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From https://en.wikipedia.org/wiki/Imaginary_unit:

The imaginary unit or unit imaginary number ($i$) is a solution to the quadratic equation $x^2+1=0$.

There are two complex square roots of $−1$:
$i$ and $-i$, just as there are two complex square roots of every real number other than zero (which has one double square root).

How can $i_1:=i$ and $i_2:=-i$ be distinguished?
It seems to me that there is no way to distinguish between $i_1$ and $i_2$ because $i_1$ is defined as a solution to the equation $x^2+1=0$ and $i_2$ is defined as the other solution to the equation $x^2+1=0$.

If we cannot distinguish between $i_1$ and $i_2$, then we also cannot distinguish between $a+bi_1$ and $a+bi_2$.

So, I think all the properties of $a+bi_1$ and $a+bi_2$ are the same.
Am I right?

佐武五郎
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    I'm pretty sure I remember reading in a textbook at some point that $-i$ and $i$ are absolutely the same, in the sense that if you swapped all of them everywhere in all of math theory and results, it would have absolutely no effect. – Trevor Dec 14 '23 at 18:44
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    Kinda. But $a+bi_1$ and $a+bi_2$ have imaginary parts with different signs (unless $b =0$). – JonathanZ Dec 14 '23 at 18:45
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    $i$ and $-i$ are different numbers which do not have identical properties , for example $z^3+i=0$ has root $i$ , but not root $-i$. We can consider one of the numbers as the "principal root of -1". – Peter Dec 14 '23 at 18:47
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    @Peter: I don't know what you were thinking, but $z^3+1$ does not have root $i$ – J. W. Tanner Dec 14 '23 at 18:48
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    @J.W.Tanner Was a typo. Fixed. – Peter Dec 14 '23 at 18:49
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    By convention $i$ corresponds to $(0,1)$ while $-i$ corresponds to $(0,-1)$, but $\mathbb R[i]$ is isomorphic to $\mathbb R[-i]$ – J. W. Tanner Dec 14 '23 at 18:49
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    @Peter True, but you’re already using $i$ in your equation. If you are doing that you might as well say $i$ satisfies $z = i$ but $-i$ does not. This hardly means there’s any actual differences between $i$ and $-i$. And the choice of principal root of -1 is also completely arbitrary between $i$ and $-i$. – David Gao Dec 14 '23 at 19:09
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    Nevertheless , $-i$ and $i$ are not the same numbers ! – Peter Dec 14 '23 at 19:12
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    Okay, but apart from the fact that they're not the same number, they're identical. – Trevor Dec 14 '23 at 19:17
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    For what is worth I don’t consider there to be any actual difference. Taking complex conjugate (that is, swapping $i$ and $-i$) is an automorphism of the field $\mathbb{C}$ fixing $\mathbb{R}$, so there isn’t much of a difference algebraically. It’s also a homeomorphism (in fact isometric under the usual Euclidean metric), so there’s no difference topologically. Analytically, if $f(z)$ is differentiable at some point $x$, then $\overline{f(\bar{z})}$ is differentiable at $\bar{z}$ with the derivative the complex conjugate of the original one, so not much difference analytically either. – David Gao Dec 14 '23 at 19:17
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    https://math.stackexchange.com/questions/13801/whats-the-thing-with-sqrt-1-i – mr_e_man Dec 14 '23 at 19:18
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    "apart from the fact that they're not the same number, they're identical." a strange formulation. The formulation with the isomorphism hits the nail on the head much better. – Peter Dec 14 '23 at 19:19
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    @Peter Not disputing the fact that they are not the same. I think the OP is also aware of that. This seems to be more about whether taking complex conjugates in all of mathematics would cause anything to break down. It does not. – David Gao Dec 14 '23 at 19:21
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    @DavidGao OK , formulated this way , it makes sense. – Peter Dec 14 '23 at 19:22
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    Admittedly this seems to hinge on what does the OP mean by “all properties”. Since they are different numbers, there are “properties” in which they are different. But all such “properties” effectively need a choice of the principal root of -1 between $i$ and $-i$ beforehand, as the isomorphism argument demonstrates. So are these parts of “all properties” in the OP’s mind? – David Gao Dec 14 '23 at 19:24
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    Of course the mp $z \mapsto \overline{z}$ is an isomorphism of fields; so (using only field properties) we cannot distinguish $i$ from $-i$. What about other (non-field) properties? If you have "orientation" of the plane, then you can distinguish $i$ and $-i$. That is, $z \mapsto \overline{z}$ does not preserve orientation. – GEdgar Dec 14 '23 at 20:27

2 Answers2

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The fact that these are, in some sense, equivalent, leads to Galois theory.

Replacing $ \mathrm i $ with $ - \mathrm i $ throughout amounts to replacing every complex number $ a + b \mathrm i $ with its complex conjugate $ a - b \mathrm i $. We can think of this as a function $ \phi $ from $ \mathbb C $ (the set of complex numbers) to $ \mathbb C $, so that $ \phi ( a + b \mathrm i ) = a - b \mathrm i $. Here are several interesting properties of this function $ \phi $:

  1. $ \phi $ is defined everywhere (on $ \mathbb C $); that is, $ \phi ( x ) $ exists for any complex number $ x $;
  2. $ \phi $ is well-defined; that is, given a complex number $ x $, there is only one value of $ \phi ( x ) $;
  3. $ \phi $ is one-to-one; that is, the only way you can get $ \phi ( x ) = \phi ( y ) $ is when $ x = y $;
  4. $ \phi $ is onto $ \mathbb C $; that is, every complex number $ y $ arises as $ y = \phi ( x ) $ for some complex number $ x $;
  5. $ \phi $ preserves addition; that is, $ \phi ( x + y ) = \phi ( x ) + \phi ( y ) $;
  6. $ \phi $ preserves multiplication; that is, $ \phi ( x y ) = \phi ( x ) \phi ( y ) $;
  7. $ \phi $ fixes $ \mathbb R $, the set of real numbers; that is, $ \phi ( x ) = x $ if the complex number $ x $ happens to be a real number.

Properties (1) and (2) are necessary for $ \phi $ to count as a function on $ \mathbb C $ in modern mathematics. Sometimes in elementary mathematics we consider partial functions that are sometimes undefined (such as the reciprocal) or multi-valued functions that can take more than one value at a single argument (such as square roots), so I mention these to make clear that that's not allowed. Adding properties (3) and (4) makes this function a permutation of $ \mathbb C $ (or a bijection from $ \mathbb C $ to itself). Adding properties (5) and (6) makes this permutation a field automorphism of $ \mathbb C $ (or just an automorphism if you know that you're talking about fields, which are algebraic structures with addition and multiplication satisfying their familiar laws, such as commutativity and the existence of inverses). Adding property (7) makes this automorphism an automorphism over the subfield $ \mathbb R $.

Now if you have any statement about complex numbers, as long as it can be stated with reference only to real numbers and the operations of addition and multiplication, it can only be true of a complex number $ x $ when it's also true of $ \phi ( x ) $, and vice versa. For example, if $ x ^ 2 + 1 = 0 $, then $ \phi ( x ) ^ 2 + 1 = 0 $ too. Or for a more complicated example, if there exist exactly two cube roots of $ x ^ 2 + 3 y $ with a positive real part, then there exist exactly two cube roots of $ \phi ( x ) ^ 2 + 3 \phi ( y ) $ with a positive real part. (This works because you can say that $ z $ is a cube root of $ w $ with reference to multiplication, using the equation $ z ^ 3 = w $; and you can define the real part using addition, multiplication, and real numbers, because $ z $ is the real part of $ w $ if and only if $ z $ is real and $ ( w - z ) ^ 2 $ is zero or a negative real number.)

This doesn't mean that any statement whatsoever is invariant under $ \phi $ like this. If I want to say that $ x $ has a positive imaginary part, then this won't be preserved by $ \phi $; in fact, if $ x $ has a positive imaginary part, then $ \phi ( x ) $ must have a negative imaginary part. But you can't define the imaginary part using only addition, multiplication, and real numbers. (If you try to copy what I did above for the real part, you can get as far as $ \mathrm i \Im x = x - \Re x $, where $ \Re x $ is the real part of $ x $ and $ \Im x $ is the imaginary part, but you still need to divide this by $ \mathrm i $ to get $ \Im x $ itself, and we have no way to distinguish $ \mathrm i $ from $ - \mathrm i $, as you know.)

In Galois theory, we generalize this massively. We consider a situation where we have any field $ L $ and any field $ K $ contained within $ L $ (so that $ K $ is a subfield of $ L $ and $ L $ is a field extension of $ K $). And then we ask if there are any automorphisms of $ L $ over $ K $. (There's technically always at least one: the trivial automorphism given by $ \phi ( x ) = x $. So the question is whether there are any others.) We also consider how some of these automorphisms may be obtained by composing others (doing one after the other) or taking an inverse (running the function backwards). Altogether, this gives us $ \operatorname { Aut } _ K L $, the automorphism group of $ L $ over $ K $. In the case of $ \mathbb C $ as an extension of $ \mathbb R $, there's not much to this group: $ \operatorname { Aut } _ { \mathbb R } \mathbb C $ has only two elements (the trivial automorphism and complex conjugation), each of which is its own inverse. But in other examples, there can be many more.

And whenever $ \phi $ is an automorphism of $ L $ over $ K $, we know that any statement about elements of $ L $, if it refers only to addition, multiplication, and $ K $, must be true about $ x $ if and only if it's also true about $ \phi ( x ) $. So in this way, $ x $ and $ \phi ( x ) $ are equivalent, just as $ \mathrm i $ and $ - \mathrm i $ are in some sense equivalent.

Toby Bartels
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Yes, they are indistinguishable in the sense that I describe below (which, I believe, is what you have in mind).

The imaginary unit or unit imaginary number ($i$) is a solution to the quadratic equation $x^2 + 1 = 0$.

is more of the intuition of what we expect from this number, and is not a definition of $i$:

  • It is not specified what $+$, $=$, and ${}^2$ are, and what the domain of $x$ is. If they are real-valued, there are no roots. If they are complex-valued, it requires defining complex numbers (and, in particular, $i$) first.
  • It is not clear that such an object exists (in this form, it's more like an axiom, similarly to how we define natural numbers $\mathbb N$). To make it a definition, a specific construction should be presented, as we do for all numbers aside from $\mathbb N$.

There are multiple ways to construct complex numbers, see e.g. here: https://en.wikipedia.org/wiki/Imaginary_unit#Matrices (there is a more elementary way to define $a + bi$ as a pair of real numbers $(a,b) \in \mathbb R^2$ with certain operations, see https://en.wikipedia.org/wiki/Complex_number#Formal_construction).

As the article says, we select $i$ to be a matrix $J$ so that $J^2 = -I$, where $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix. So, a "complex number" $a + bi$ is defined as a matrix $a I + bJ$. As long as you select $J$ like this, all desired properties of complex numbers hold. The "standard" choice is $J_1 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. But $J_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ is another perfectly valid choice, and $J_2 = -J_1$, as you pointed out in the question.

So, there are a lot of arbitrary choices which you can make when constructing complex numbers. After you are done with the construction and fixed $i$, the objects $i$ and $-i$ are different.

Dmitry
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