Symbolic definition for $i$?

Question

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?

Answer 1

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.

• 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.

Answer 2

An idea: using the "identification" (in fact isomorphism, but making it a simple identification) of $\;\Bbb C\;$ with $\;\Bbb R^2\;$, we have that $\;z=x+iy \stackrel{\text{Ident.}}\sim (x,y)\;$, and thus

$$i\sim(0,1)\;,\;\;\text{whereas}\;\;-i\sim (0,-1)$$

and that's one way (a rather simple and algebraic one) to distinguish completely $\;i\;$ from $\;-i\;$ .

Answer 3

As @DanielFischer commented, $\mathbb C$ has the complex-conjugation automorphism that interchanges the two square roots of $-1$ (however we decide to label them). Thus, any algebraic relation (with real coefficients) that holds for one holds for the other. In different words, $\mathbb R(i)$ is constructed algebraically as $\mathbb R[x]/\langle x^2+1\rangle$, and the image of $x$ is a canonical square root of $-1$ in that model. (But/and, also, the image of $-x$ is another.) But/and I think this is not the type of distinction you want.

It gets worse in the Hamiltonian quaternions: there are infinitely-many square roots of $-1$, and they are all conjugate to each other in the quaternions.

Yes, if we choose to represent complex numbers as the real plane, we can label/name the square root of $-1$ that is in the upper half-plane "$i$". But, as in my first remark, flipping/interchanging upper and lower half-planes is an isomorphism of $\mathbb C$ to itself, and it is essentially impossible to distinguish... so this hasn't really accomplished anything.

For contrast, the case of $\sqrt{2}$ is somewhat different. It is the same, in the sense that the field $\mathbb Q[x]/\langle x^2-2\rangle$ is abstractly made by adjoining a square root of $2$ to $\mathbb Q$, because the image of $x$ in the quotient is a sort of canonical $\sqrt{2}$ in that model. BUT the distinction that matters in practice is that $\mathbb Q(\sqrt{2})$ admits two different imbeddings into $\mathbb R$, and in one the abstract square root of $2$ goes to $1.414...$ while in the other it is the negative of that. The "standard/canonical" square root of $2$ we usually refer to is actually the real number $1.414...$, rather than an abstract one.

Answer 4

You could construct the field of complex numbers $\mathbb{C}$ as the real vector space $\mathbb{R}^2$ with an operation $$ (a_1,b_1)\cdot (a_2,b_2) = (a_1a_2 - b_1b_2, a_1b_2 + a_2b_1). $$ It can "easily" be seen that $(\mathbb{R}^2,\cdot)$ is a field and has the same properties as the $\mathbb{C}$ you know and love. Now you could define the imaginary unit as $$ i := (0,1). $$ Then $i$ and $(1,0)$ form a basis of $\mathbb{R}^2$ and we can write each element $$ (a,b) = a + bi $$ by abuse of notation (by identifying $a=(1,0)a$).

Answer 5

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.