Arclength under Stereographic Projection Let $G$ be an open connected set in $\Bbb C$, $f\in H(\Bbb C)$, and $\gamma$ a closed rectifiable curve in $\Bbb C$.  According to Ahlfors, stereographic projection maps the curve $f\circ\gamma$ to a curve with arclength $\int_\gamma \rho(f) |dz|$ where $$\rho(f)=\frac{2|f'(z)|}{1+|f(z)|^2}.$$
I'm not seeing how to derive this from the metric
$$d(z_1,z_2)=\frac{2|z_1-z_2|}{\sqrt{(1+|f(z_1)|^2)(1+|f(z_2)|^2)}}.$$

The formula for spherical distance gives the straight-line (chord) distance between the images of $z_1$ and $z_2$ on the sphere: $$d(z_1,z_2)=\frac{2|z_1-z_2|}{\sqrt{(1+|z_1|^2)(1+|z_2|^2)}} \tag{1}$$
As $z_2$ approaches $z_1$, we have 
$$
\frac{d(z_1,z_2)}{|z_1-z_2|} \to \frac{2}{1+|z_1|^2} \tag{2}
$$
Geometrically, (2) means that in a small neighborhood of $z_1$,  the distances are distorted by the factor on the right hand side of (2). 
There are two ways to use (2), depending on the definition of length you use. If length is defined via sums of $|\gamma(t_j)-\gamma(t_{j-1})|$ over partitions, then  you compare 
$$
\sum_j |\gamma(t_j)-\gamma(t_{j-1})| \quad \text{and}\quad \sum_j d(\gamma(t_j), \gamma(t_{j-1}))
$$
for sufficiently fine partitions. 
If the length is understood as the integral of $|\gamma'|$, then (2) can be used to show 
$$
|(\phi\circ \gamma)'(t)|  = \frac{2}{1+|\gamma(t)|^2} | \gamma '(t)|
$$
where $\phi:\mathbb C\to S^2$ is the stereographic projection.
