Square roots of $j$ and $ε$

Question

I know how to find the square root of the imaginary unit $i$, but I'm still learning about split-complex and dual numbers. I can't find any info anywhere about the square roots of $j$ and $ε$, if they have them.

Answer 1

There are no square roots for $j$. Say that $(a+bj)^2 = j$. Then $$a^2+2abj+b^2=j$$leads to $a^2+b^2 = 0$ and $2ab=1$, which has no solution. Similarly, if $(a+b\epsilon)^2=\epsilon$, then $$a^2+2ab\epsilon = \epsilon$$leads to $a^2 = 0$ and $2ab=1$, which has no solution.

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This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if $\alpha$ and $\beta$ are real numbers, the behavior of the set $$\Bbb C_{\alpha,\beta} = \{a+b\mathfrak{u} \mid a,b \in \Bbb R\mbox{ and }\mathfrak{u}^2=\alpha+\beta\mathfrak{u}\},$$equipped with the obvious operations, can be controled by the discriminant $\Delta = \beta^2+4\alpha$.

• If $\Delta < 0$, then $\Bbb C_{\alpha,\beta}$ is a field.
• If $\Delta=0$, the zero divisors of $\Bbb C_{\alpha,\beta}$ are precisely the elements $a+b\mathfrak{u}$ with $a+\beta b/2=0$, while the others have inverses.
• If $\Delta>0$, the zero divisors are precisely the elements $a+b\mathfrak{u}$ such that $$a+(\beta+\sqrt{\Delta})b/2 = 0 \quad\mbox{or}\quad a+(\beta-\sqrt{\Delta})b/2=0,$$while the others have inverses.

Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!

Answer 2

Square root of $j$ is not a split-complex or complex number, it is a tessarine, a combination of complex and split-complex numbers. Actually, there are 4 roots, according the fundamental theorem of tessarine algebra:

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

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

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

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

You can square these expressions by hand so to verify the roots.