Past exam of the mathematics course of the University of Cambridge / 2001 / ib / Paper 4 / 8F / Solution

Multiplying numerator and denominator by the same nonzero constant does not change a Möbius transformation. Since $b\ne 0$, divide by $b$ and relabel the coefficients so that $b=1$. On $|z|=1$, preservation of the circle givesEquating the constant and first Fourier coefficients yieldsIf $c=0$, the second equality forces $a=0$, contradicting the nonzero determinant. Hence $d=\bar{a}/\bar{c}$. With $C=|c{|}^2>0$, the first equality becomesThe root $C=|a{|}^2$ would imply $ad=|a{|}^2/\bar{c}=c$, making the numerator and denominator proportional. It is excluded by nondegeneracy. Thus $|c|=1$ and $d=c\bar{a}$. Writing $1/c=e^{i\psi }$ givesHere $a$ is the normalized coefficient after division by the original $b$. The inequality $|a|\ne 1$ is again the determinant condition. For $|a|<1$ the map preserves the disc; for $|a|>1$ it interchanges its inside and outside. Circle preservation alone does not justify restricting to automorphisms of the unit disk.