User:Paul Murray/Complex Numbers as a 3 Vector

One useful way to express complex numbers is by stereographically projecting them onto a sphere &Gamma;, and treating it as the Riemann sphere - the extended complex plane C. Doing this produces some nice results for arithmetic and Möbius Transformations that transparently handles the point at infinity.

The Transformations
This pair of transformations is one way of performing a steriographic projection, and it will be assumes for the rest of the article:


 * $$\mathbb C = \{ z \}, \Gamma = \{ \hat z \} :$$
 * $$\hat z = \begin{pmatrix} z_0 \\ z_1 \\ z_2 \end{pmatrix} =

\begin{pmatrix} \frac {2 a} {1 + \bar z z} \\ \frac {2 b} {1 + \bar z z} \\ \frac {1 - \bar z z} {1 + \bar z z} \end{pmatrix} $$
 * $$z = a + b i = \frac {z_0 + z_1 i} {1 - z_2} $$

Although there are many ways of projecting C onto &Gamma;, the transformation above has useful propperties. The sphere &Gamma; is the unit sphere. The pair (z0, z1) has the same argument as z. z2 reflects the modulus of z (although not in a direct way). All |z| < 1 get mapped to z2 < 0, and all |z| > 1 get mapped to z2 > 0. This gives us a simple set of transformations:

$$- \hat z = \begin{pmatrix} - z_0 \\ - z_1 \\ z_2 \end{pmatrix}$$, $$i \hat z = \begin{pmatrix} - z_1 \\ z_0 \\ z_2 \end{pmatrix}$$, $$\bar \hat z = \begin{pmatrix} z_0 \\ - z_1 \\ z_2 \end{pmatrix}$$, $${\hat z} ^{-1} = \begin{pmatrix} z_0 \\ z_1 \\ - z_2 \end{pmatrix}$$

Circles
A circle on &Gamma; may be defined by the combination of a unit vector v normal to the plane of the circle, and a distance d along that vector where the center of the circle lies. The radius of the circle (in 3-space) will be &radic; (1-d&sup2;). Note that this is not, in general, the radus of the circle when transformed onto the complex plane, nor is the centre of the circle (where v intersects &Gamma;) the centre of the circle on the complex plane.

Using the expression for a circle on User:Pmurray_bigpond.com/Geometry of Complex Numbers:


 * $$A z \bar z + B z + \bar B \bar z + D = 0$$, $$\mathfrak C =

\begin{pmatrix} A & B \\ C & D \end{pmatrix} = \bar \mathfrak C^T $$

A circle defined by v and d on &Gamma; will take the form:


 * todo: work this out

geodesics
A point to note is that geodesics on the sphere are always great circles. That is, they lie on planes passing through the center. Thus (&alpha;, &beta;, &gamma;) defines a set of points &alpha; z0 + &beta; z1 + &gamma; z2 = 0 unique up to multiplication by a constant &lambda; which are geodesic on &Gamma;. Projecting this back onto the complex plane:


 * todo: solve the above an convert to our form for the equation of a circle

Möbius transformations
The whole point of this exercise is that Möbius transformations are particularly pretty on a sphere mapped in this manner.

geometry
A general möbius transformation has two fixed points &gamma;1 and &gamma;2 on &Gamma;. There will be a line L1 between these two points.


 * $$L1 = \lambda \hat {\gamma_1} + (1 - \lambda) \hat {\gamma_2}$$

These two points will also each define a plane tangent to the sphere at each point. These two planes will intersect in a line L2.


 * todo - derive L2

Note that if the two fixed points are diametrically opposite, then L2 will be at infinity. L1 and L2 are always perpendicular. They will be skew lines, except at the limit when &gamma;1 and &gamma;2 are the same point - a "parabolic" transformation.

L1 and L2 each define a "sheaf" of planes S1 and S2 which intersect the lines radially. If L2 is at infinity, then this S2 will be a set of parallel planes orthogonal to L1.


 * todo - derive S1 and S2

Each plane in S1 and a subset of the planes in S2 intersect &Gamma; in a "pencil" of circles P1 and P2.


 * todo - derive P1 and P2

Each circle in P1 is orthogonal to each circle in P2. This, P1 and P2 are orthogonal pencils.

I suspect that the stuff above may be totally wrong, and that the pencils will be the intersection of a pebncil of spheres with the unit sphere.

mobius
The Möbius transformation $$\mathfrak H$$ can be derived from the fixed points &gamma;1, &gamma;2, and a characteristic constant k. If $$\mathfrak H$$ is an "elliptical" transformation, then it will transform each circle in P1 onto itself and each circle in P2 onto some other one. If it is "hyperbolic", the converse is true.

If $$\mathfrak H$$ is "loxodromic", then the lokodrome traced out by a point under continuous iteration of $$\mathfrak H$$ will join &gamma;1 and &gamma;2 in a spiral having some angle &theta; to all cicles in P1 and &pi;/2-&theta; to all circles in P2.