Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 3 / 2D / Solution

A Möbius transformation is $M_A(z)=(az+b)/(cz+d)$ with $\Delta =ad-bc\ne 0$. On the Riemann sphere, a zero denominator represents infinity, and $M_A(\infty )=a/c$ when $c\ne 0$; when $c=0$, infinity is fixed. The numerator and denominator cannot vanish together because $\Delta \ne 0$.

The matrix $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ acts on homogeneous coordinates $[z:1]$, so composition satisfies $M_A\circ M_B=M_{AB}$. The product is invertible since $\det (AB)=\det A\det B\ne 0$, proving closure. Composition is associative because these are maps of the sphere. The identity is $z\mapsto z$, andis again a Möbius transformation. Thus the transformations form a group. Nonzero scalar multiples of a matrix give the same transformation, but the identity and inverse assertions concern the transformations themselves.

Write $T_q(z)=z+q$, $S_k(z)=kz$ with $k\ne 0$, and $H(z)=1/z$. If $c=0$, then $a,d\ne 0$ and $M_A=T_{b/d}\circ S_{a/d}$. If $c\ne 0$, division givesHence the Möbius translation-scaling-inversion factorization isThe scaling coefficient is nonzero. At $z=-d/c$ the reciprocal sends the translated zero to infinity; at infinity it sends infinity to zero. Thus the factorization holds on the entire Riemann sphere, including both exceptional points, and proves the stated generating property.