Let $D$ be the open unit disk in the complex plane and let $f: D \rightarrow \mathbb{C}, z \mapsto \frac{i+z}{1+iz}$. We equip the upper half plane $H = \{z \in \mathbb{C} \mid \operatorname{Im}(z) > 0\}$ with the Riemannian metric $$m_H(X, Y) = \frac{1}{\operatorname{Im}(z)^2}\langle X, Y \rangle_{\mathbb{R}^2}$$ and $D$ with the Riemannian metric $$m_D(X,Y) = \frac{4}{(1-|z|^2)^2}\langle X, Y \rangle_{\mathbb{R}^2}.$$
How should I proceed to show that $f:(D, m_D) \rightarrow (H, m_H)$ is an isometry ? I really have no idea in this case.
Thanks in advance.
Things will be easier if you write all in differential notation. Denote the complex coordinates in $H$ and $D$ by $w$ and $z$, respectively. Then, write $$m_H = \frac{{\rm d}w\,{\rm d}\overline{w}}{{\rm Im}(w)^2}\quad\mbox{and}\quad m_D = \frac{4\,{\rm d}z\,{\rm d}\overline{z}}{(1-|z|^2)^2}.$$The pull-back $f^*(m_H)$ can be computed by writing $$w = \frac{i+z}{1+iz}$$and substituting everything in the expression for $m_H$. That is, we have $$\overline{w} = \frac{-i+\overline{z}}{1-i\overline{z}}, \quad {\rm d}w = \frac{{\rm d}w}{{\rm d}z}\,{\rm d}z, \quad {\rm d}\overline{w} = \frac{{\rm d}\overline{w}}{{\rm d}\overline{z}}\,{\rm d}\overline{z}.$$Compute the derivatives, and check that$$\frac{{\rm d}w\,{\rm d}\overline{w}}{{\rm Im}(w)^2} = \frac{4\,{\rm d}z\,{\rm d}\overline{z}}{(1-|z|^2)^2}.$$