... where $z, w \in \mathbb{D}$, $\mathbb{D}$ the open unit disc. I realise that the way to start working on this is to express $T$ as $e^{i\alpha}S_a$ for some $\alpha \in [0, 2\pi]$ and $a \in \mathbb{D}$, where $S_a (z) = \frac{z-a}{1-\overline{a}z}$. Then we find that the $e^{i\alpha}$ factors disappear from the modulus on the left-hand side with handy cancellation. But after that the fractions seem to make everything a mess. I tried to prove the identity by showing $|T(z)-T(w)||1-z\overline{w}| = |z-w||1-T(z)\overline{T(w)}|$, but the fractions and terms got out of hand and I doubt it's that complicated.
Is there a simple way of proving this?
Fix a point $w \in \mathbb D$ and use some additional transformations.
$Sz = \dfrac {z+w}{1 + z \bar w}$ is a Möbius transformation on $\mathbb D$ that takes $0$ to $w$.
$Rz = \dfrac{z - Tw}{1 - z \overline{Tw}}$ is a Möbius transformation on $\mathbb D$ that takes $Tw$ to $0$.
The composition $R \circ T \circ S$ is a Möbius transformation on $\mathbb D$ that fixes $0$. The only such transformations are rotations, and thus $$|R(T(S(z)))| = |z|$$ $$|R(T(z))| = |S^{-1}(z)|$$ $$ \frac{|Tz - Tw|}{|1 - Tz \overline{Tw}|} = \frac{|z-w|}{|1 - z \bar w|}.$$