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?
How to tell $i$ from $-i$?
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0Square of 6 and $-6$ are 36 but both are not same. – 2016-09-25
6 Answers
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.
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1Of course, since we can arbitrarily define coordinates increasing, we’re back where we started. – 2013-11-14
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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0@I.J.Kennedy: That's true, but what is your point? – 2018-10-27
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.
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1"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. – 2012-08-02
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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0@sarnold: Thanks, fixed. – 2012-08-02