On Complex Inversion

22 Dec, 2023

A student in my A1 class asked me about the geometric significance of $\frac{1}{z}$ and I responded that it would be an circular inversion composite with a reflection in real axis. But then that feels very unsatisfactory. In order not to increase the complexity for highschool students, I stick to that explanation.

However, there is a much more satisfying explanation (avoiding circular inversion) if we adopt stereographic projection on $ℂ$.

Take a unit sphere centered at the origin, the north pole denoted as $N$, then it is possible to map any point on the $x-y$ plane on to the sphere with one extra point $\infty$ to $N$.

Therefore, $z=r{e}^{i\theta }$ would be mapped to $Z_1(1,\theta ,\varphi {)}_s$ in spherical coordinates, where $\varphi =2\arctan r-\frac{\pi }{2}$. Similarly, $\frac{1}{z}=\frac{1}{r}{e}^{-i\theta }$ corresponds to $Z_2(1,-\theta ,-\varphi {)}_s$. Now it is clear that the inversion in $ℂ$ corresponds to two reflection in the stereographic projection: one about $ℂ$ plane, the other $\Re -z$ plane.

I made a Geogebra demo so that you can play with it.

The intuition is quite straightforward, a reflection about the $ℂ$ plane maps everything on the north hemisphere to the south, thus put everything outside the unit circle inside.

On the other hand, it is very interesting then to see what a Mobius transformation looks like on that unit sphere.

#"hs-maths"