Isometry in Hyperbolic space Let $\mathbb{H}^2=\{ (x,y)\in\mathbb{R}|\ y>0 \}$ the hyperbolic space with the metric $g=(dx^2+dy^2)/y^2$. Let $$\psi(x,y)=(\psi_1(x,y),\psi_2(x,y))=\left(\mathrm{Re}\frac{az+b}{cz+d},\mathrm{Im}\frac{az+b}{cz+d}\right),$$ where $z=x+iy$.
The numbers are chosen such that $ad-bc=1$.
I want to show that $\psi$ is an isometry, that is 
\begin{equation}
\psi^*g=g\quad (1)
\end{equation}. 
In coordinates the above equation is
\begin{equation}
g_{\mu\nu}(x,y)=g_{\alpha\beta}(\psi(x,y))\frac{\partial \psi_\alpha(x,y)}{\partial x_\mu}\frac{\partial \psi_\beta(x,y)}{\partial x_\nu},\quad (2)
\end{equation}
where $x_1=x$ and $x_2=y$.
I have trouble with the calculation. I'm sure that I can use (1) to show the isometry property. Unfortunately I failed. So I used the coordinate definition.
First I noted, that
\begin{align*}
g_{\alpha\beta}(\psi(x,y))&=\begin{pmatrix}
\frac{1}{\psi_2}&0\\0&\frac{1}{\psi_2}
\end{pmatrix}\\
\frac{\partial \psi_\alpha(x,y)}{\partial x_\mu}&=\begin{pmatrix}
\mathrm{Re}\frac{1}{(cz+d)^2}&\mathrm{Re}\frac{i}{(cz+d)^2}\\
\mathrm{Im}\frac{1}{(cz+d)^2}&\mathrm{Im}\frac{i}{(cz+d)^2}
\end{pmatrix}
\end{align*}
By doing the matrix calculation for the right hand side of the equation (2) I don't get the final result $g_{\mu\nu}(x,y)$.
Hence the following questions arise:


*

*How can I use the coordinate free equation to prove the claim?

*Where is the mistake in the above calculations or in other words: Is my way absolutely nonsense?

 A: As noted in the comment, it is easier to work with complex coordinate. It will be even easier if we break down the general transformation to simpler ones: Note that when $a, c\neq 0$, 
$$\begin{split}
\frac{az + b}{cz+d} &=  \frac ac \frac{cz + \frac{cb}{a}}{cz+ d}\\
& =  \frac ac - \frac{d - \frac{cb}{a}}{cz + d} \\
&= \frac ac - \frac 1a \frac{ad-cb}{cz+ d} \\
&= \frac ac - \frac 1a \frac{1}{cz+ d}.
\end{split} $$
If $a=0$, 
$$ \frac{b}{cz+d} = b\frac{1}{cz+d},$$
and when $c=0$,
$$\frac{az+ b}{d} = \frac{1}{d} (az+b).$$
So the mapping are composition of translation (by a real number), scaling (by a positive constant) and inversion $I(z) = -\frac{1}{z}$ . As composition of isometries is still an isometry, it suffices to consider only translation, scaling and inversion. The metric is given by 
$$ g =\frac{1}{y^2} (dx^2 + dy^2),$$
so it is obvious that translations (by a real number) are isometries. For a scaling, let $A >0$ and $L_A (z) = Az$. Then
$$ L_A^* g = \frac{1}{(L_A(y))^2} \big( d (L_A x)^2 + d(L_A y)^2\big) = \frac{1}{A^2 y^2} (A^2 dx^2 + A^2 dy^2) = g. $$
Lastly, using complex coordinate $z =x+ iy$, then 
$$ dx = \frac 12 (dz + d\bar z), \ \ dy = \frac 1{2i}(dz - d\bar z)\Rightarrow g = \frac{1}{(\text{Im}z)^2} dzd\bar z. $$
Using 
$$ -\frac 1z = -\frac{\bar z}{|z|^2} \ \ \ \Rightarrow \text{Im}\big( I(z)\big) = \frac{1}{|z|^2} \text{Im} z, $$
\begin{split} \Rightarrow I^* g &= \frac{1}{(\text{Im} I(z)\big)^2} d(I(z)) d\overline{(I(z))} \\
&= \frac{|z|^4}{(\text{Im} z)^2} \bigg( \frac{1}{z^2} dz\bigg)\bigg( \frac{1}{\bar z^2} d\bar z\bigg) \\
& =  \frac{1}{(\text{Im}z)^2} dzd\bar z = g
\end{split}
Thus $I$ is also an isometry. 
