Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2005 / 6

A Möbius transformation has the form $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, extended to the Riemann sphere by its limiting values at its pole and at infinity. If $c\ne 0$, it sends $-d/c$ to infinity and infinity to $a/c$; if $c=0$, infinity maps to infinity.

Represent $M$ by the invertible matrix $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, with nonzero scalar multiples representing the same map. Substitution shows that composition corresponds to matrix multiplication. The product determinant is the product of two nonzero determinants, so the composition is again a Möbius transformation. Solving $w=M(z)$ giveswith the same extended-value convention. This inverse proves that $M$ is a bijection of the extended complex plane.

Use the Cayley transformFor $\mathrm{Im}z>0$, its denominator is nonzero, and direct subtraction gives $1-|w{|}^2=4\mathrm{Im}z/|z+i{|}^2>0$. Thus the image lies in the open unit disk. Conversely, for $|w|<1$ the inverse is $z=i(1+w)/(1-w)$, with $\mathrm{Im}z=(1-|w{|}^2)/|1-w{|}^2>0$. Hence it maps the upper half-plane bijectively onto the disk.