 Solution

Write a Möbius transformation as $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, interpreted on the Riemann sphere. Let $T_u(z)=z+u$, $D_{\lambda }(z)=\lambda z$ with $\lambda \ne 0$, and $I(z)=1/z$.

If $c=0$, then $a,d\ne 0$ and $M(z)=(a/d)z+b/d=T_{b/d}\circ D_{a/d}(z)$. If $c\ne 0$, algebraic division givesConsequentlyThe scaling coefficient is nonzero because $ad-bc\ne 0$. Thus translations, nonzero complex scalings and reciprocal inversion generate every Möbius map. The formulas extend over their poles and infinity by the sphere interpretation.