Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 4 / 3G / Solution

For a Euclidean circle with centre $c$ and radius $r$, two distinct finite points are inverse when they lie on the same ray from $c$ and their distances from $c$ have product $r^2$. Points on the circle are fixed, and $c$ and infinity are inverse to each other. In complex coordinates this uniquely defines inversion in a circle byUniqueness follows from the ray and product conditions, including the specified centre–infinity pair; the formula also gives $I_{\Gamma }^2=\mathrm{id}$. A circle through infinity is a straight line in the complex plane; its inversion is reflection in that line, fixing infinity. For a line through $c$ with direction angle $\theta$, its formula is $c+e^{2i\theta }\overline{z-c}$.

Thus every inversion is an anti-Möbius transformation, of the form $(a\overline{z}+b)/(c\overline{z}+d)$ with nonzero determinant. If $J(z)=\overline{z}$, write such an inversion as $M\circ J$ with $M$ a Möbius transformation. Since $J\circ M\circ J$ is the Möbius transformation with conjugated coefficients, the product of two inversions is $M_1\circ (J\circ M_2\circ J)$, hence Möbius. Pairing consecutive inversions proves every composition of an even number of inversions is a Möbius transformation.