 Solution

For $|\lambda |<1$, let $F_{\lambda }(\zeta )=(\zeta -\lambda )/(1-\overline{\lambda }\zeta )$. Direct calculation givesIts denominator does not vanish in the disc, and this identity shows that it maps the disc into itself. Its inverse $(w+\lambda )/(1+\overline{\lambda }w)$ has the same property, so $F_{\lambda }$ is a Poincare disc automorphism. Multiplication by a complex number of modulus one is another such automorphism.

Conversely let $T$ be a Möbius transformation mapping the disc onto itself, and let $\lambda =T^{-1}(0)$. Then $T\circ F_{\lambda }^{-1}$ is a disc automorphism fixing zero, say $S(\zeta )=a\zeta /(c\zeta +d)$ with $d\ne 0$. Since it maps the boundary circle to itself, $|a{|}^2=|c\zeta +d{|}^2=|c{|}^2+|d{|}^2+2\mathrm{Re}(c\overline{d}\zeta )$ for every $|\zeta |=1$. This forces $c\overline{d}=0$, hence $c=0$ and $|a/d|=1$. Thus $S$ is a rotation $\eta \zeta$, and $T=\eta F_{\lambda }$. Absorb the denominator's minus sign into $\omega =-\eta$ to obtain precisely the paper's convention:Here mapping the disc to itself means bijectively onto itself; there are nonautomorphic Möbius transformations whose image is a proper subdisc.