# A simple theorem on the zero’s of polynomials on the space of 3D complex numbers.

In this post we look in detail at a very simple yet important polynomial namely

p(X) = X (X – 1).

Why does it have four zero’s in the space of 3D complex numbers? Well if you solve for the zero’s of p so try to solve p(X) = 0, that is you are looking for all numbers such that X^2 = X.
These numbers are their own square, on the real line or on the complex plane there are only two numbers that are their own square namely 0 and 1.
On the space of 3D complex numbers we also have an exponential circle and the midpoint of that circle is the famous number alpha. It is a cakewalk to calculate that alpha is it’s own square just like (1 – alpha).

This post is four pictures long in the size 550×825 pixels so it is not such a long read this time. In case you are not familiar with this number alpha, use the search function on this website and search for the post “Seven properties of the number alpha”. Of course since it is math you will also need a few days time of thinking the stuff out, after all the human brain is not very good at mathematics…
Well have fun reading it.

The last crazy calculation shows that a polynomial in it’s factor representation is not unique. Those zero’s at zeta one and zeta two are clearly different from 0 and 1 but they give rise to the same polynomial.

At last I want to remark that unlike on the complex plane there is no clean cut way to tell how many zero’s a given polymial will have. On the complex plane it is standard knowledge that an n-degree polynomial always has n roots (although these roots can all have the same value). But on the complex 3D numbers it is more like the situation on the real line. On the real line the polynomial p(x) = x^2 + 1 has no solution just like it has no solution on the space of 3D complex numbers.
That was it for this post, thanks for your attention.

# Is this the most simple proof for a more general version of the theorem of Pythagoras? The inner product proof.

Last week I started thinking a bit about that second example from the pdf of Charles Frohman; the part where he projected a parallelogram on the three coordinate planes. And he gave a short calculation or proof that the sum of squares of the three projected areas is the square of the area of the original object.

In my own pdf I did a similar calculation in three dimensional space but that was with a pyramid or a simplex if you want. You can view that as three projections too although at the time I just calculated the areas and direved that 3D version of Pythagoras.

Within the hour I had a proof that was so amazingly simple that at first I laid it away to wait for another day or for the box for old paper to be recycled. But later I realized you can do this simple proof in all dimensions so although utterly simple it has absolutely some value.
The biggest disadvantage of proving more general versions of the theorem of Pythagoras and say use things like a simplex is that it soon becomes rather technical. And that makes it hard to read, those math formula’s become long can complex and it becomes harder to write it out in a transparant manner. After all you need the technicalities of your math object (say a simplex or a parallelogram) in order to show something is true for that mathematical object or shape.

The very simple proof just skips that all: It works for all shapes as long as they are flat. So it does not matter if in three dimensional real space you do these projections for a triangle, a square, a circle, a circle with an elleptical hole in it and so on and so on. So to focus the mind you can think of 3D space with some plane in it and on that plane is some kind of shape with a finite two dimensional area. If you project that on the three coordinate planes, that is the xy-plane, the yz and xz-plane, it has that Pythagoras kind of relation between the four areas.

I only wrote down the 3D version but you can do this in all dimensions. The only thing you must take in account is that you make your projections along just one coordinate axis. So in the seven dimensional real space you will have 7 of these projections that each are 6 dimensional…

This post is four pictures long, I did not include a picture explaining what those angles alpha and theta are inside one rectangular triangle. Shame on me for being lazy. Have fun reading it.

So all in all we can conclude the next: You can have any shape with a finite area and as long as it is flat it fits in a plane. And if that plane gets projeted on the three coordinate planes, the projected shapes will always obey the three dimensional theorem of Pythagoras.

Ok, thanks for your attention and although this inner product kind of proof is utterly simple, it still has some cute value to it.