Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2004 / 6 / Solution

A Möbius transformation is $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, extended to the Riemann sphere at its pole and at infinity. Wherever its finite coordinate is defined,At a pole, the ordinary complex derivative is undefined, rather than zero; using the target coordinate $1/M$ gives a nonzero derivative there as well. Thus the source's nowhere-zero assertion is interpreted in local sphere coordinates, or on the finite domain with the pole excluded.

Composition corresponds to multiplication of the invertible matrices $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, with scalar multiples giving the same map. The identity matrix gives the identity, and the inverse matrix gives an inverse Möbius map. This proves the group property on the Riemann sphere.

The two boundary circles meet at $\pm i$. PutFor $w=u+iv$, direct substitution givesThis gives a lens-to-half-plane map through the intersection points. Both lens inequalities hold exactly when $u<-|v|$, the open sector with continuous argument between $3\pi /4$ and $5\pi /4$. This proves bijectivity, since the inverse is defined throughout that sector and the excluded endpoint $-i$ is not in the lens.

Rotate the sector through $-3\pi /4$ and square it, obtaining a bijection onto the upper half-plane. Therefore a conformal map isSquaring is injective on this sector because it contains no pair of opposite points. Its derivative and the Möbius derivative are nonzero in the interior, so the bijection is conformal.