= Solution
Write a <Möbius transformation> as $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne0$, interpreted on the <Riemann sphere>. Let $T_u(z)=z+u$, $D_\lambda(z)=\lambda z$ with $\lambda\ne0$, and $I(z)=1/z$.
If $c=0$, then $a,d\ne0$ and $M(z)=(a/d)z+b/d=T_{b/d}\circ D_{a/d}(z)$. If $c\ne0$, algebraic division gives
$$
M(z)=\frac ac+\frac{bc-ad}{c^2}\frac1{z+d/c}.
$$
Consequently
$$
\boxed{M=T_{a/c}\circ D_{(bc-ad)/c^2}\circ I\circ T_{d/c}.}
$$
The scaling coefficient is nonzero because $ad-bc\ne0$. Thus translations, nonzero complex scalings and reciprocal inversion generate every <Möbius map>. The formulas extend over their poles and infinity by the sphere interpretation.
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