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Homework Help: Fun with a sphere

  1. Mar 13, 2006 #1
    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!
    Last edited: Mar 13, 2006
  2. jcsd
  3. Mar 13, 2006 #2
    You sure about that?
  4. Mar 13, 2006 #3
    perhaps im missing something here??
  5. Mar 13, 2006 #4
    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


    I get a different magnitude than yours.
  6. Mar 13, 2006 #5
    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]

    i know sily math errors everywhere!!
  7. Mar 14, 2006 #6
    is what i did in the above post correct now??
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