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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.

Anixx
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William Ryman
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    Are you talking about bicomplex numbers here? Zero divisors exist in the set of bicomplex numbers, so "$ab=0$" does not imply "$a=0$ or $b=0$". – MPW May 25 '22 at 14:28
  • @MPW Oh, okay. So it seems that hyperbolic numbers themselves are not closed under roots, so bicomplex numbers are needed. I was thinking that the logical conclusion if hyperbolic numbers themselves are not closed under roots is there is another number system that combines ordinary complex numbers and hyperbolic numbers. Thanks for the comment. – William Ryman May 25 '22 at 14:33
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    "I was thinking... there is another number system that combines ordinary complex numbers and hyperbolic numbers" There is, e.g. this one that associates but doesn't commute. – J.G. May 25 '22 at 14:46
  • @J.G. the system that combines complex and hyperbolic numbers is called tessarines, it is commutative and associative. – Anixx May 30 '22 at 01:09
  • @Anixx a system. People are welcome to compare multiplication tables. – J.G. May 30 '22 at 05:35
  • @J.G. Well. So, tessarines/bicomplex is when you have two imaginary units and one hyperbolic, split-quaternions is when you have two hyperbolic units and one imaginary. It also has a subring isomorphic to duals, but not commutative. – Anixx May 30 '22 at 09:06

2 Answers2

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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 $-1$).

Oscar Lanzi
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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).

Anixx
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