Reading about the cyclic values returned by integral powers of $i$: $$\begin{align*} i^0&=\hphantom{-}1\\ i^1&=\hphantom{-}i\\ i^2&=-1\\ i^3&=-i\\ & \,\,\,\vdots \end{align*}$$

and the pattern continues. Now, a simple question popped up in my mind: $i^2 = -1\implies i\notin\mathbb{R}$. So, how did we come to the conclusion that: $$i^3=i^2\cdot i=-1\cdot i=-i$$ because $i$ does not necessarily have the properties of reals. Is this just an assumption for the construction of the complex numbers, or is there a proof?

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    $\begingroup$ Your assumption is that the "exponent rules" for example ($k^3 = k^2\cdot k$) are "real number rules." They aren't. The whole field of abstract algebra developed from this observation: very different sets, with very different operations, have similar properties. Often the elements of those sets don't look like numbers at all, and the operations don't look like arithmetic, and yet the same rules apply. $\endgroup$ – mathguy Aug 26 '16 at 18:25
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    $\begingroup$ It does depend on what framework you invented complex numbers for. If your definition involves complex numbers being a field, then $x^{n}x^{m}=x^{n+m}$ is a property of fields. $\endgroup$ – fleablood Aug 26 '16 at 19:07

It does indeed have a proof. But this involves going back to exactly how the complex numbers are defined in the first place.

Let's say for the moment that we're happy with the real numbers and their properties. (Of course, there's a separate question - "how do we construct the reals?" - but let's leave that aside for now.) There are now a number of ways to rigorously define the complex numbers. Here's probably the simplest (although not the most useful): we let $\mathbb{C}$ be the set of ordered pairs $(a, b)$ where $a, b\in\mathbb{R}$, and we define the operations $+$ and $\times$ on $\mathbb{C}$ as follows (intuitively, the pair $(a, b)$ represents the number $a+bi$):

  • $(a, b)+(c, d)=(a+b, c+d)$.

  • $(a, b)\times (c, d)=(a\times c-b\times d, a\times d+b\times c)$.

Note that on the right hand side of each of those expressions, I'm using the notion of $+$ and $\times$ (and $-$) for real numbers. Arguably, a better way to write this would be

  • $(a, b)+_\mathbb{C}(c, d)=(a+_\mathbb{R}b, c+_\mathbb{R}d)$.

  • $(a, b)\times_\mathbb{C} (c, d)=(a\times_\mathbb{R} c-_\mathbb{R}b\times_\mathbb{R} d, a\times_\mathbb{R} d+_\mathbb{R}b\times_\mathbb{R} c)$.

That is, I've already defined operations $+_\mathbb{R}, \times_\mathbb{R}, -_\mathbb{R}$ for the reals, and now I'm defining new operations $+_\mathbb{C}, \times_\mathbb{C}$ for the complex numbers. In general, it will always be clear from context which is meant; but it's a good habit when learning this material to get into to clarify things with subscripts whenever we're using both types of operation.

We can use this explicit definition to verify basic claims about complex numbers. For example,

  • Check that $(0, 1)\times(0, 1)=(-1, 0)$, and that $(0, 1)$ and $(0, -1)$ are the only complex numbers with this property. We call the former "$i$" (but it doesn't really matter which one we pick).

  • Now, we can compute $(0, 1)^3=(0, 1)\times (0, 1)\times (0, 1)=(0, -1)$. This shows that indeed $i^3=-i$.

What about more general claims? For example, I implicitly used the associativity of $\times$ above, in writing out $(0, 1)^3$. We can also use the definition of $\mathbb{C}$ to prove that $\times$ is associative, and more generally that $\mathbb{C}$ is a field). This gets a bit messy; let me give as an example the proof that addition of complex numbers is commutative, which is simpler:

$$(a, b)+_\mathbb{C}(c, d)=(a+_\mathbb{R}c, b+_\mathbb{R}d)=(c+_\mathbb{R}a, d+_\mathbb{R}b)=(c, d)+_\mathbb{C}(a, b).$$

Note how we use (at the second "$=$" sign) the commutativity of $+_\mathbb{R}$; we're building on facts we already know about the reals, to prove facts about the complex numbers, and this is justified because of how the complex numbers are constructed from the reals.

EDIT: Just to tie up a loose end, what did I mean above when I said that this was "not the most useful" way to construct the complex numbers?

Well, it turns out that there is a much more abstract approach we can use, that - while much harder at first - ultimately turns out to be much more mathematically useful. The key point is to notice that essentially the whole point of the complex numbers is to solve a single equation, $x^2=-1$. In some sense, they are what you get when you start with the reals and "fill in" the hole left by that equation.

This suggests that we should look for some tool for doing the following: if $\mathbb{S}$ is some number system (reals, complexes, integers, rationals, quaternions, hyperbolic sedenions, or thing-I-just-made-upions) and $E$ is some equation in $\mathbb{S}$, there should be a way to build a thing like $\mathbb{S}$ but that can also solve $E$. This is really really vague, but ultimately leads to a bunch of ideas in abstract algebra; for example, algebraic extensions of fields. In this context, there's a precise sense in which the complex numbers are the "smallest" number system containing the reals where $x^2=-1$ has a solution, and the relevant notation is $\mathbb{R}[x]/\langle x^2+1\rangle$ - which is probably gibberish right now but down the road you'll see how to view this as a "recipe" for building the complex numbers from the real numbers.

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    $\begingroup$ Great Explanation, thnx $\endgroup$ – codetalker Aug 26 '16 at 18:17
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    $\begingroup$ Wouldn't it be simpler to define $\mathbb C$ as $\mathbb R[x] / (x^2 + 1)$ and then deduce the formula for a product of two complex numbers? $\endgroup$ – Santiago Aug 26 '16 at 18:19
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    $\begingroup$ @Santiago I tend not to think so, although that is certainly the more mathematically useful way to define $\mathbb{C}$ (and I've added a bit about it very very briefly); I find that passing to polynomials, then modding out by an ideal, etc. winds up introducing a lot of unneeded confusion for most people (myself included) seeing it for the first time. But of course this varies person to person, and what's simpler for me may not be simpler for anyone else. $\endgroup$ – Noah Schweber Aug 26 '16 at 18:24
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    $\begingroup$ @BillDubuque In my experience, many students find even congruences more complicated than the pair construction. Of course, preference varies student to student, but I tend to think that the pair construction is the easiest for the majority of students. One possible reason why is that it doesn't involve "type confusion" - I've often found that students find it weird that an element of the "new" algebra is on the same level as a subset of the old algebra, and canonical representatives don't help here as much as with $\mathbb{Z}/m$. But we can agree to disagree. $\endgroup$ – Noah Schweber Aug 26 '16 at 20:41
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    $\begingroup$ @BillDubuque if you think there's a better presentation, the place for it would be as an alternate answer. The people to convince are the visitors to this question, not Noah Schweber, who already gave his answer. $\endgroup$ – Spike0xff Aug 27 '16 at 4:32

Although those are excellent answers I think the answer is even simpler.

$b^{m}b^{n} = b^{m+n}$ is not a "real number rule" but an "associative binary operation rule".

If we define/construct/invent any set of elements (these could be gumdrops or chess pieces for all we care) and create any operation between to elements ($a*b$ could be "melt the two gumdrops together an make a new one" or "always pick the second chess piece; always") so that:

--$a * b$ for any two elements of the set will result in $c$, also a member of the set and the same $a,b$ combined will always results in the same member $c$ (The term for this is $*$ is a "binary operation".

--If we evaluate $a*b*c*d$ it doesn't matter how we group them. $(a*(b*c))*d$ (where we first do $b*c= e$ and then we do $a*e = f$ and then we do $f*d = g$) will give the same result as $(a*b)*(c*d)$ (where we do $a*b$ and get $h$ and then we do $c*d$ and get $i$ and then we do $h*i$ and get, amazingly, $g$). Or more simply $(a*b)*c = a*(b*c)$. (such an operation is called "associative". An example of something that isn't associative is $(2^3)^2 \ne 2^{(3^2)}$.)

Then if we define as a matter of notation (it's just notation) that if $n\in \mathbb N$ than $a^n := a*a*a*....*a$ where a is operated $n$ times.

If we create such a system, (whether gumdrops, chess pieces, or numbers), then we will always have:

$b^mb^n = (b*.....*b)*(b*.....*b) = (b*....................*b) = b^{m+n}$.


So this is true for complex numbers. (Assuming we defined what $c*d$ means and that $c*(d*e) = (c*d)*e$).


Just to be perverse, let's do this with chess pieces. And define $a*b = b$. Then $(knight*pawn)*bishop = pawn*bishop = bishop$ which is equal to $knight*(pawn*bishop) = knight*bishop = bishop$.

Then $pawn^2*pawn = (pawn*pawn)*pawn=pawn*pawn= pawn; pawn*pawn^2 = pawn*(pawn*pawn) = pawn*pawn = pawn; pawn^3 = pawn*pawn*pawn = pawn*pawn = pawn$.

So $pawn^2pawn=pawn*pawn^2 = pawn^3$.

That was kind of boring.


To go to Noah Schweber's excellent answer. If $i = (0,1)$ and $(a,b)*(c,d)= (ac - bd, ad + bc)$ is the definition of complex numbers then:

-- yes, it is binary. ($(a,b)*(c,d)= (ac - bd, ad + bc)$ results in a real value pair.)

-- yes it is associative. ($(a,b)*[(c,d)*(e,f)] = (a,b)*(ce-df,cf+de)=(a(ce-df)-b(cf+de), b(ce-df) + a(cf+de))=(ace-adf-bcf-bde,bce-bdf+acf+ade)$ while ($[(a,b)*(c,d)]*(e,f) = (ac - bd,bc+ad)*(e,f)=((ac-bd)e-(bc+ad)f,(bc+ad)e + f(bc+ad))=(ace-adf-bcf-bde,bce-bdf+acf+ade)$ so $(a,b)*[(c,d)*(e,f)] = [(a,b)*(c,d)]*(e,f)$.


$(a,b)^3 = (a,b)(a,b)(a,b)=(a^2 - b^2,2ab)(a,b) = (a^3 - 3ab^2,3a^b-b^3)$

$(a,b)^2(a,b) = [(a,b)(a,b)](a,b)=(a^2 - b^2,2ab)(a,b) = (a^3 - 3ab^2,3a^b-b^3)$

$(a,b)(a,b)^2 = (a,b)(a^2 - b^2,2ab)=(a^3 - 3ab^2,3a^b-b^3)$

all equal.

  • $\begingroup$ +1, at least for positive integer exponents, there's no need to go beyond checking associativity. $\endgroup$ – celtschk Aug 27 '16 at 11:23
  • $\begingroup$ I'm not sure where I did go beyond checking associativity. But we need binary and associativity otherwise $b^n$ isn't well defined. As $n$ is integer and $b^n$ is notation it is impossible for $b^{n+m} \ne b^nb^m$ but I do feel this will seem like "magic" if it's simply stated as fact. So I feel a physical demonstration was in order. $\endgroup$ – fleablood Aug 27 '16 at 17:46
  • $\begingroup$ Oh, I see. you are talking about my display that (a,b)^3 is (a,b)^2(a,b). I wasn't actually trying to "prove" that but demonstrate that my argument really does hold in the hopes that by seeing it concretely and looking at it it will be made apparent and obvious the my argument must be true and hold for any associative binary operation. $\endgroup$ – fleablood Aug 27 '16 at 17:49
  • $\begingroup$ You misunderstood me. I didn't criticize you, I just summarized. However for the $i^0$ also present in the question, you actually do need a bit more: Namely the neutral element ($1$). Well, strictly speaking, you only need an idempotent element that when multiplied by $i$ from either side gives $i$. For your chess pieces example, $a^0=a$ would work, definins some piece as "1"and writing $a^0=1$ would not, as for $a\ne 1$ you'd get $a*1=1\ne a$ $\endgroup$ – celtschk Aug 27 '16 at 17:52
  • $\begingroup$ Oh, I did skip the $i^0$. The question holds without it. If you want to be picky I always felt"$b^1=b$" times itself 1 time" could be be considered ambiguous. Anyway, IF the system has an identity element. (gumdrops and chessmen do not) $a^0 = $ identity *by definition* is the one value that would extend the $b^{n+m}=b^{n}b^m$. Likewise if the system has inverses $b^{-m} := {b^{m}}^{inv}$ consistently extends to preserve the rule. $\endgroup$ – fleablood Aug 27 '16 at 18:06

Are properties of the imaginary unit assumed or proved?

Both. The intuitive approach of assuming a solution to a previously unsolvable equation (such as an $i$ solving $i^2= -1$) that satisfies other algebraic properties of the number system, came into use in the 1400-1700's and was later proved to create a consistent number system that works as expected (such as any element of the system with $i$ being uniquely expressible as $x+yi$).

Today the sequence is similar: it is assumed to work in high school, and proved to work at university.


There are various logically rigorous ways to prove that the set of all complex numbers is a field, i.e. it satisfies certain axioms about addition, subtraction, multiplication, and division, but there is also another point of view worth knowing about:

  • Multiplication by any complex number $z = x+iy$ amounts to rotating through the angle whose sine is $y/|z|$ and whose cosine is $x/|z|$ and then multiplying by the real number $|z|$.
  • In particular, multiplying by $i$ means rotating $90^\circ$ counterclockwise. Thus multiplying by $i^3$ amounts to rotationg $270^\circ$ counterclockwise, whcih gives you $-i$.

As you point out, any integral power of $i$ is $i$ or $-1$, or $-i$ or $1$.

Complex 'multiplication' is defined by $(a,b).(c,d)=(ac-bd,ad+bc)$ where $a,b,c,d\in\mathbb{R}$. Also note that the complex number $(0,1)=0+i.1$ is defined by the notation $i$, the imaginary unit.

So, $i^2=(0,1)(0,1)=(-1,0)=-1$ and we can similarly show that $i^3=(0,-1)=-i$.

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    $\begingroup$ Complex addition is defined as you have above. Complex multiplication on the other hand is defined as $(a,b)\cdot (c,d) = (ac-bd, ad+bc)$ $\endgroup$ – JMoravitz Aug 26 '16 at 18:13
  • $\begingroup$ @JMoravitz Sorry, my bad. $\endgroup$ – StubbornAtom Aug 26 '16 at 18:15

Define $ i $ as the root of $ x^2 + 1 $, define $ \mathbb C = \mathbb R(i) $, a finite field extension of the reals. From this definition, the field axioms are preserved, and you can thus apply associativity to obtain your statement.

  • $\begingroup$ Haha, of course. $\endgroup$ – user347499 Aug 28 '16 at 8:32

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