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I am interested in showing that isometries in $\mathbb{D}$ are either conformal self-maps in $\mathbb{D}$ or they are compositions of conformal self-maps with $z\mapsto \bar{z}$.

It is given that conformal self-maps in $\mathbb{D}$ are isometries. I think, to finish this, I just need to show that $z\mapsto \bar{z}$ is an isometry.

So far, I have considered the fact that the real interval $(-1,1)\subset \mathbb{D}$ is a geodesic in $\mathbb{D}$ that is fixed under $z\mapsto \bar{z}$. This means we are free to consider two points, find the geodesic that goes through them (say $\gamma$), map $\gamma$ conformally to $(-1,1)$, perform $z\mapsto \bar{z}$, and there will be no change in two initial points along $\gamma$ throughout this process. I mean, if $\rho(z_1,z_2)$ represents hyperbolic distance between $z_1$ and $z_2$ then we are allowed, by the previous argument, to say $\rho\left(\overline{\phi(z_1)},\overline{\phi(z_2)}\right)=\rho(z_1,z_2)$. But... this isn't quite what I want. What I want to show is that $\rho\left(\overline{z_1},\overline{z_2}\right)=\rho(z_1,z_2)$. Any suggestions? Thanks guys.

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    $\begingroup$ Don't you have an explicit formula for the hyperbolic distance? $\endgroup$ Apr 1, 2014 at 22:41
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    $\begingroup$ The hyperbolic length of a curve $\gamma$ in $\mathbb{D}$ is given by the integral $$2\int_\gamma \frac{|dz|}{1-|z|^2}.$$ The hyperbolic distance between two points $a$ and $b$ is thus the infimum of the hyperbolic lengths over all curves joining $a$ and $b$. This simplifies to a formula when $a$ and $b$ are real. $\endgroup$
    – J. Moeller
    Apr 1, 2014 at 22:57

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It is straightforward to verify that the curve of shortest hyperbolic length connecting $0$ to $s > 0$ is the line segment $[0,s]$. One computes

$$\rho(0,s) = \int_0^s \frac{2}{1-t^2}\,dt = \log \frac{1+s}{1-s}.$$

Since holomorphic automorphisms of the unit disk leave the hyperbolic length invariant, given two distinct $z,w \in \mathbb{D}$ we find

$$\rho(z,w) = \rho\left(0,\frac{w-z}{1-\overline{z}w}\right) = \rho\left(0,\left\lvert\frac{w-z}{1-\overline{z}w}\right\rvert\right) = \log \frac{1+\left\lvert\frac{w-z}{1-\overline{z}w}\right\rvert}{1-\left\lvert\frac{w-z}{1-\overline{z}w}\right\rvert}.$$

With this explicit formula for the hyperbolic distance, it is easily seen that the conjugation is an isometry.

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