Past exam of the metric spaces and complex analysis course of the University of Oxford / 2014 / 6 / a

The line through the north pole and $p=(a,b,c)$ isIts intersection with the equatorial complex plane has third coordinate zero, giving $t=1/(1-c)$ when $p$ is not the north pole. The first two coordinates are therefore $a/(1-c)$ and $b/(1-c)$, henceThe north pole maps to the point at infinity. This is stereographic projection onto the Riemann sphere.

For $z=x+iy$ and $r^2=|z{|}^2$, solving the sphere and projection equations givesFor $z\ne 0$, the reciprocal has real and imaginary parts $x/r^2$ and $-y/r^2$, and squared modulus $1/r^2$. Substituting these into the inverse formula producesThusThe reciprocal exchanges zero and infinity, corresponding to exchanging the south and north poles, so the formula holds there too. On the sphere this is a rotation through $\pi$ about the first coordinate axis, not the antipodal map.