= Solution
For the unit sphere, project from its north pole $N=(0,0,1)$ onto the equatorial plane, identified with the <complex plane>. The line through $N$ and $(X,Y,Z)$ meets the plane at the <stereographic projection>
$$
u=\frac{X+iY}{1-Z},\qquad (X,Y,Z)=\frac{(2\operatorname{Re}u,2\operatorname{Im}u,|u|^2-1)}{1+|u|^2}.
$$
The north pole is omitted from the plane chart; adding its image infinity gives the <Riemann sphere>.
For a point whose image and antipodal image are finite, stereographic projection of $(-X,-Y,-Z)$ gives $v=-(X+iY)/(1+Z)$. Therefore
$$
u\overline v=-\frac{X^2+Y^2}{1-Z^2}=-1,
$$
using $X^2+Y^2+Z^2=1$. Conversely $u\overline v=-1$ implies $v=-1/\overline u$. Substituting this into the inverse <stereographic projection> changes the signs of all three coordinates, proving the converse. Thus \b[finite chart images correspond to antipodes exactly when $u\overline v=-1$]. On the extended chart the exceptional antipodal pair is $0,\infty$, for which the displayed ordinary complex product is not defined.
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