Find isometry group of the Poincaré disc I cannot show that any isometry of the Poincaré disc $\mathbb{H}^2$ is given by $e^{i\theta} \frac{z + c}{cz + 1}$ where $\theta$ and $c$ are constants. I know that they are indeed isometries, however I don't know how to prove that any isometry is of this kind. Any hint or answer?
Thanks in advance
 A: Every isometry of the disk $\mathbb{H}^2$ is either a holomorphic automorphism of the disk, or a holomorphic automorphism composed with complex conjugation.
Let $f$ be an isometry of the disk. If $w = f(0)$, then composing with the Möbius transformation
$$T_w \colon z \mapsto \frac{z-w}{1-\overline{w}z}$$
produces an isometry $f_1$ with $f_1(0) = 0$. Since $f_1$ is an isometry, $\left\lvert f_1\left(\frac12\right)\right\rvert = \frac12$ (Euclidean circles with centre $0$ are also the loci of points equidistant from $0$ with respect to the hyperbolic metric), and composing with a rotation $\rho \colon z \mapsto e^{i\varphi}z$, we obtain an isometry $f_2$ with $f_2(0) = 0$ and $f_2\left(\frac12\right) = \frac12$.
There are two possibilities for $f_2\left(\frac i2\right)$, either $f_2\left(\frac{i}{2}\right) = \frac{i}{2}$, then we have $f_2 = \operatorname{id}$ and $f = T^{-1}\circ \rho^{-1}$, which can be written in the form
$$f(z) = e^{i\theta}\frac{z+c}{1+\overline{c}z},$$
or $f_2\left(\frac{i}{2}\right) = -\frac{i}{2}$, in which case $f_2$ is the complex conjugation, and $f(z) = T^{-1}\left(\rho^{-1}(\overline{z})\right)$, which can be written in the form
$$f(z) = e^{i\theta}\frac{\overline{z}+c}{1+\overline{cz}}.$$
The crucial property is that every point is uniquely determined by its distances from three points that don't lie on an orthocircle, like in the Euclidean plane, every point is determined by its distance from three points not on a straight line, so an isometry that fixes three such points must be the identity.
