= Solution
For $z=x+iy$ and $r^2=|z|^2$, solving the <sphere> and projection equations gives
$$
\pi^{-1}(z)=\left(\frac{2x}{1+r^2},\frac{2y}{1+r^2},\frac{r^2-1}{r^2+1}\right).
$$
For $z\ne0$, 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 produces
$$
\left(\frac{2x}{1+r^2},-\frac{2y}{1+r^2},-\frac{r^2-1}{r^2+1}\right).
$$
Thus
$$
\boxed{\pi^{-1}\circ(z\mapsto1/z)\circ\pi:(a,b,c)\longmapsto(a,-b,-c).}
$$
The 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.
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