Let $S=\{(x_1,x_2,x_3):x_1^2+x_2^2+x_3^2=1\}$ be the unit sphere in ${\mathbb R}^3$, and $\phi: {\mathbb C}\rightarrow S$ the stereographic map $$\phi(x+iy)=\frac{1}{x^2+y^2+1}(2x,2y,x^2+y^2-1).$$ Then if $z=x+iy, z'=x'+iy'$, $\phi(z)=(x_1,x_2,x_3)$, $\phi(z')=(x_1',x_2',x_3')$, I worked out the formulas for $\phi(z+z')=(y_1,y_2,y_3)$ to be $$y_1=\frac{2(x_1+x_1'-x_1x_3'-x_3x_1')}{3+2x_1x_1'+2x_2x_2'-x_3-x_3'-x_3x_3'}$$
$$y_2=\frac{2(x_2+x_2'-x_2x_3'-x_3x_2'}{3+2x_1x_1'+2x_2x_2'-x_3-x_3'-x_3x_3'}$$
$$y_3=\frac{1-3x_3x_3'+x_3+x_3'+2x_1x_1'+2x_2x_2'}{3+2x_1x_1'+2x_2x_2'-x_3-x_3'-x_3x_3'}$$ My question is: can these formulas be simplified or re-written in a better way? For example it would be nice to be able to check directly that $y_1^2+y_2^2+y_3^2=1$ but I have not been able to check that by hand or even with Mathematica.