 Solution

An inversion in a circle on the Riemann sphere is the anti-Möbius involution fixing that generalized circle pointwise. For a Euclidean circle with centre $a$ and radius $R>0$, it isIt reverses the argument relative to the centre and sends radius $r$ to $R^2/r$, so exactly the points of the circle are fixed. For a straight line, regarded as a circle through infinity, inversion is the ordinary reflection in that line. These formulas prove existence. For uniqueness, the composition of two anti-Möbius transformations fixing the same circle pointwise is a Möbius transformation fixing at least three distinct points, hence the identity. Therefore the two inversions agree.

Every inversion has the form $(a\overline{z}+b)/(c\overline{z}+d)$ with $ad-bc\ne 0$. Composing two conjugate-linear fractional expressions cancels the complex conjugations and gives $(Az+B)/(Cz+D)$ with nonzero determinant. Induction proves that an even number of inversions gives a Möbius transformation.

Conversely translations are compositions of reflections in parallel lines, rotations are compositions of reflections in two lines through their centre, and positive dilations are compositions of two concentric circle inversions: $J_R{J}_1(z)=R^{2}z$. Thus every nonconstant affine map $z\mapsto az+b$ is an even composition. The map $z\mapsto 1/z$ is $J_{|z|=1}$ composed with reflection in the real axis, also an even composition. If $c\ne 0$, factor a general Möbius transformation asThis is a composition of affine maps and $1/z$. If $c=0$, it is already affine. Every Möbius transformation is therefore a composition of an even number of inversions, with the identity represented by zero inversions or by any inversion repeated twice.