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Fun with a sphereby stunner5000pt
Tags: sphere 
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#1
Mar1306, 05:19 PM

P: 1,439

Show that the stereographic projections of the points z and [itex] 1/\overline{z} [/itex] are reflections of each other in teh equatorial plane of the Reimann sphere
ok so let z = x + iy then [tex] \frac{1}{\overline{z}} = \frac{x  iy}{x^2 + y^2} [/tex] so the magnitude of [tex] \frac{1}{\overline{z}} [/tex] is [tex] \frac{1}{x^2 + y^2} [/tex] the stereogrpahic projection of z is [tex] x_{1} = \frac{2x}{x^2 + y^2 +1} [/tex] [tex] y_{1} = \frac{2y}{x^2 + y^2 +1} [/tex] [tex] z_{1} = \frac{x^2 + y^2 1}{x^2 + y^2 +1} [/tex] for 1/ z bar is [tex] x_{2} = \frac{2x(x^2 + y^2)^2}{1 + (x^2 + y^2)^2} [/tex] [tex] y_{2} = \frac{2y(x^2 + y^2)^2}{1 + (x^2 + y^2)^2} [/tex] [tex] z_{2} = \frac{1(x^2 + y^2)^2}{1 + (x^2 + y^2)^2} [/tex] i fail to see how these are reflections of each other, then shouldnt the x1 =  x2?? and so on?? please help! 


#2
Mar1306, 05:36 PM

P: 116




#3
Mar1306, 05:38 PM

P: 1,439




#4
Mar1306, 05:48 PM

P: 116

Fun with a sphere
[tex]\frac{1}{\overline{z}}=\frac{xiy}{x^2+y^2}=\frac{x}{x^2+y^2}i\frac{y}{x^2+y^2}[/tex] I get a different magnitude than yours. 


#5
Mar1306, 07:52 PM

P: 1,439

ys i got what u got ( i made a sign error)
so the magnitude would be... [tex] \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}} [/tex] the projection is then [tex] x_{2} = \frac{2x(x^2 + y^2)}{1 + (x^2 + y^2)} [/tex] right? i know sily math errors everywhere!! 


#6
Mar1406, 03:40 PM

P: 1,439

is what i did in the above post correct now??



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