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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.

Math101
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  • For a much more simpler representation see https://math.stackexchange.com/questions/4016091/stereographic-projection-of-unit-sphere-in-a-banach-space – Michael Hoppe Aug 17 '21 at 07:30
  • @Michael Hoppe Yes there is a much simpler way to represent the stereographic projection, but what I am interested in is a formula to express the coordinates $(y_1,y_2,y_3)$ in $S$ corresponding to the sum $z+z'$ in $\mathbb C$ in terms of the coordinates $(x_1,x_2,x_3)$ and $(x_1',x_2',x_3')$ of $z$ and $z'$. So my question is about how addition in $\mathbb C$ gets represented as an operation on $S$. – Math101 Aug 17 '21 at 21:38
  • Why do you come to the idea that there's an easy formula for $\phi(z+z')$? Even if $z\perp z'$ and both have equal lengths, i.e., $z'=iz$, it's quite ... uncomfortable. – Michael Hoppe Aug 18 '21 at 10:34
  • This is an exercise from "Functions of One Complex Variable" from John B. Conway so I'd expect there's a better answer than just injecting the sum in the formulas. But I can't find a good solution either... – Colin Pitrat Jul 01 '23 at 08:44

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