Distance formula for inverse stereographic projection

Statement

Consider the inverse stereographic projection map:

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

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

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

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

Facts used

We use here the formula for inverse stereographic projection:

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

Proof

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