In his book "Normal Families", Schiff mentions that the spherical arc length element ds on the Riemann sphere $\Sigma = \{(x_1, x_2, x_3)\in\mathbb{R}^3 \colon\ x_1^2+x_2^2+(x_3-\frac{1}{2})^2 = \frac{1}{4}\}$ works out to be $$ds=\frac{|dz|}{1+|z|^2}.$$ I tried to prove it as follows:
Suppose that we have a curve $\gamma$ on $\Sigma$. Let $\hat{\gamma}=p(\gamma)$ be its image in the (extended) complex plane under the stereographic projection $p$. Taking a parametrization $z(t)=x(t)+iy(t),\ t\in I$ for $\hat{\gamma}$ and composing back with $p^{-1}$, we obtain a parameterization $\gamma(t)=(a(t),b(t),c(t))$ for $\gamma.\ $ Then a straightforward calculation yields
$$ds=\sqrt{a'(t)^2+b'(t)^2+c'(t)^2}dt=\frac{|z'(t)|dt}{1+|z(t)|^2}.$$ Is this the idea?
In addition, I am a bit confused with the formula for the spherical length given later in the book: $$L(\gamma) = \int_{\gamma}{\frac{|dz|}{1+|z|^2}}.$$ Shouldn't it be written as $$L(\gamma) = \int_{\gamma}ds=\int_{p(\gamma)}{\frac{|dz|}{1+|z|^2}}\ ?$$
There is no real problem here. It's just about the intended meaning of $\gamma$ and $ds$. It seems that one considers a curve $\gamma$ in the extended complex plane (not on $S^2$), whereby the usual euclidean $ds:=|dz|$ in ${\mathbb C}$ is replaced by the pullback of the line element on $S^2$ via stereographic projection. Your curve $\gamma\subset\bar{\mathbb C}$ has a stereographic image $\hat\gamma:=p^{-1}(\gamma)\subset S^2$. The sphere $S^2$ inherits its Riemannian metric, i.e., its line element ${\rm d}\sigma$, from the imbedding $S^2\subset {\mathbb R}^3$, and pulling back this ${\rm d}\sigma$ to the $z$-plane results in $${\rm d}\sigma={|dz|\over1+|z|^2}\ .$$ The "spherical length" $L_{\rm sph}(\gamma)$ of your curve $\gamma\subset\bar{\mathbb C}$ then appears as $$L_{\rm sph}(\gamma)=\int_\gamma{|dz|\over1+|z|^2}=\int_a^b{|z'(t)|\over 1+|z(t)|^2}\>dt\ .$$