The map from the hyperboloid to the Poincaré disk

Starting with a point $(x,y,z)$ on $x^2+y^2+1=z^2$, we follow these maps to $(\frac{x}{z},\frac{y}{z})$, and then to the point on the lower hemisphere

$$\left(\frac{x}{z},\frac{y}{z},-\sqrt{1-\frac{x^2}{z^2}-\frac{y^2}{z^2}}\right) =\left(\frac{x}{z},\frac{y}{z},-\frac{1}{z}\right),$$

and then ultimately to the point

$$F(x,y,z)= r+is= \frac{x+iy}{1+z}$$

in the Poincaré disk.

To invert this map, note that $x=(1+z)r$ and $y=(1+z)s$. Squaring, we get $(r^2+s^2)(1+z)^2=x^2+y^2=z^2-1$. Thus, $r^2+s^2=\frac{z^2-1}{(z+1)^2}=\frac{z-1}{z+1}$. That gives

$$z=\frac{1+r^2+s^2}{1-r^2-s^2}.$$

If we set $\zeta=r+is$, we then obtain

$$F^{-1}(\zeta)=\left( \frac{2\Re\zeta}{1-\vert\zeta\vert^2},\frac{... ...}{1-\vert\zeta\vert^2}, \frac{1+\vert\zeta\vert^2}{1-\vert\zeta\vert^2}\right)$$

It is worth checking that $F^{-1}\circ F(x,y,z)=(x,y,z)$ and $F\circ F^{-1}(\zeta)=\zeta$, but we leave this to the reader.