Distance formula for inverse stereographic projection

Statement

Consider the inverse stereographic projection map:

$\mathrm{\Phi }:\mathbb{C}\to S^2\setminus \{N\}$

where $N$ is the north pole in $S^2$.

Then the distance in ${\mathbb{R}}^3$ between $\mathrm{\Phi }(z)$ and $\mathrm{\Phi }(w)$ is given by:

$\frac{|z-w|}{(1+|z{|}^2)(1+|w{|}^2)}$

Facts used

We use here the formula for inverse stereographic projection:

$\mathrm{\Phi }(z)=(\frac{z+\overline{z}}{1+|z{|}^2},\frac{z-\overline{z}}{i(1+|z{|}^2)},\frac{|z{|}^2-1}{1+|z{|}^2})$

Proof

We simply plug in the formula for $\mathrm{\Phi }(z),\mathrm{\Phi }(w)$ in the formula to calculate the distance between two points in Euclidean space. Fill this in later