TITLE: Symbolic definition for $i$?
QUESTION [5 upvotes]: Please help!
I am trying to find a definition for $i$ that doesn't work for $-i$.
let $j$ be either $i$ or $-i$.

saying $j^2=-1$ doesn't help since $(i)^2=-1$ and $(-i)^2 = -1$
saying $j=\sqrt{-1}$ doesn't halp since $(-1)^{\frac{1}{2}}$ is multivalued $i$ and $-i$
saying $j=e^{i\pi/2}$ doesnt help since $e^{i\pi/2}=i$ but $e^{(-i)\pi/2}=-i$ too (yeah I know I said j=$e^{i\pi/2}$ and not $j=e^{j\pi/2}$ I did that to hopefully make it more intuitive)
saying $j=ln(i)/(\pi/2)$ doesn't work since $i=ln(i)/(\pi/2)$ and $-i=ln(-i)/(pi/2)$ (check with a phase plotter, ln is multivalued but includes i and -i in their respective functions shown above http://davidbau.com/conformal/#log(z)%2F(pi%2F2)-z)
saying $Im(j) > 0$ doesn't work because $Im(z)=Re(z/i)$ and $Re(i/i)>0$ but $Re((-i)/(-i))>0$

Is this parity some sort of law? And yet $i\neq-i$ since $i=(-1)\cdot(-i)$. Unless...
Post Scriptum: Irionically, it is very easy to symbolically represent $1$ vs $-1$ even though $(1)^2=1$ and $(-1)^2=1$, we can just say $x$ such that $x=x^2$
EDIT: can quaternions help? Is $i$ a set of 2 numbers? Are they always equal?

REPLY [2 votes]: Interestingly, a similar problem exists in dual numbers. Algebraically one cannot distinguish between $\varepsilon$ and $-\varepsilon$. But it is possible to augment their definition analytically so to distinguish them.
In dual numbers there is a common equality: for differentiable at $x=a$ function $f(x)$, $f(a+b\varepsilon)=f(a)+b\varepsilon f'(a)$.
Now, one can define that if at point $x=a$ $f(x)$ has right and left derivatives $f'_r(a)$ and $f'_l(a)$, and they are not equal, then
$f(a+\varepsilon)=f(a)+\varepsilon f'_r(a)$ and $f(a-\varepsilon)=f(a)-\varepsilon f'_l(a)$. In other words, $\varepsilon$ is defined as a positive infinitesimal, and $-\varepsilon$ is defined as negative infinitesimal. This provides an optional analytic definition distinguishing $\varepsilon$ from $-\varepsilon$, but the algebraic structure can work just well without such additional analytic property (but with it one can evaluate more functions at more dual numbers).
Still, in dual numbers one cannot distinguish $\varepsilon$ from $a \varepsilon$ when $a>0$ even with this analytic addition.