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Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is

a. a circle if $ T $ is a circle.
b. a circle minus (0, 0, 1) if $ T $ is a line.

Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$

To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:

$$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$

We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:

$ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.

I really am at a loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?

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You really need to fix the LaTeX here by using $ signs - it is all but unreadable. –  Ron Gordon Mar 19 '13 at 2:43
    
How do I fix the LaTeX using $ signs? This is my first time using this stack exchange, sorry. –  Max Mar 19 '13 at 2:43
    
See meta.math.stackexchange.com/questions/5020/… -- That or just click edit on enough entries until you can figure it out –  muzzlator Mar 19 '13 at 2:45
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wrap the tex code between dollar signs. So, for instance, to get $\zeta = \xi$, write the text code: \zeta = \xi in between dollar signs. –  Ittay Weiss Mar 19 '13 at 2:46
    
I fixed that TeX. @muzzlator I don't think that theorem applies to this problem. –  Max Mar 19 '13 at 2:54
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1 Answer

The type of map you describe is called a stereographic projection:

http://en.wikipedia.org/wiki/Stereographic_projection

There are many resources out there which proves the thing you are looking for:

See http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/node33.html for a proof that it maps circles to circles for a slightly different sphere, this will give you the basic idea

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I understand this proof, but it is going in the opposite direction that I am trying to solve. This proof shows the circles on the sphere are circles in the plane. I am trying to show that circles on the plane or circles on the sphere. –  Max Mar 19 '13 at 11:05
    
Try to reverse engineer the process, the map is $1-1$ so it shouldnt be too hard undoing it. Otherwise just google, I remember finding a gazillion results when I googled last time. Hope that helps (otherwise I have actual books I can recommend) –  muzzlator Mar 19 '13 at 11:14
    
I can't find any any proofs going in the opposite direction, and I can't figure out how to go backwards either. If you could provide some books that you know solve this, that would be helpful. –  Max Mar 19 '13 at 11:28
    
Ponnusamy "Complex Variables with applications" devotes an entire subchapter to this matter. Thinking about it a bit more, reverse engineering it should be easy. Aha, here is a proof of the converse. As you can see, it's just a matter of playing with the coordinates you get in the other direction: people.maths.ox.ac.uk/earl/G2-lecture5.pdf –  muzzlator Mar 19 '13 at 11:51
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