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Wikipedia gives this form of the stereographic Projection from $S^{2} \rightarrow \hat{\mathbb C}$:$$ (1) : z=\frac{x_{1}+ix_{2}}{1-x_{3}}$$ and for the inverse projections the points are supposedly:$$(2): x_{1}=\frac{\overline{z}+z}{z\overline{z}+1}, x_{2}=\frac{z-\overline{z}}{i(z\overline{z}+1)}, x_{3}=\frac{z\overline{z}-1}{z\overline{z}+1}$$

How does he go from $(1)$ to $(2)$ ? I tried calculating the inverse matrix and reading the coefficients from it, looking at $\overline{z}, z$ in $(1)$ and solving it for $x_{1},x_{2},x_{3}$ but I don't get anything of this form

(f.e. I get: $\displaystyle x_{1}=\frac{z(1-x_{3})-2ix_{2}}{\overline{z}(1-x_{3})})$

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Try doing it geometrically. See $\mathbb{C}$ as the plane $\mathbb{R}^2$, and if you take a point $(x,y)\in\mathbb{R}^2$, write down the equation of the line that connects it to $(0,0,1)$. Then find the point (not $(0,0,1)$) on the line where the norm is 1. – rfauffar Dec 14 '11 at 14:16
Why search on the line for a point where the norm is 1? – VVV Dec 14 '11 at 14:29
That's where the line intersects the sphere, and that's the point you send your point on the plane to. – rfauffar Dec 15 '11 at 13:06
up vote 1 down vote accepted

If $z=\dfrac{x_{1}+ix_{2}}{1-x_{3}}$, then $$|z|^2=z\overline{z}=\frac{x_{1}+ix_{2}}{1-x_{3}}\cdot\frac{x_{1}-ix_{2}}{1-x_{3}}=\frac{x_{1}^2+x_{2}^2}{(1-x_{3})^2}=\frac{1-x_{3}^2}{(1-x_{3})^2}=\frac{1+x_{3}}{1-x_{3}},$$ since $x_{1}^2+x_{2}^2+x_{3})^2=1$ for $(x_1,x_2,x_3)\in S^2$. From the above equality, we have $|z|^2(1-x_{3})=1+x_{3}$, or equivalently $(|z|^2+1)x_{3}=|z|^2-1$, which implies $$x_{3}=\frac{|z|^2-1}{|z|^2+1}=\frac{z\overline{z}-1}{z\overline{z}+1}.$$

From $z=\dfrac{x_{1}+ix_{2}}{1-x_{3}}$ again, we have the real part of $z$ $$\Re(z)=\frac{x_1}{1-x_{3}}.$$ Since $\Re(z)=(z+\overline{z})/2$, we get $$x_1=\frac{z+\overline{z}}{2}\cdot(1-x_{3})=\frac{z+\overline{z}}{2}\cdot(1-\frac{z\overline{z}-1}{z\overline{z}+1})=\frac{\overline{z}+z}{z\overline{z}+1}.$$

I will let you figure out the expression for $x_2$. Here is the hint for $x_2$: $\Im(z)=\dfrac{x_{2}}{1-x_{3}}$.

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I get it. Thank you. – VVV Dec 14 '11 at 14:28

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