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 $\mathbb{C}$. $\endgroup$ – André Nicolas Aug 1 '12 at 14:38
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    $\begingroup$ $-1$ and $1$ both square to $1.$ Does this means they're indistinguishable? $\endgroup$ – user2468 Aug 1 '12 at 14:40
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    $\begingroup$ @J.D. Ah - but there is no automorphism taking 1 to -1. $\endgroup$ – Old John Aug 1 '12 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 Aug 1 '12 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 Aug 1 '12 at 17:24

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 $\mathbb{C}$? $\endgroup$ – Nils Matthes Aug 1 '12 at 15:02
  • $\begingroup$ @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. $\endgroup$ – Andrea Mori Aug 1 '12 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 $-X+(X^2+1) \mathbb{R}[X]$. $\endgroup$ – David E Speyer Aug 1 '12 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 Aug 1 '12 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 Aug 2 '12 at 1:36

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 Aug 1 '12 at 14:47
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    $\begingroup$ With the proviso that, once you've made your decision, you must stick with it thereafter. $\endgroup$ – Mark Beadles Aug 1 '12 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. $\endgroup$ – I. J. Kennedy Oct 24 '18 at 13:46
  • $\begingroup$ @I.J.Kennedy: That's true, but what is your point? $\endgroup$ – celtschk Oct 27 '18 at 6:27

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 beause 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 Aug 1 '12 at 20:40
  • $\begingroup$ @sarnold: Thanks, fixed. $\endgroup$ – Ben Crowell Aug 2 '12 at 3:46

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 $\endgroup$ – Michaela Light Aug 11 '12 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 Aug 11 '12 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. $\endgroup$ – Michaela Light Aug 11 '12 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 Nov 14 '13 at 8:14

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