= Solution
\b[A suitable <Poincare disc automorphism>] is the <Möbius transformation>
$$
\boxed{f(z)=\frac{z-a}{1-\overline a z}.}
$$
It sends $a$ to zero and has inverse $(w+a)/(1+\overline a w)$. To verify the <isometry> directly, note that
$$
1-|f(z)|^2=\frac{(1-|a|^2)(1-|z|^2)}{|1-\overline a z|^2},\qquad |f'(z)|=\frac{1-|a|^2}{|1-\overline a z|^2}.
$$
These identities show both that $f$ maps the <unit disc> onto itself and that $2|df|/(1-|f|^2)=2|dz|/(1-|z|^2)$, preserving the <Riemannian metric>.
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