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Fun with a sphere |
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| Mar13-06, 05:19 PM | #1 |
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Fun with a sphere
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! |
| Mar13-06, 05:36 PM | #2 |
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| Mar13-06, 05:38 PM | #3 |
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| Mar13-06, 05:48 PM | #4 |
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Fun with a sphere[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. |
| Mar13-06, 07:52 PM | #5 |
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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!! |
| Mar14-06, 03:40 PM | #6 |
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is what i did in the above post correct now??
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