Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 5E (Groups)

Write the Möbius transformation asA finite fixed point of a Möbius transformation satisfiesIf three distinct points of the Riemann sphere are fixed, this quadratic is identically zero, with the usual interpretation when one point is infinity. Hence $c=b=0$ and $d=a$, so $f(z)=z$. A nonidentity transformation's fixed-point equation is a nonzero polynomial of degree at most two on the sphere, and therefore it has one repeated fixed point or two distinct fixed points.

True. Conjugate the unique fixed point to infinity. The resulting map lies in the Affine subgroup of the Möbius group, so it has the form $g(z)=az+b$. A unique fixed point forces $a=1$ and $b\ne 0$; otherwise $g$ would have a finite fixed point as well. Conjugation by the scaling $z\mapsto z/b$ changes $z\mapsto z+b$ into $z\mapsto z+1$. Thus a nonidentity Möbius transformation with one fixed point is conjugate to a translation.

True. A Möbius transformation fixes infinity exactly when $c=0$, so it has the affine formThis is the composition of the scaling $z\mapsto \alpha z$ and the translation $z\mapsto z+\beta$.

False. Compositions of scalings $z\mapsto \mu z$ and inversion $z\mapsto 1/z$ reduce to one of the forms $\nu z$ or $\nu /z$; they preserve the unordered pair $\{0,\infty \}$. Butis a Möbius transformation fixing $0$ and sends infinity to $1$, so it cannot be such a composition.