# Symbolic definition for $i$?

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?

• The two solutions to $i^2=-1$ are not equal: if you add one to the other you get $0$ rather than twice one of them. But they do have the same properties as each other, as Favst points out in an answer. Quaternions would not help and face a similar issue in identification. Oct 29, 2020 at 17:45
• Algebraically, $\pm i$ are indistinguishable.
– anon
Oct 30, 2020 at 7:46
• Does this count? $\ln(-1)=i\pi$, not $-i\pi$? This is a function that behaves differently for $i$ and $-i$ (by convention). May 19, 2022 at 21:46
• @runway44 a question remains, whether they are distinguishable analytically (see my post). May 19, 2022 at 22:48
• @Anixx : This convention requires that you have previously fixed the complex unit. So it would be a circular definition. May 20, 2022 at 6:41

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.

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 $$\;-i\;$$ .

• That's great, but if there is any function that ultimately behaves differently to i then it does to -i? Because then we would want to SURE to map the "right" i to (0, 1) Oct 29, 2020 at 17:10
• how do I know the i you use in x+iy isnt my x+(-i)y if we only say i²=-1. Is i actually a set of 2 numbers? There is always 2 complex planes? Can quaternions fix this? Oct 29, 2020 at 17:12
• I suspect that your identification is slightly circular: you are saying that $i$ is the complex number where the real part is $0$ and the imaginary part is $+1$. But you have to know which of the two solutions to $z^2=-1$ has the positive imaginary part to do the identification. Oct 29, 2020 at 17:14
• Since conjugation is an automorphism of $\mathbb{C}$, one cannot distinguish algebraically between $i$ and $-i$. Oct 29, 2020 at 17:28
• @Henry No, I'm not saying anything like that at all. I'm just identifying $\;i\;$ with the point $\;(0,1)\;$ on the real plane. That you interpret this as it having real part zero and etc. is something that isn't included in this identification. Nothing circular here, as far as I can see..All the rest you write about the solutions to $\;z^2=-1\;$ is irrelevant about what I did and also about what the OP wrote in his question. Oct 29, 2020 at 17:41

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.

• Adding to your comparison with $\mathbb Q(\sqrt2)$: The conjugation $\sqrt2\mapsto-\sqrt2$ is an automorphism of fields, but not an automorphism of ordered fields. $\mathbb Q(\sqrt2)$ has an ordering which is not preserved under conjugation. $\mathbb Q(\mathrm i)$ has no such ordering which lets us distinguish $\mathrm i$ from $-\mathrm i$. Oct 29, 2020 at 20:35
• you lost me there. What is ℚ[x]⟨x² − 2⟩ Oct 30, 2020 at 2:15
• @cmarangu, $\mathbb Q[x]/\langle x^2-2\rangle$ is meant to denote the quotient of the ring \mathbb Q[x]$of polynomials with coefficients in$\mathbb Q$by the ideal generated by$x^2-2$. This way of rigorously and abstractly creating field extensions was initiated by Kronecker in the mid 19th century, to avoid "begging the question" of whether numbers existed "out there" waiting to be "adjoined to" fields. Oct 30, 2020 at 2:36 • @paulgarrett I see. I have heard of this "Ring theory" but have not studiet it yet. thanks for help though Oct 30, 2020 at 14:14 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 $$a=(1,0)a$$). • So basically, note i in ℂ to be homogeneous to the (0, 1) vector in ℝ? and linear combinations of 1 and i (with real scalars) to be linear combinations of (0, 1) and (1, 0)? Dec 21, 2020 at 16:26 • should note - I'm actually used to defining conformal ℝ²→ℝ² transforms in terms of the complex numbers - my question was whether {1, i} homogenius to {1, -i} something like that Dec 21, 2020 at 16:29 • Well yes, it can be helpful to consider$\mathbb{R}^2\to\mathbb{R}^2$transformations as transformations$\mathbb C\to\mathbb C$, but$\mathbb C$can be defined in the way I presented. I'm not quite sure what you mean by two elements being homogenius to each other? Dec 23, 2020 at 18:46 • Stefan Hante ONE: it is homogenious if it is a homomorphism TWO: it is a homomorphism iff it is a linear map THREE: it is a linear map iff f(a⃗)+f(b⃗) = f(a⃗+b⃗) AND f(ka⃗) = kf(a⃗) Dec 28, 2020 at 8:50 • Of course there is a homomorphism$f\colon \mathbb C \to \mathbb C$such that$f(1)=1$and$f(i)=-i$. It is called the conjugation. Dec 30, 2020 at 17:32 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. • How can right and left derivatives not be equal when the dual numbers are commutative? – anon May 20, 2022 at 0:31 • @runway44 please explain what do you mean. May 20, 2022 at 0:32 • @runway44 yes, left and right limits. When they are equal, they are called just the derivative. May 20, 2022 at 0:35 • @cmarangu as you can see, the formula (any of the above) to be evaluated at$\varepsilon$requires$f(0)$defined. It is the real part of$f(\varepsilon)$. So, your function has to be continuous at$0$to be defined at$\varepsilon$. May 20, 2022 at 19:39 • @cmarangu Right side derivative of a function is$f_r'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$, so$f(x)\$ should be defined for the right-hand-side derivative to exist. May 20, 2022 at 19:45