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Fun with a sphere

by stunner5000pt
Tags: sphere
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stunner5000pt
#1
Mar13-06, 05:19 PM
P: 1,440
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!
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assyrian_77
#2
Mar13-06, 05:36 PM
P: 116
Quote Quote by stunner5000pt
so the magnitude of [tex] \frac{1}{\overline{z}} [/tex] is [tex] \frac{1}{x^2 + y^2} [/tex]
You sure about that?
stunner5000pt
#3
Mar13-06, 05:38 PM
P: 1,440
Quote Quote by assyrian_77
You sure about that?
perhaps im missing something here??

assyrian_77
#4
Mar13-06, 05:48 PM
P: 116
Fun with a sphere

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

[tex]\frac{1}{\overline{z}}=\frac{x-iy}{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.
stunner5000pt
#5
Mar13-06, 07:52 PM
P: 1,440
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!!
stunner5000pt
#6
Mar14-06, 03:40 PM
P: 1,440
is what i did in the above post correct now??


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