What are the uses of split-complex numbers?
The set of Complex numbers can be used in lots of domains like geometry, vectorial calculations, solving equation with no real solution etc. But what are the uses of split-complex number that can't be done with complex numbers? I think you could do the same works in geometry or vectorial calculation in a "split-complex" plane but what advantages gives us to know that j is a solution of the equation $x^2=1$?
What I've thought so far is that using complex numbers and split-complex numbers together, we can have numbers of the form $a+bi+cj$ so everything that can be done in the complex-plan could be extended to 3 dimensional space by adding a split-complex part.
$\endgroup$ 92 Answers
$\begingroup$I think the known uses of split-complex numbers are probably going to be addressed by the Wiki page which MJD linked in the comments above, and other "fan pages" on the internet. So, I wanted to address this question in the post:
But what are the uses of split-complex number that can't be done with complex numbers?
In Clifford algebra (or geometric algebra, as called by a small segment of the population that uses them) these two algebras are used to encode the geometry of $\Bbb R$ under two different geometries.
The long story short is that a bilinear form gives rise to a geometry on a vector space. The "signature" of a real bilinear form determines its basic character, and since there are lots of forms with different signatures, you get different geometries.
The complex numbers study $\Bbb R$ with a bilinear form $B(x,y)=-xy$.
For the split-complex numbers, the bilinear form on $\Bbb R$ is just $B(x,y)=xy$.
The quaternions study $\Bbb R\oplus \Bbb R$ with the bilinear form $B((x_1,x_2),(y_1,y_2))=x_1y_1-x_2y_2$.
$\endgroup$ 1 $\begingroup$What I've thought so far is that using complex numbers and split-complex numbers together
You will get exactly tessarines, a commutative associative algebra. But it is 4-dimensional: $ij$ is a separate unit, similar to complex unit. $i^2=-1$, $j^2=1$, $(ij)^2=-1$.
So, your number will look $a+bi+cj+dij$ or $(a+bi)+j(c+di)$ or $(a+bj)+i(c+dj)$.
$\endgroup$
