###### Quantized Academy

# The Simple Math Behind the Mighty Roots of Unity

If you’ve ever taken an algebra or physics class, then you’ve met a parabola, the simple curve that can model how a ball flies through the air. The most important part of a parabola is the vertex — its highest or lowest point — and there are many mathematical techniques for finding it. You can try vertex form, or the axis of symmetry, or even calculus.

But last week one of my students located the vertex of a parabola in a particularly elegant way. “The vertex is at *x* = 4,” she said, “because the roots are *x* = 1 and *x* = 7, and the roots are symmetric about the vertex.” She used the fact that the parabola is the graph of a quadratic polynomial, and that the roots of that polynomial — the values where it becomes 0 — have a certain structure she could take advantage of.

There is a structure to the roots of every polynomial, and mathematicians study these structures and look for opportunities to capitalize on them, just as my student did with her parabola. And when it comes to the roots of polynomials, none have more structure than the “roots of unity.”

Roots of unity are the roots of the polynomials of the form *x ^{n}* – 1. For example, when

*n*= 2, this gives us the quadratic polynomial

*x*– 1. To find its roots, just set it equal to 0 and solve:

^{2} *x ^{2}* – 1 = 0.

You might remember factoring expressions like this using the “difference of squares” formula, which says that *a ^{2}* –

*b*= (

^{2}*a*–

*b*)(

*a*+

*b*). Here

*x*– 1

^{2}*= (*

^{2}*x*– 1)(

*x*+ 1), which gives you

(*x* – 1)(*x* + 1) = 0.

Now that you’ve got a product equal to 0, you can invoke one of the most underappreciated rules from algebra class: the “zero product property.” This says that the only way two real numbers can multiply to 0 is for one of them to be 0. So if (*x* – 1)(*x* + 1) = 0, then either *x* – 1 = 0 or *x* + 1 = 0. The first equation is true when *x* = 1, the second when *x* = −1. So 1 and −1 are the two “second roots of unity,” which might be more familiar to you as the two square roots of 1.

For any *n* you can find the *n*th roots of unity, which are the solutions to the equation *x ^{n}* – 1 = 0. These roots of unity possess a remarkably rich structure that connects to high school math like trigonometry and rotations of the plane as well as ongoing research that involves some of the great unanswered questions in modern math.

When *n* = 2, the two roots 1 and −1 have a symmetric structure that is related to how my student found her vertex. You can see even more structure in the fourth roots of unity. These are the solutions to the equation *x*^{4 }= 1. You might recognize two of the fourth roots of unity right away: Since 1^{4 }= 1 and (−1)^{4 }= 1, *x *= 1 and *x *= −1 both satisfy the equation, so they are fourth roots of unity. But there are actually two more, and you can find them using algebra as we did above: Just put the equation into standard form and factor:

*x*^{4 }= 1

*x*^{4 }– 1 = 0

Since *x*^{4 }and 1 are both perfect squares, you can use the difference of squares formula here as well:

*x*^{4 }– 1 = (*x*^{2})^{2 }– 1^{2 }= (*x*^{2 }– 1)(*x*^{2 }+ 1). This turns the equation *x*^{4 }– 1 = 0 into

(*x*^{2 }– 1)(*x*^{2 }+ 1)=0.

The *x*^{2 }– 1 should look familiar: We factored that when we found the square roots of unity. This gives us

(*x*^{ }– 1)(*x*^{ }+ 1)(*x*^{2 }+ 1) = 0.

We can’t factor any more right now. The expression *x*^{2 }+ 1 is “irreducible” over the real numbers, which means it can’t be broken down into simpler multiplicative factors that involve only real numbers. But we can still apply the zero product property. If these three numbers multiply to 0, then one of them must be zero. That is, either *x*^{ }– 1 = 0, *x*^{ }+ 1 = 0, or *x*^{2 }+ 1 = 0.

The first two equations tell us what we already knew: *x *= 1 and *x *= −1 are solutions to the equation *x*^{4 }= 1 and are therefore fourth roots of unity. But what can we do with *x*^{2 }+ 1 = 0?

Well, if you know about complex numbers, then you know that *i*, the “imaginary unit,” satisfies this equation because it is defined by the property that *i*^{2 }= −1. It isn’t a real number — no real number squared is negative — but it turns out that most roots of unity are complex numbers, and since *x *= *i *satisfies *x*^{2 }+ 1 = 0, it must be one of the fourth roots of unity. You can easily verify this with some rules of exponents: Since *i*^{2 }= −1, then *i*^{4 }= (*i*^{2})^{2 }= (−1)^{2 }= 1. And since complex numbers follow most rules that real numbers follow, it’s true that (−*i*)^{2 }= *i ^{2 }*, and so

*x = −i*also satisfies

*x*

^{2 }+ 1 = 0 and is also a fourth root of unity.

These four numbers, 1, −1, *i*, and −*i*, are all fourth roots of unity, and the matching fours are not a coincidence. The fundamental theorem of algebra says that every *n*th-degree polynomial has *n* complex roots. Therefore the equation *x ^{n }*= 1 has

*n*complex solutions, and these are all the

*n*

^{th}roots of unity. (Since real numbers are also complex numbers, the real solutions, like 1 and −1, are included in the count of complex solutions.)

For a given *n* the *n*th roots of unity possess some remarkable properties. Geometrically, if you graph the *n*th roots of unity in the complex plane you’ll find that they are equally spaced around the unit circle centered at the origin.

This geometric structure is closely connected to important ideas in trigonometry, like the angle sum and difference formulas for sine and cosine, the theory of rotations of the plane, and *e*, the base of the natural logarithm function. This geometry is also connected to an interesting algebraic property: For any *n*, the sum of the *n*th roots of unity is 0.

For *n* = 2 this is immediately obvious: The sum of both square roots of unity is 1 + (−1) = 0. It’s clear, too, for the four fourth roots of unity:

1 + *i *+ (−1) + (−*i*) = 0.

In both cases it’s easy to see why the sum is 0: The roots of unity come in opposite pairs, which cancel out when you add them up.

However, the result holds even when the roots of unity don’t come in opposite pairs. For example, the three third roots of unity are 1, $latex -\frac{1}{2}+i \frac{\sqrt{3}}{2}$, and $latex -\frac{1}{2} – i \frac{\sqrt{3}}{2}$. The two non-real roots don’t cancel out, but they do sum to −1, which then cancels out with the remaining root of unity, giving you 0 in the end:

1 + $latex \left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)+\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)$ = 1 + (−1) = 0.

You can establish this property geometrically, but there’s an elegant algebraic argument that shows this is true. Let’s call the three third roots of unity 1, α and β. All three of these numbers satisfy the cubic equation

*x*^{3 }– 1 = 0.

Because you know the roots of this cubic equation, you know that the polynomial on the left factors as

(*x* – 1)(*x* – α)(*x* – β) = 0.

If you multiply this expression out using the distributive property a few times, you get the following:

*x*^{3} – (1 + α + β)*x*^{2} + (α + β + αβ)*x* – αβ = 0.

But we already know what cubic polynomial we should get when we multiply this out: *x*^{3} – 1. So *x*^{3} – (1 + α + β)*x*^{2} + (α + β + αβ)*x* – αβ is really *x*^{3} – 1, which means that the coefficient of *x*^{2} on the left side, 1 + α + β, has to equal the coefficient of *x*^{2} on the right side, which is 0. Thus 1 + α + β =0, and so the three third roots of unity sum to 0.

This argument generalizes and produces one of “Vieta’s formulas,” which are famous results that relate the roots of a polynomial to its coefficients. One of Vieta’s formulas says that, in a polynomial that begins with *x*^{n}, the sum of the roots of the polynomial will always be the negation of the coefficient of *x*^{n-1}. Since roots of unity come from polynomials of the form *x*^{n}– 1, where the coefficient of *x*^{n-1} is always 0, Vieta’s formula tells us that the sum of the *n*th roots of unity is 0 for any *n*.

There’s an even more remarkable algebraic result when it comes to roots of unity. For a given *n*, if α and β are two *n*th roots of unity, then α × β is also an *n*th root of unity! And if α and β are both *n*th roots of unity, then α^{n }= 1 and β^{n }= 1. So what is (α × β)^{n}?

In general you have to be careful raising complex numbers to a power, but since the *n* in the *n*th roots of unity is assumed to always be an integer, the basic rules of exponents still apply, like this one:

(α × β)^{n }= α^{n} × β^{n}^{ }.

So (α × β)^{n }= α^{n} × β^{n }= 1 × 1 = 1. This means α × β satisfies the equation *x*^{n }= 1 and so is an *n*th root of unity. For example, when *n* = 4, if you multiply the two roots of unity *i* and −1, you get another fourth root of unity: *i* × (−1) = −*i*. And when *n* = 3, you can also verify by multiplication that the two non-real roots of unity multiply to the real one: $latex \left(-\frac{1}{2}+i \frac{\sqrt{3}}{2}\right)×\left(-\frac{1}{2}-i \frac{\sqrt{3}}{2}\right)$ = 1.

This property gives rise to an incredibly rich algebraic structure on the *n*th roots of unity: a “group” structure. A group is a set of elements (here, the *n*th roots of unity) and an operation (here, normal multiplication) that satisfies some familiar properties. One of those properties is “closure,” which we just demonstrated. This means the product of two *n*th roots of unity is always another *n*th root of unity. Another important property of groups is that inverses always exist. This means that for any *n*th root of unity there is another *n*th root of unity such that their product is 1, the multiplicative identity. For example, when *n* = 4, the inverse of *i* is −*i *since *i* × (−*i*) = −*i*^{2} = −(−1) = 1, and among the third roots of unity the inverse of $latex -\frac{1}{2}+i \frac{\sqrt{3}}{2}$ happens to be $latex -\frac{1}{2}-i \frac{\sqrt{3}}{2}$.

The study of groups is fundamental to Galois theory, an advanced field of mathematics built to study abstract algebraic structures associated with polynomials and their roots. You probably know the quadratic formula and possibly know of the cubic and quartic formulas, but there is no general formula for finding the roots of a polynomial of degree 5 or higher, and Galois theory helps unravel this mystery by studying the groups associated with the roots of polynomials.

Because the *n*th roots of unity have their own group structure, they occupy an important place in Galois theory, especially because that structure is so easy to work with. Roots-of-unity groups are always “abelian,” meaning that the order in which you multiply objects doesn’t change the result, and they are always “cyclic,” meaning that you can always generate the entire group by multiplying a single element by itself over and over.

In Galois theory, being associated with an abelian group is a very nice property for a polynomial, and the impact of the roots of unity extend well beyond just polynomials of the form *x ^{n}* − 1. It turns out that any polynomial associated with an abelian group in Galois theory has roots that can be expressed as sums of different roots of unity. In a sense, the roots of unity form the foundation of all the nice polynomials in a particular mathematical world, and generalizing the role of roots of unity to other mathematical worlds has been the goal of Hilbert’s 12th problem, one of the 23 math problems posed by David Hilbert in 1900 to guide the course of mathematical discovery for the next 100 years. Now, more than a century later, people are still working on the 12th problem, and progress is being made, but mathematicians aren’t all the way there yet. Perhaps soon they’ll get to the root of it all.

## Exercises

1. Show that the four fourth roots of unity are also eighth roots of unity.

2. Find the other four eighth roots of unity. (Hint: $latex \sqrt{i}$ is one of them, but you need to write it in *a* + *bi* form.)

3. When is an *n*th root of unity also an *m*th root of unity?

4. A “primitive *n*th root of unity” is an *n*th root of unity whose powers include all the *n*th roots of unity. For example, *i* is a primitive fourth root of unity, since the powers of *i* are* i*, -1, *-i*, and 1, all four of the fourth roots of unity. But −1 is not a primitive fourth root of unity, since the powers of −1 are just −1 and 1.

Which of the eighth roots of unity are primitive?

Challenge: What is the product of all *n* of the *n*th roots of unity? (Hint: Take a closer look at Vieta’s formulas.)

## Answers

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