# Fun with a sphere

by stunner5000pt
Tags: sphere
 P: 1,439 Show that the stereographic projections of the points z and $1/\overline{z}$ are reflections of each other in teh equatorial plane of the Reimann sphere ok so let z = x + iy then $$\frac{1}{\overline{z}} = \frac{x - iy}{x^2 + y^2}$$ so the magnitude of $$\frac{1}{\overline{z}}$$ is $$\frac{1}{x^2 + y^2}$$ the stereogrpahic projection of z is $$x_{1} = \frac{2x}{x^2 + y^2 +1}$$ $$y_{1} = \frac{2y}{x^2 + y^2 +1}$$ $$z_{1} = \frac{x^2 + y^2 -1}{x^2 + y^2 +1}$$ for 1/ z bar is $$x_{2} = \frac{2x(x^2 + y^2)^2}{1 + (x^2 + y^2)^2}$$ $$y_{2} = \frac{2y(x^2 + y^2)^2}{1 + (x^2 + y^2)^2}$$ $$z_{2} = \frac{1-(x^2 + y^2)^2}{1 + (x^2 + y^2)^2}$$ i fail to see how these are reflections of each other, then shouldnt the x1 = - x2?? and so on?? please help!
P: 116
 Quote by stunner5000pt so the magnitude of $$\frac{1}{\overline{z}}$$ is $$\frac{1}{x^2 + y^2}$$
P: 1,439
 Quote by assyrian_77 You sure about that?
perhaps im missing something here??

P: 116
Fun with a sphere

 Quote by stunner5000pt perhaps im missing something here??
Well, the magnitude of a complex number $z=x\pm iy$ is $|z|=\sqrt{x^2+y^2}$, but you probably know this already. You have the complex number

$$\frac{1}{\overline{z}}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}$$

I get a different magnitude than yours.
 P: 1,439 ys i got what u got ( i made a sign error) so the magnitude would be... $$\sqrt{\left(\frac{x}{x^2 + y^2}\right)^2 +\left(\frac{y}{x^2 + y^2}\right)^2} = \sqrt{\frac{1}{x^2 + y^2}}$$ the projection is then $$x_{2} = \frac{2x(x^2 + y^2)}{1 + (x^2 + y^2)}$$ right? i know sily math errors everywhere!!
 P: 1,439 is what i did in the above post correct now??

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