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2008-05-18T19:12:25Z

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Vipul: New page: ==Statement== Consider the inverse stereographic projection map: $\Phi:\mathbb{C} \to S^2 \setminus \{ N \}$ where $N$ is the north pole in $S^2$ . ...

2008-04-27T14:36:25Z

New page: ==Statement== Consider the inverse stereographic projection map: <math>\Phi:\mathbb{C} \to S^2 \setminus \{ N \}</math> where <math>N</math> is the north pole in <math>S^2</math>. ...

New page

==Statement==

Consider the inverse [[stereographic projection]] map:

<math>\Phi:\mathbb{C} \to S^2 \setminus \{ N \}</math>

where <math>N</math> is the north pole in <math>S^2</math>.

Then the distance in <math>\R^3</math> between <math>\Phi(z)</math> and <math>\Phi(w)</math> is given by:

<math>\frac{|z - w|}{(1 + |z|^2)(1 + |w|^2)}</math>

==Facts used==

We use here the formula for inverse stereographic projection:

<math>\Phi(z) = \left(\frac{z + \overline{z}}{1 + |z|^2}, \frac{z - \overline{z}}{i(1 + |z|^2)}, \frac{|z|^2 - 1}{1 + |z|^2}\right)</math>

==Proof==

We simply plug in the formula for <math>\Phi(z), \Phi(w)</math> in the formula to calculate the distance between two points in Euclidean space. {{fillin}}

Distance formula for inverse stereographic projection - Revision history

Vipul: 1 revision

Vipul: New page: ==Statement== Consider the inverse stereographic projection map: \Phi:\mathbb{C} \to S^2 \setminus \{ N \} where N is the north pole in S^2. ...

Vipul: New page: ==Statement== Consider the inverse stereographic projection map: $\Phi:\mathbb{C} \to S^2 \setminus \{ N \}$ where $N$ is the north pole in $S^2$ . ...