About the Poincaré metric isometry and automorphism on the unit disc Consider two domains $\Omega_1$ and $\Omega_2$ in $\mathbb{C}$ with the corresponding metrics $g|d z|^2$ and $h|d w|^2$ respectively. Let $d_1$ be the distance function induced from $g|d z|^2$, and $d_2$ that from $h|d w|^2$.
We say a map $f$ smooth if it is $C^{\infty}$ as a map from $\Omega_1 \subset \mathbb{R}^2$ to $\Omega_2 \subset \mathbb{R}^2$. Given a smooth bijection $f: \Omega_1 \rightarrow \Omega_2$, we call $f$ is an isometry if $f$ is distance-preserving in the sense that $d_1(z, w)=d_2(f(z), f(w))$ for any two points $z, w \in \Omega_1$.
Consider the unit disk $\mathbb{D}$ with its standard Poincaré metric.
$$
d s_P^2=\frac{4}{\left(1-|z|^2\right)^2}|d z|^2 .
$$
In the lecture we checked that any automorphism (holomorphic bijection) of $\mathbb{D}$ must be a holomorphic isometry. As a corollary we solved the corresponding distance function $d$ (called the Poincaré distance) induced from the Poincaré metric.
Question:
Given any isometry $f$ of $\mathbb{D}$ with respect to the Poincare distance, then either $f$ is an automorphism of $\mathbb{D}$ or its conjugate $\bar{f}$ is so}(why?). If we drop the assumption that an isometry is smooth, can we get the same conclusion?
I can only figure out $|f(z)|$ which is equal to $|\frac{z-\beta}{1-\bar{\beta}z}|$(where $\beta=f^{-1}(0)$).S ince $d(0,z)=\ln^{\frac{1-|z|}{1+|z|}}$ ，f is an isometry , thus $d(\beta,z)=d(0,f(z))$, and any automorphism of D is a holomorphic isometry. We have $d(\beta,z)=d(0,\frac{z-\beta}{1-\bar{\beta}z})=d(0,f(z))$, so this is all I can get. But how to proof either $f$ or $\bar{f}$ is actually equal to $\frac{z-\beta}{1-\bar{\beta}z}e^{i\theta}$?
This is my first time to ask questions in this community. Is there anyone who can help me?
 A: By comments from @ziqishen, its enough to consider $f(0) = 0$. So lets assume $f(0) = 0$.
Since we have,
$d(0,z) = d(f(0), f(z))$, we get, $|z| = \frac{|f(z)-f(0)|}{|1-f(0)^* f(z)|} \implies z = \frac{(f(z)-f(0))e^{j A(z)}}{1-f(0)^* f(z)} $.
Hence if $f(0) = 0$, we have that, $|f(z)| = |z|$. Hence $f(z) = ze^{j A(z)}$.
Prove that $\frac{|r_1 - r_2 e^{j \theta}|}{|1-r_1 r_2 e^{j \theta}|} = constant < 1$ has a unique solution $\gamma$ with $\theta = + \gamma$ or $\theta = - \gamma$ for fixed $r_1,r_2 \neq 0$.
$ d(f(r_1),f(r_2)) = d(r_1,r_2) = d(r_1 e^{j \theta},r_2 e^{j \theta}) =  d(f(r_1 e^{j \theta}),f(r_2 e^{j \theta}))$
Note that $f(z) = z e^{j A(z)}$. Using this in above equation we get:
$d(r_1,r_2) = d(f(r_1),f(r_2)) = d(f(r_1 e^{j \theta}),f(r_2 e^{j \theta})) \implies \frac{|r_1 - r_2|}{|1-r_1 r_2|} = \frac{|r_1 - r_2 e^{j (A(r_1)-A(r_2)}|}{|1-r_1 r_2 e^{j (A(r_1)-A(r_2)}|} = \frac{|r_1 - r_2 e^{j (A(r_1 e^{j \theta})-A(r_2 e^{j \theta})}|}{|1+r_1 r_2 e^{j (A(r_1 e^{j \theta})-A(r_2 e^{j \theta})}|} $.
Hence $A(r_1 e^{j \theta})-A(r_2 e^{j \theta}) = A(r_1)-A(r_2) = \gamma = 0 \text{ or } -\gamma = 0$ for fixed $r_1$ and $r_2$.
Hence for a fixed $r^*$ and for every other $r, \theta$, $A(r e^{j \theta}) = A(r^* e^{j \theta})$.
Now by similar argument i.e., with $r_1 = r_2 = r$, $A(r e^{j 2\theta}) - A(r e^{j \theta}) = +/-(A(r e^{j \theta}) - A(r))$. Now using continuity of derivative of $g(t) = A(re^{j \theta(t)})$ (try this), we have  $A(re^{j \theta})$ is linear in $\theta$ for a fixed $r$ i.e., of the form $s(r) \theta + g(r)$.
But by $A(r e^{j \theta}) = A(r^* e^{j \theta})$, we have that $s(r) = s(r^*)$ and $g(r) = g(r^*)$.
we have that $f(r e^{j \theta}) = r e^{j(s \theta + g)}$ for some constants $s,g$.
