Summary
The explanation below seems longer than I would like. I have tried to answer any possible criticism.
Briefly, what I am saying is that the complex numbers can be given put into three classes when compared to another complex number. They can either be smaller, larger or equal to the original complex number. In order to do this we only need to use the existing properties of complex numbers. That is their modulus and argument. The complex numbers are ordered in three dimensions not two.
Order of the Complex Numbers
Until now the Ordering of Complex Numbers has been considered impossible. If I want to establish a mechanism and system for
complex vectors I need to overcome this problem too. The real numbers are order in two dimensions but to order the complex numbers we need to place then in three dimensions (see image below). Our aim to to establish which complex numbers are larger, smaller or equal to any given complex number.
Consider first the real component of z = x + iy. Values of x belong to the real set and can be ordered. So we have a subset
of complex numbers which can be ordered by their real components. That is 3x + iy is smaller than 4x + iy in my new
interpretation of complex mathematics. The reason I give is that complex numbers are directional.
Now consider next the imaginary component of z = x + iy. Valuesof y belong to the real set and can be ordered. So we have
another subset of complex numbers which can be ordered by their imaginary components. That is x + 3iy is smaller than x + 4iy.
Next suppose we keep the modulus of the complex numbers constant (assume it has no effect) and the order the complex set of
numbers solely by their modulus (a bigger modulus gives a bigger complex number in this subset) and changing only their argument.
Changing their argument has no effect on the size or order of numbers in this set. See figure 5-^2. (sorry this is an old file
from year 2000 or 8 years ago I have not got all my data intact but I think you can still follow my argument)
If we draw a locus of points where these complex numbers are equal we get a circle.
The interior of the circle represents complex numbers which are smaller and the exterior of the circle represents complex
numbers which are larger.
So we are beginning to see that the complex set of numbers, supposedly random, have a large degree of order. Let us carry on.
So note that in this subset of complex numbers we ordered them by their modulus which is given by ™ (x^2 + y^2) and the circle
(given by the equation z^2 = x^2 + y^2).
For our next subset of complex numbers I am going to order them by their argument alone and changing only their modulus. So
changing their modulus will not effect their order now (a bigger argument gives a bigger complex number in this subset) . See
figure below. Note the argument is given by tan q = y/x and the modulus by square root of (x^2 + y^2).
In fact it is important to note that this relationship is also dependant upon the perimeter (and half of it, too) of the
enclosing rectangle shown shaded.
If we take values of x and y which give a perimeter of 4 (that is a half perimeter of given by ^2 = x+y) We get a locus of equal
points shown in figure below by the square or diamond shape.
Now there is only one option left by which we can order complexnumbers and that is by both their argument alone. See figure 1
below. The numbers in this subset can be compared and ordered by resolving them into their component parts. If we are to order
them in any particular direction then we resolve them in the appropriate directions according to vector mechanics. So if we
have two complex numbers in this subset then we are interested in their particular effect in a particular direction. So we
resolve the numbers in that direction of interest to see if one is larger than the other.
Now comes the pulling together of the theory! There is no random element left in the set of complex numbers!
The area of the circle (pi *r^2) is always given by a real number.So this subset represents numbers on the interior surface of the
complex plane. That is the adjacent surface to our 3D!
The area of the square (measured in complex i units on one side and real numbers on the other side) is always given by an
imaginary number. So this subset represents numbers or the exterior surface of the complex plane. That is the adjacent
surface to the 4D!
So that leaves the numbers resolved by vectors alone as the surface shape distorts so much. This subset defines the space and numbers between the two surfaces. That is this subset of complex numbers describes complex space. I suggest that you can actually use this information to map the
interior of a Black Hole.
So to recap: from the 3D interior to the 4D exterior we use the properties of 3 complex numbers. That is the modulus, the
argument, then finally the modulus AND the argument depending upon your vantage point. The ordering in 4D is equivalen to that in 3D in that the 4D numbers around a square are all ordered as equal (although their argument does have the advantage of being able to trace them if they translate along the zunnel) just as they wre for the circle in 3D.
This matches perfectly the behaviour of the spatial geometries. In 3D space the subset which is defined by the modulus is
present, while in complex space the subset which is defined by the argument is present, and finally in 4D space the subset
needs to be defined by both the modulus and the argument. So we have a match with the spherical and cubical geometries of 3D and
4D space.
The reason why i^2 = -1 is therefore shown to be because this represents a transition from the interior to the exterior
surfaces of the complex plane.
The reason how i^2 = -1 is therefore shown to be in the ordering of the three complex number subsets comprising the interior and
the exterior surfaces of the complex plane; and the subset representing the complex space between these two surfaces.
