I'm studying geometry and am studying geometry when one calculation in the book had me confused. Specifically, I'm studying about stereographic projection and how it relates to complex numbers.
The specific image in question is:
It's a unit sphere in $\Bbb{E}^3$, with point $N$ being the north pole and $O$ the origin. The coordinates for points $P$ and $Q$ are $(x, y, h) = (z, h)$ and $(u, v, 0) = (\omega, 0)$, respectively, with $z = x + iy$ and $\omega = u + iv$.
A series of calculations take starting with:
$$ (1 - h) : \vert z \vert = 1 : \vert \omega \vert \\ h^2 + \vert z \vert^2 = 1 $$
This gives us:
$$ h^2 + (1 - h)^2 \vert \omega \vert^2 = 1 $$
which in turn is factored into:
$$ (h - 1) \left( (1 + \vert \omega \vert^2)h - \left( \vert \omega \vert^2 - 1 \right) \right) = 0 $$
giving us:
$$ \begin{align} h & = \dfrac{\vert \omega \vert^2 - 1}{\vert \omega \vert^2 + 1} \\ 1 - h & = \dfrac{2}{\vert \omega \vert^2 + 1} \\ z & = \dfrac{2\omega}{\vert \omega \vert^2 + 1} \end{align} $$
and in conclusion:
$$ (x, y, h) = \left( \dfrac{2u}{1 + u^2 + v^2}, \dfrac{2v}{1 + u^2 + v^2}, \dfrac{-1 + u^2 + v^2}{\phantom{-}1 + u^2 + v^2} \right) $$
This probably stems from my lack of background knowledge, but my main question is how were these calculations obtained? Specifically:
- How was the factorization of the third equation into the fourth equation obtained? I noticed that expanding the fourth equation gives us an extra $h - h$ term (i.e. $h^2 + \vert \omega \vert^2 h^2 - \vert \omega \vert^2 h + h - h - \vert \omega \vert^2 h + \vert \omega \vert^2 - 1 = 0$) but how do we know this factors into the final equation?
- After factorization, how do we obtain that $h = \dfrac{\vert \omega \vert^2 - 1}{\vert \omega \vert^2 + 1}$?
- How are the final values of the coordinates $(x, y, h)$ obtained? I tried replacing the $\omega$ for the equation of $z$ with $u + iv$, but am not sure if that is the correct approach.
Thanks in advance.
The 3rd equation can be written :
$$h^2 -1 + (1-h)^2 \vert \omega \vert^2 = 0$$
i.e.,
$$(h-1)(h+1) + \underbrace{(h-1)^2}_{\text{note the change}} \vert \omega \vert^2 = 0$$
giving
$$(h-1)[(h+1) + (h-1) \vert \omega \vert^2] = 0$$
then
$$(h-1)[h(\vert \omega \vert^2+1) -(\vert \omega \vert^2-1)] = 0$$
As $h \neq 1$ (the case $h=1$ would correspond to $Q$ at infinity), it is the content of the square brackets which is equal to zero, giving :
$$h=\dfrac{\vert \omega \vert^2-1}{\vert \omega \vert^2+1}$$
Regarding your 3rd point, you are right in writing $\omega=u+iv$ (which can be seen as issued from the rotation of the 2D figure around the horizontal axis in order to generate a 3D figure).