Prove that the unique metric that makes the map from the complex plane with infinity the complex projective line an isometry has this formula:

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Firstly let us define the map $\iota: \mathbb{C}_\infty \to P^1(\mathbb{C})$ by $\iota(z) = [z:1]$, $\;\iota(\infty) = [1:0]$.

Here the equivalence class of $(z_1,z_2)$ is denoted $[z_1:z_2]$ where $(z_1,z_2) \sim (\tilde{z}_1,\tilde{z}_2)$ if and only if there is a $\lambda> 0$ such that $z_1 = \lambda \tilde{z}_1$ and $z_2 = \lambda \tilde{z}_2$.

$P^1(\mathbb{C})$ denotes the set of equivalence classes.

Now let $\tilde{d}$ be the unique metric on $P^1(\mathbb{C})$ such that $\iota$ is an isometry (where $\mathbb{C}_\infty$ has the usual metric $d$).

How do I show that

$\tilde{d}([z_1:w_1],[z_2:w_2]) = 2\sqrt{1-\dfrac{|\langle v_1,v_2 \rangle|^2}{||v_1||^2||v_2||^2}}$,

where $v_1 = (z_1,w_1), v_2 = (z_2,w_2)\in \mathbb{C}^2\setminus \{0\}$ and $\langle , \rangle$ denotes the Hermitian inner product on $\mathbb{C}^2$?