: If we want to make that fraction look even more like the first fraction, we'd want the numerator's $z$ to have $1$ for a coefficient. That's easily accomplished by forcibly factoring-out an $a$:
$$\frac{a \; ( z - b/a )}{\overline{a}-\overline{b}z}$$
That worked so well that we should do the same thing in the denominator, to change its constant term to $1$, and see where that takes us:
$$\frac{\;a\;}{\;\overline{a}\;}\;\frac{z-b/a}{1-(\overline{b}/\overline{a})z}
=
\frac{\;a\;}{\;\overline{a}\;}\;\frac{z-b/a}{1-\overline{(b/a)}z} = \frac{a}{\;\overline{a}\;}\;\frac{z-c}{1-\overline{c}z} \qquad \text{where} \quad c := b/a$$
Recall that, if $a = |a| e^{i\alpha}$, then $\overline{a} = |a| e^{-i\alpha}$, so that $a/\overline{a} = e^{2i\alpha}$; we need only take 
$$\alpha = \theta/2$$
to get $a/\overline{a}$ to match the multiplied exponential of the original fraction. To complete the conversion of the fraction, we only need to replace $c$ with $z_0$, which we can do by setting
$$b = a z_0$$
At this stage, however, we haven't completely identified $a$, only its argument. To get its modulus, observe that $|b| = |a z_0| = |a||z_0|$. Now,
$$1=|a|^2-|b|^2= |a|^2 - |a|^2|z_0|^2 = |a|^2 \left( 1 - |z_0|^2 \right) \quad \implies \quad |a| = \frac{1}{\sqrt{ 1 - |z_0|^2}}$$
Consequently,
$$a = |a|e^{i\alpha} = \frac{e^{i\theta/2}}{\sqrt{1-|z_0|^2}} \qquad\qquad b = a z_0 = \frac{z_0 e^{i\theta/2}}{\sqrt{1-|z_0|^2}}$$
