Suppose now we are trying to explain to students who do not know complex numbers, how do we distinguish $i$ and $-i$ to them? They will object that they both squared to $-1$ and thus they are indistinguishable. Is there a way of explaining this in an elementary way without go into introductory things in complex analysis?

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    $\begingroup$ Roughly speaking you can't. The map that takes $a+bi$ to $a-bi$ is an automorphism of $\mathbb{C}$. $\endgroup$ Commented Aug 1, 2012 at 14:38
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    $\begingroup$ $-1$ and $1$ both square to $1.$ Does this means they're indistinguishable? $\endgroup$
    – user2468
    Commented Aug 1, 2012 at 14:40
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    $\begingroup$ @J.D. Ah - but there is no automorphism taking 1 to -1. $\endgroup$
    – Old John
    Commented Aug 1, 2012 at 14:50
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    $\begingroup$ Does the relationship between $i$ and $-i$ the same with "left" and "right",or "clockwise" and "anticlockwise"? $\endgroup$
    – bigeast
    Commented Aug 1, 2012 at 15:18
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    $\begingroup$ First, $1$ and $-1$ are distinguishable because $1$ is its own square, while $-1$ isn't. From this, you also can easily distinguish $2$ and $-2$ because $2=1+1$ and $-2=(-1)+(-1)$. Indeed, it's not hard to extend this to $\mathbb{Z}$, and then to $\mathbb{Q}$. Finally you get all of $\mathbb{R}$ by considering sequences of rational numbers converging to that real number. Thus for each real number $r$ you can uniquely determine which one of $r$ and $-r$ is positive and which one negative. But this doesn't work for $i$. $\endgroup$
    – celtschk
    Commented Aug 1, 2012 at 17:24

7 Answers 7


Simple: there are two distinct values whose square is $-1$, so you choose one of them and call it $i$. The other one, then, is $-i$. It doesn't matter which of them you choose, just as it doesn't matter which direction in space you call $x$.

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    $\begingroup$ Upvoted because OP asked for "explaining this in an elementary way." $\endgroup$
    – user2468
    Commented Aug 1, 2012 at 14:47
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    $\begingroup$ With the proviso that, once you've made your decision, you must stick with it thereafter. $\endgroup$ Commented Aug 1, 2012 at 19:14
  • $\begingroup$ There are two distinct values whose square is +1, but it does matter which we call 1, and which we call −1. $\endgroup$ Commented Oct 24, 2018 at 13:46
  • $\begingroup$ @I.J.Kennedy: That's true, but what is your point? $\endgroup$
    – celtschk
    Commented Oct 27, 2018 at 6:27

If you construct $\Bbb C$ as $\Bbb R[X]/(X^2+1)$ the roots are indistinguishable. You just choose one and identify it with the point $(0,1)$ in the Gauss-Argand plane. That chosen one will be $i$.

If, on the other hand you construct $\Bbb C$ as the set of pairs $(a,b)\in\Bbb R^2$ with suitable addition and multiplication, then $\pm i=(0,\pm1)$.

The two constructions are of course isomorphic but not canonically isomorphic, due to the arbitrarity of the choice of a root which is the effect of existence of a trivial $\Bbb R$-automorphism, namely complex conjugation.

  • $\begingroup$ You mean non-trivial $\mathbb{C}$-automorphism instead of trivial $\mathbb{R}$-automorphism, right? Or do you mean one which is trivial on $\mathbb{R}$, though non-trivial on $\mathbb{C}$? $\endgroup$ Commented Aug 1, 2012 at 15:02
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    $\begingroup$ @NilsMatthes : No, I mean $\Bbb R$-automorphism! If $F\subset L$ are fields, a $F$-automorphism (of $L$) is an automorphism that leaves the elements of $F$ fixed. $\endgroup$ Commented Aug 1, 2012 at 15:22
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    $\begingroup$ If you construct $\mathbb{C}$ as $\mathbb{R}[X]/(X^2+1)$, then $i$ is the coset $X+(X^2+1) \mathbb{R}[X]$ and $-i$ is the coset $-X+(X^2+1) \mathbb{R}[X]$. $\endgroup$ Commented Aug 1, 2012 at 16:08
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    $\begingroup$ If you construct $\mathbb{C}$ as pairs, then there's initially no $i$,. but only the pairs $(0,1)$ and $(0,-1)$. Therefore you still have the freedom which of them you call $i$. $\endgroup$
    – celtschk
    Commented Aug 1, 2012 at 17:33
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    $\begingroup$ "If you construct $\,\Bbb C\,$ as $\,\Bbb R[X]/(X 2 +1)\,$ the roots are indistinguishable" ---This isn't right and David Speyer already gave a reason, but we could also argue that $\,i\,,\,-i\,$ would be indistinguishable iff, in some algebraic structure, we'd have $\,i=-i\Longrightarrow 2i=0\,$ , and since the construction $\,\Bbb R[x]/(x^2+1)\,$ is a field of characteristic $\,\neq 2\,$ the above is impossible. $\endgroup$
    – DonAntonio
    Commented Aug 2, 2012 at 1:36

If you construct $\mathbb{C}$ purely algebraically then there is no distinction until you've chosen what $i$ is, and then the distinction is made by the observation that if $x=i$ is one solution to $x^2=-1$ then $x=-i$ is the other.

This is because we construct $\mathbb{C}$ algebraically as the field $\mathbb{R}(i)$ obtained by adjoining $i$ to $\mathbb{R}$, where $i$ is an abstract root of the polynomial $x^2+1$. But once you've done this, you can declare that $i$ is the point on the Argand diagram lying at unit length 'above' $0$, and this distinguishes it from $-i$. As André Nicolas says in the comments, if you'd chosen $-i$ instead of $i$ then it doesn't matter, since the (unique!) map fixing $\mathbb{R}$ and sending $i \mapsto -i$ is a $\mathbb{R}$-automorphism of $\mathbb{C}$. (Intuitively: choosing the other root changes nothing except labelling.)

In more human terms, you say $i$ is 'something' which squares to give $-1$, and then $-i$ is 'the other thing' which does the same.

We can then make the translation to the Argand diagram by identifying this new field $\mathbb{C}=\mathbb{R}(i)$ with the plane $\mathbb{R}^2$ by using the following correspondence: $$\mathbb{C} \ni a+bi \leftrightarrow (a,b) \in \mathbb{R}^2$$ with operations $$(a,b)+(c,d)=(a+c,b+d)$$ $$(a,b)\cdot(c,d)=(ac-bd, ad+bc)$$


It doesn't matter which one you call $i$; whichever it is, the other one is called $-i$.

I suspect it's not necessary to say more than that. And in a way, to say much more in answer to that question might amount to making the answer seem more complicated than it really is. That doesn't mean you shouldn't say more after saying that; e.g. that the field $\mathbb{C}$ has exactly one automorphism that's not hideously ill-behaved, etc.


First off, you don't need to know anything about complex analysis in order to understand this. Complex analysis refers to doing calculus with complex numbers, but this is just a question about the basic complex-number system itself.

By way of analogy, let's think for a moment about the way the ancient Greeks conceived of ordinary numbers. Euclid's Elements is actually not just about what is taught today as geometry; it's full of material that today would be considered to be algebra and number theory. For example, what we would represent as multiplication $xy$, Euclid would represent as the area of a rectangle with sides of certain lengths. An identity like $x(y+z)=xy+xz$ would be represented as cutting a rectangle into two smaller rectangles. In Euclid's system, there is no such thing as the real number system, and in fact there is not even a number 1. If I say I'm 6 feet tall, in Euclid's language we would say that we have two different line segments, and their ratio is 6. The small one is arbitrarily chosen as the unit of measurement. But the choice of the unit segment is completely arbitrary, and it can be different if you finish one calculation and start another one.

Similarly, it's completely arbitrary how to label $i$ versus $-i$. For example, a typical application of complex numbers is to describe oscillations. Suppose two cars' windshield wipers are both running. They have the same frequency, say 1 Hz, but they're out of phase with one another by 90 degrees. We can represent the two oscillations as complex numbers whose magnitudes represent the frequency and whose arguments represent the phase. The magnitudes both have to be 1, but it's completely arbitrary whether we use a bigger argument to represent an oscillation with a leading phase, or a lagging one. We have to make an arbitrary choice. There is nothing obliging us to be consistent about this choice from one calculation to the next.

So just as Euclid's system "doesn't care" which length we consider a unit, the complex number system "doesn't care" which root of -1 we call $i$ and which $-i$.

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    $\begingroup$ area of a triangle -- did you mean rectangle? $\endgroup$
    – sarnold
    Commented Aug 1, 2012 at 20:40
  • $\begingroup$ @sarnold: Thanks, fixed. $\endgroup$
    – user13618
    Commented Aug 2, 2012 at 3:46

This question relates directly to issues that are often discussed in the philosophy of mathematics.

According to one of the standard accounts of structuralism, what mathematical objects are at bottom are the structural roles that they play within a mathematical system. One might define the equivalence relation whereby element $a$ in structure $A$ is equivalent to element $b$ in structure $B$ exactly when there is an isomorphism of $A$ to $B$ taking $a$ to $b$. This results in what I have called the isomorphism orbit of the object $a$ in $A$.

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According to some accounts of the philosophy of abstract structuralism, what a mathematical object is, is the structural role that it plays in abstract mathematical structures.

The problem for this view arises with nonrigid mathematical structures such as the complex field $\newcommand\C{\mathbb{C}}\C$.

Here is how I describe the situation is my book, Lectures on the Philosopy of Mathematics (MIT Press)

Although one conventionally describes $i$ as "the square root of negative one," nevertheless one might reply to this, ``Which one?'' in light of the fact that $\newcommand\unaryminus{-}\unaryminus i$ also is such a root: $$(\unaryminus i)^2=(\unaryminus 1\cdot i)^2=(\unaryminus 1)^2i^2=i^2=\unaryminus 1.$$ Indeed, the complex numbers admit an automorphism, an isomorphism of themselves with themselves, induced by swapping $i$ with $\unaryminus i$---namely, complex conjugation: $$z=a+bi\qquad\mapsto\qquad\bar z= a-bi.$$ The conjugation map preserves the field structure, since $\overline{y+z}=\bar y+\bar z$ and $\overline{y\cdot z}=\bar y\cdot\bar z$, and therefore the complex field is not a rigid mathematical structure. Since conjugation swaps $i$ and $\unaryminus i$, it follows that $i$ can have no structural property in the complex numbers that $\unaryminus i$ does not also have. So there can be no principled, structuralist reason to pick one of them over the other. Is this a problem for structuralism? It does seem to be a problem for singular terms, since how do we know that the $i$ appearing in my calculations this week is the same number as what will appear in your calculations next week? Perhaps my $i$ is your $\unaryminus i$, and we do not even realize it.

If one wants to understand mathematical objects as abstract positions within a structure, as in abstract structuralism, then one must grapple with the fact that in light of the conjugation automorphism, the numbers $i$ and $\unaryminus i$ play exactly the same roles in this structure (see Shapiro, 2012). The numbers $i$ and $\unaryminus i$ have the same isomorphism orbit with respect to the complex field, and so in this sense, although distinct, they each play exactly the same structural role in $\C$. This would seem to undermine the idea that mathematical objects are abstract positions in a structure, since we want to regard these as distinct complex numbers.

The point is that one cannot understand the mathematical objects as identical to the structural roles that they play within a system, since $i$ and $\unaryminus i$ play exactly the same role in $\C$ as each other, yet are to be regarded as distinct complex numbers.

In fact, there is nothing special about the numbers $i$ and $\unaryminus i$ in this argument. For example, the numbers $\sqrt{2}$ and $\unaryminus\!\sqrt{2}$ also happen to play the same structural role in the complex field $\C$, because there is an automorphism of $\C$ that swaps them (although one uses the axiom of choice to prove this). Contrast this with the real field $\newcommand\R{\mathbb{R}}\R$, where $\sqrt{2}$ and $\unaryminus\!\sqrt{2}$ are of course discernible, since one is positive and the other is negative, and the order is definable from the field operations in $\R$ via $x\leq y\iff\exists u\ x+u^2=y$. It follows that the real number field is not definable in the complex field by any assertion in the language of fields. In fact, there is an enormous diversity of automorphisms of the complex field; one may move $\sqrt[3]{2}$, for example, to one of the nonreal cube roots of $2$, such as $\sqrt[3]{2}(\sqrt{3}i-1)/2$. Therefore, the numbers $\sqrt[3]{2}$ and $\sqrt[3]{2}(\sqrt{3}i-1)/2$ are indiscernible in the complex field---there is no property expressible in the language of fields that will distinguish them. Indeed, except for the rational numbers, every single complex number is part of a nontrivial orbit of automorphic copies, from which it cannot be distinguished in the field structure. So the same issue as with $i$ and $\unaryminus i$ occurs with every irrational complex number. For this reason, it is problematic to try to identify complex numbers with the abstract positions or roles that the numbers play in the complex field.

Meanwhile, one may recovers the uniqueness of the structural roles simply by augmenting the complex numbers with additional natural structure. Specifically, once we augment the complex field $\C$ with the standard operators for the real and imaginary parts: $$\text{Re}(a+bi)=a\qquad\qquad\text{Im}(a+bi)=b,$$ then the expanded structure $\langle\C,+,\cdot,\text{Re},\text{Im}\rangle$ becomes rigid, meaning that it has no nontrivial automorphisms. Thus, every complex number plays a unique structural role in this new structure, which is Leibnizian. This additional structure is implicit in the complex plane conception of the complex numbers, which is part of why the number $i$ appears fine as a singular term---it refers to the point $(0,1)$ in the complex plane---whereas $\unaryminus i$ refers to $(0,-1)$. The complex plane is not merely a field, for it carries along its coordinate information by means of the real-part and imaginary-part operators, making it rigid. In the complex plane, every complex number plays a different role.

This answer was copied from a related post at meta.mathoverflow.

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    $\begingroup$ I think this ought to be the accepted answer! $\endgroup$
    – Doradus
    Commented Sep 21, 2021 at 21:08

You asked for an elementary explanation of the difference between i and -i. Multiplying a complex number by i rotates it by 90 degrees anti-clockwise, multiplying by -i rotates it by 90 degrees clockwise. Sure if I repeat either one twice I have rotated by 180 degrees (or multiplied by -1) but I can still distinguish one from the other.

  • $\begingroup$ Yes there is an automorphism (a mirror reflection of the complex plane) that swaps i to -i and changes clockwise to anti-clockwise. I don't think that affects the above way to explain the difference between i and -i $\endgroup$ Commented Aug 11, 2012 at 5:17
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    $\begingroup$ If you could only use words, and could not use physical expression, could you describe clockwise or counter-clockwise? right or left? $\endgroup$
    – robjohn
    Commented Aug 11, 2012 at 5:49
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    $\begingroup$ That is an interesting question - is there an algebraic definition of orientation in the (complex) plane. For the question that was asked for an elementary explanation of difference between i and -i and I think it is reasonable to use the complex plane and the natural notions of geometry that students usually already know. Here is my attempt at describing clockwise in words - if you multiply a complex number by another complex number of length 1 that is close to 1 then if the y coordinate increases it is an anti-clockwise rotation and if it decreases it is clockwise. $\endgroup$ Commented Aug 11, 2012 at 19:16
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    $\begingroup$ Of course, since we can arbitrarily define coordinates increasing, we’re back where we started. $\endgroup$
    – kinokijuf
    Commented Nov 14, 2013 at 8:14

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