What is $\sqrt{-1}$ (or $\sqrt{-j}$) in the Hyperbolic (Split-complex) Numbers?

Question

Given a number system such that $j^2 = 1, j \ne \pm 1$, what would be the solution to $z^2 + 1 = 0$? Are the hyperbolic numbers not closed under taking roots unlike the complex numbers?

My assumption would be that there is no hyperbolic solution, because if you have a hyperbolic number $a + bj$, then $(a + bj)^2 = a^2 + b^2 + 2abj$, meaning for there to not be a $j$ component in the result of squaring, either $b$ or $a$ must be $0$. If $b$ is $0$, then there is no solution for the same reason there is no solution for the real numbers, and if $a$ is $0$, there is no solution as $bj$'s multiplication works very similarly to $a$'s.

Answer 1

You are right, there is no solution.

The numbers $a+bj$ described here are actually isomorphous with real symmetric $2×2$ matrices having the form

$$ \begin{pmatrix} a & b \\ b & a \end{pmatrix}. $$

Such matrices have real eigenvalues $a\pm b$. So upon squaring they can give only positive semidefinite matrices, not $-I_2\equiv-1$.

We can say a little more about the squaring operation. Since the squared matrix must be positive semidefinite, having two nonnegative eigenvalues, only $1/2^2=1/4$ of the "split-complex" plane is covered by quantities that are squares. The boundaries of this favored quadrant are defined by the condition where the trace is nonnegative, thus $a\ge0$, and an eigenvalues is zero therefore the determinant is zero, $b=\pm a$. So the boundaries are given by a pair of 45° rays in the $a,b$ plane, starting from the origin and directed into the first and fourth coordinate quadrants. The region of squares, requiring $a\ge|b|$, lies to the right of the pair of 45° rays, a 90° sector overall. Their square roots hog up four square-root values apiece, leaving no square roots for anything in the 270° sector lying outside (including $-1$).

Answer 2

If you add imaginary unit $i$ to split-complex numbers with hyperbolic unit $j$, you will get tessarines (basically the same as bicomplex numbers but with different notation). In tessarines the equation $z^2 + 1 = 0$ has four roots: $z_1=i$,$z_2=-i$,$z_3=ij$,$z_4=-ij$.

The equation $z^2 + j = 0$ also has four solutions:

$z_1=\frac{i}{2}+\frac{1}{2}+j(\frac{i}{2}-\frac{1}{2})$

$z_2=-\frac{i}{2}-\frac{1}{2}-j(\frac{i}{2}-\frac{1}{2})$

$z_3=\frac{i}{2}-\frac{1}{2}+j(\frac{i}{2}+\frac{1}{2})$

$z_4=-\frac{i}{2}+\frac{1}{2}-j(\frac{i}{2}+\frac{1}{2})$

Any polynomial with non-zero-divisor coefficients of power $n$ has $n^2$ roots (this is the main theorem of tessarine/bicomplex algebra).

The both equations have no purely hyperbolic solutions.

The split-complex numbers were initially born as a subset of tessarines and were called "real tessarines" (that is, those which did not involve complex unity).