= Solution
For $|\lambda|<1$, let $F_\lambda(\zeta)=(\zeta-\lambda)/(1-\overline\lambda\zeta)$. Direct calculation gives
$$
1-|F_\lambda(\zeta)|^2=\frac{(1-|\lambda|^2)(1-|\zeta|^2)}{|1-\overline\lambda\zeta|^2}.
$$
Its 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 $d\ne0$. Since it maps the boundary circle to itself, $|a|^2=|c\zeta+d|^2=|c|^2+|d|^2+2\operatorname{Re}(c\overline d\zeta)$ 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:
$$
\boxed{T(\zeta)=\omega\frac{\zeta-\lambda}{\overline\lambda\zeta-1},\qquad |\lambda|<1,\quad|\omega|=1.}
$$
Here mapping the disc to itself means bijectively onto itself; there are nonautomorphic <Möbius transformations> whose image is a proper subdisc.
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