How do you explain intuitively that in split-complex numbers, $0^{\frac12+\frac{j}2}=\frac12-\frac{j}2$ and $0^{\frac12-\frac{j}2}=\frac12+\frac{j}2$?
Usually zero to any power is zero, except when the power is zero itself, or otherwise cannot be defined at all.
But in split-complex numbers, when we rise zero to a power of some zero divisors, we obtain other zero divisors. But this does not work with all zero divisors, for instance, $0^{\frac{j}2-\frac12}$ would be infinite.
Is there an intuitive, for instance, geometric explanation for the equality in the title?
Note that for real $x$, we have: $$e^x\cosh x-e^x\sinh x=e^x\left(\dfrac{e^x+e^{-x}}2-\dfrac{e^x-e^{-x}}2\right)=e^xe^{-x}=1\tag{$\star$}$$
Now let $a$ be a positive real number, and set $t=(\ln a)/2$ for convenience. We have \begin{align} &\phantom{=,,}a^{(1\pm j)/2}\\&=\left(\exp(2t)\right)^{(1\pm j)/2}\\&=\exp\left(t\pm jt\right)\\&=\exp(t)\exp(\pm jt)\\&=\exp(t)\left(\cosh(t)\pm j\sinh(t)\right)\\&=\exp(t)\sinh(t)+1\pm j\exp(t)\sinh(t)\text{ by }(\star)\\&=1+\exp(t)\sinh(t)(1\pm j)\text{.}\end{align}
Geometrically, this means that for positive $a$, $a^{(1\pm j)/2}$ always lies on the line $y=\pm(x-1)$ (parallel to the "null cone").
In the limit as $a$ approaches $0$ from above, we have $t\to -\infty$. But $e^{t}\sinh t=\dfrac{e^{2t}-1}2\to-1/2$, so that $a^{(1\pm j)/2}\to 1-(1\pm j)/2=(1\mp j)/2$. This is a good reason to define $0^{(1\pm j)/2}=(1\mp j)/2$.