Symbolic definition for $i$? Please help!
I am trying to find a definition for $i$ that doesn't work for $-i$.
let $j$ be either $i$ or $-i$.

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*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?

It sounds like you have picked up on the fact that conjugation $$f: a+bi \mapsto a-bi$$ is a ring isomorphism where $i$ and $-i$ correspond to each other.

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*Addition preserving: $f((a+bi)+(c+di))=f((a+c)+(b+d)i)=(a+c)-(b+d)i=(a-bi)+(c-di)=f(a+bi)+f(c+di)$

*Multiplication preserving: $f((a+bi)(c+di))=f((ac-bd)+(ad+bc)i)=(ac-bd)-(ad+bc)i=(a-bi)(c-di)=f(a+bi)f(c+di)$

*$f(1)=1$
Conjugation is clearly bijective too.
