While studying stereographic projection of extended complex on unit sphere $S$ in $\mathbb{R^3}$ we get two metrics one is chordal metric and second one is spherical metric. The spherical metric $d_s$ is given by
\begin{align} d_s(z,w) &= \inf_{\gamma} L[\gamma] \end{align} where $\gamma$ is curve joining $z$ and $w$ on sphere $S$ and $L[\gamma]$ is the length of curve given by $\int_\gamma \frac{2|dz|}{1+ |z|^2}$.
I am continuously using this spherical metric during the study of complex dynamics for my Master's project. The proof and the idea behind the proof is important because the change in scale (element of arc length) under the function is useful in some of the theorem's proof.
Although I have some intuitive idea from the differential geometry point of view that is at each point on the surface $S$ we have tangent plane and each vector of tangent plane has length. So if we consider curve on sphere then at each point of curve we have tangent vector and its length (i.e. element of arc length) and then we integrate over it. But till now I didn't study proof of this metric in differential geometry.
In the case of complex analysis, I also don't know how to proof it. Therefore I need to know the proof that how element of arc length is $\frac{2|dz|}{1+ |z|^2}$.
We start from the parametrization of the sphere given by the stereographic projection $$ \matrix{{\bf C} & \rightarrow & S^2 \subset {\bf R}^3 \cr z=x+iy & \mapsto & v = (v_1, v_2, v_3)} $$ $$ \pmatrix{v_1 \cr v_2 \cr v_3} = {1 \over 1+x^2+y^2} \pmatrix{2x \cr 2y \cr x^2+y^2-1}. $$ To simplify notation, let us define $r^2 = |z|^2 = x^2+y^2$ and $\lambda = 1+|z|^2 = 1+r^2$.
The images of the two vectors from the canonical basis of ${\bf C} \simeq {\bf R}^2$ are given by ${\partial v \over \partial x}$ and ${\partial v \over \partial y}$. The metric on ${\bf C}$ obtained by pulling back the metric on the sphere is thus given by $$ \left\|{\partial v \over \partial x}\right\|^2 dx^2 + \left\|{\partial v \over \partial y}\right\|^2 dy^2 + 2 \langle{\partial v \over \partial x}, {\partial v \over \partial y}\rangle dx dy $$ where the norm and the scalar product refer to the standard euclidean norm on ${\bf R}^3$. We are left with computing the partial derivatives of $v$. $$ {\partial v \over \partial x} = -\lambda^{-2}\pmatrix{2(x^2-y^2-1) \cr 4xy\cr -4x}, \quad {\partial v \over \partial y} = -\lambda^{-2}\pmatrix{4xy \cr 2(y^2-x^2-1)\cr -4y} $$ From these expressions, we deduce after some computations $$ \langle{\partial v \over \partial x}, {\partial v \over \partial y}\rangle = 0, \quad \left\|{\partial v \over \partial x}\right\|^2 = 4\lambda^{-2}, \quad \left\|{\partial v \over \partial x}\right\|^2 = 4\lambda^{-2}. $$ The Riemannian metric obtained is equal to $$ 4\lambda^{-2} (dx^2+dy^2) = {4 |dz|^2 \over (1+|z|^2)^2}. $$ The associated arc length is ${2 |dz| \over (1+|z|^2)}$.