Formula for Poincare metric on the upper half plane Let $\mathbb{H^2}$ denote the open upper half plane in $\mathbb{C}$ and $z$ and  $w$ be any two elements belonging it. The unique circle centered on the real line that passes through both $z$ and $w$ intersects the real line at two points (in case $z$ and $w$ lie on a straight line by abuse of language call the line passing through them as a circle centered on the real line and it intersects the real line at two points , one of which is at $\infty$)  .Of these two points let the one nearer to $z$ be called $z^*$ and the other one be $w^*$ . If $\rho(.,.)$ is the hyperbolic metric on $\mathbb{H^2}$ then we know that
$$ \rho(z,w) =|\log(z^*,w^*,w,z)|,$$
where $(z_1,z_2,z_3,z_4)$ is the cross-ratio of the four points.
Now they've given it as as exercise in my book that $\rho$ as above can also be expressed by the formula
$$ \rho(z,w) =\log\frac{|z-\overline{w}|+|z-w|}{|z-\overline{w}|-|z-w|},$$
but I don't know how to prove it. Can someone please help me on this?
 A: We wish to prove $$\frac{|z^*-w|\cdot|w^*-z|}{|w^*-w|\cdot|z^*-z|} = \frac{|z-\overline{w}|+|z-w|}{|z-\overline{w}|\cdot|z-w|}.$$
Note that $z^*,z,w,w^*$ is a cyclic quadrilateral, so Ptolemy implies $(1)$: $$|z^*-w|\cdot|z-w^*| = |z-z^*|\cdot|w-w^*|+|z-w|\cdot|z^*-w^*|.$$
Using this and doing a bit of algebra leads us to want to prove $$2|z-z^*|\cdot|w-w^*| = |z^*-w^*|\cdot|z-\overline{w}|-|z^*-w^*|\cdot|z-w|.$$
Now, the fact that $z^*,z,w^*,\overline{w}$ is a cyclic quadrilateral gives (via Ptolemy) $$|z^*-w^*|\cdot|z-\overline{w}| = |z-z^*|\cdot|w^*-\overline{w}|+|z^*-\overline{w}|\cdot|z-w^*|.$$
So we just wish to prove $$|z-z^*|\cdot|w-w^*| = |z^*-\overline{w}|\cdot|z-w^*|-|z^*-w^*|\cdot|z-w|.$$Now just note $|z^*-\overline{w}| = |z^*-w|$, and so we're done by $(1)$.
A: This is solved  here for the case that your riemaniann metric slightly changes from $\begin{pmatrix} 1/\sigma^2 & 0 \\ 0 & 1/\sigma^2 \end{pmatrix}$ to $\begin{pmatrix} 1/\sigma^2 & 0 \\ 0 & 2/\sigma^2 \end{pmatrix}$. The procedure is with details explained and you should come up with the same result up to some constant multiplication (maybe somewhere a $\sqrt{2}$).
