Consider the unit disk $\mathbb{D}=\{z: |Z| < 1\} \subset \mathbb{C}$ equipped with the hyperbolic metric $g$ induced by $1$ form $ds=\frac{|dz|}{(1-|z|^2)}$
I am trying to show that every Mobius transformation $T(z)=\frac{\alpha z+ \beta}{\bar \beta z+ \bar \alpha}$
with $|\alpha|^2 -|\beta|^2=1$ acts as an isometry of the hyperbolic disk.
This is the proof:
The main idea is to show the pull back of the metric in the $w$ plane is identical in the $z$ plane.
$$dw=\frac{\alpha(\bar \beta z- \bar \alpha )-\bar \beta(\alpha z + \beta)}{(\bar \beta z+ \bar \alpha)^2} dz = \frac{|\alpha|^2 - |\beta|^2}{(\bar \beta z+ \bar \alpha)} dz=\frac{dz}{(\bar \beta z+ \bar \alpha)^2}$$
We can see that $$|w|^2=\frac{\alpha z+ \beta}{\bar \beta z + \bar \alpha} \times \frac{\overline{\alpha z + \beta}}{\bar \beta z + \bar \alpha}$$
How do we get from this step to $1-|w|^2=\frac{(|\alpha|^2 -|\beta |^2)(1-|z|^2)}{(\bar \beta z + \bar \alpha )^2}$ ?
$$\frac{\alpha z+\beta}{\overline\beta z+\overline\alpha}\cdot\frac{\overline\alpha\overline z+\overline\beta}{\beta \overline z+\alpha}=\frac{|\alpha|^2|z|^2+\alpha\overline\beta z+\overline\alpha\beta\overline z+|\beta|^2}{|\beta\overline z +\alpha|^2}=\frac{|\alpha|^2|z|^2+\alpha\overline\beta z+\overline\alpha\beta\overline z+|\beta|^2}{|\beta\overline z +\alpha|^2}$$
Also, observe that
$$|\beta\overline z+\alpha|^2=|\beta|^2|z|^2+2\text{ Re}\,(\overline\alpha\beta\overline z)+|\alpha|^2\implies(\beta\overline z+\alpha)^2-|\alpha|^2|z|^2-\alpha\overline\beta z-\overline\alpha\beta\overline z-|\beta|^2=$$
$$=(|\beta|^2-|\alpha|^2)|z|^2+|\alpha|^2-|\beta|^2=1-|z|^2$$
and you have your equality