Projections on the Riemann Sphere are antipodal

Solution 1:

We could of course just compute it using the explicit formula for the stereographic projection. But let's argue geometrically.

Let $z \in \mathbb{C}\setminus\{0\}$ and $\zeta$ the image of $z$ on the sphere under stereographic projection. Let $C$ be the great circle on the sphere passing through $\zeta$ and the north and south pole.

Let $\varphi$ be the angle between the plane and the line connecting $z$ with the north pole $N$. Since the radius of the sphere is $1$, the distance of $z$ from the intersection of the plane and the axis of the sphere (that is, $0 \in \mathbb{C}$) is $\cot \varphi$, i.e. $\lvert z\rvert = \cot \varphi$.

The line connecting $\zeta$ and its antipodal point $\alpha := -\zeta$ is a diameter of $C$, hence, by Thales' theorem, the triangle $\alpha N\zeta$ has a right angle at $N$. $\angle z 0N$ is a right angle, hence $\angle 0N\zeta = \angle 0Nz = \frac{\pi}{2}-\varphi$, and therefore $\angle 0N\alpha = \varphi$. But that means the line $N\alpha$ intersects the plane at a distance of $\tan\varphi$ from $0$, on the opposite side of $z$, and that means the point projected to $\alpha$ is $$w = -\frac{z}{\lvert z\rvert^2},$$

whence $z\overline{w} = -1$.

Conversely, if $z\overline{w} = -1$, and $w$ is projected to $\beta$ on the sphere, then $\lvert w\rvert = \tan \varphi$, whence $\angle \beta N0 = \varphi$, and since $w$ (and $\beta$) lie on the opposite side of $0$ from $z$, we have

$$\angle \beta N\zeta = \angle \beta N0 + \angle 0N\zeta = \varphi + \left(\frac{\pi}{2} - \varphi\right) = \frac{\pi}{2},$$

and by Thales, the line segment connecting $\beta$ and $\zeta$ is a diameter of the great circle $C$, hence $\beta = -\zeta$ is the antipodal point of $\zeta$.

Analytically, with the explicit formulae of the stereographic projection

$$\begin{align} \varphi \colon z = x+iy &\mapsto \left(\frac{2x}{x^2+y^2+1},\,\frac{2y}{x^2+y^2+1},\,\frac{x^2+y^2-1}{x^2+y^2+1} \right)\\ \varphi^{-1} \colon (\alpha,\,\beta,\,\gamma) &\mapsto \frac{1}{1-\gamma}(\alpha + i \beta) \end{align}$$

we obtain that the point being projected to the antipodal point of $z$'s projection is

$$\begin{align} w &= \varphi^{-1} \left(\frac{-2x}{x^2+y^2+1},\,\frac{-2y}{x^2+y^2+1},\, \frac{1-x^2-y^2}{x^2+y^2+1}\right)\\ &= \frac{x^2+y^2+1}{2(x^2+y^2)}\frac{-2(x+iy)}{x^2+y^2+1}\\ &= \frac{-(x+iy)}{x^2+y^2}\\ &= -1/\overline{z}. \end{align}$$

Solution 2:

For points $z$, $w\in{\mathbb C}$ with antipodal stereographic projections to $S^2$ the triangle with vertices $z$, $N$, $w$ has a right angle at $N$. By a Pythagorean theorem ("Höhensatz") it follows that $$|z|\>|w|=1^2\ .\tag{1}$$ Furthermore, looking at the situation from high up on the $x_3$-axis we see that $$\arg(w)=\arg(z)+\pi\ \tag{2}$$ The equations $(1)$ and (2) together easily imply $z\>\bar w=-1$.