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…

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 xn – 1. For example, when n = 2, this gives us the quadratic polynomial x2 – 1. To find its roots, just set it equal to 0 and solve: x2 – 1 = 0. You might remember factoring expressions like this using the “difference of squares” formula, which says that a2 – b2 = (a – b)(a + b). Here x2 – 12 = (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 nth roots of unity, which are the solutions to the equation xn – 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 x4 = 1. You might recognize two of the fourth roots of unity right away: Since 14 = 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: x4 = 1 x4 – 1 = 0 Since x4 and 1 are both perfect squares, you can use the difference of squares formula here as well: x4 – 1 = (x2)2 – 12 = (x2 – 1)(x2 + 1). This turns the equation x4 – 1 = 0 into (x2 – 1)(x2 + 1)=0. The x2 – 1 should look familiar: We factored that when we found the square roots of unity. This gives us (x – 1)(x + 1)(x2 + 1) = 0. We can’t factor any more right now. The expression x2 + 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 x2 + 1 = 0. The first two equations tell us what we already knew: x = 1 and x = −1 are solutions to the equation x4 = 1 and are therefore fourth roots of unity. But what can we do with x2 + 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 i2 = −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 x2 + 1 = 0, it must be one of the fourth roots of unity. You can easily verify this with some rules of exponents: Since i2 = −1, then i4 = (i2)2 = (−1)2 = 1. And since complex numbers follow most rules that real numbers follow, it’s true that (−i)2 = i2 , and so x = −i also satisfies x2 + 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 nth-degree polynomial has n complex roots. Therefore the equation xn = 1 has n complex solutions, and these are all the nth 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 nth roots of unity possess some remarkable properties. Geometrically, if you graph the nth 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 nth 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, -\frac{1}{2}+i \frac{\sqrt{3}}{2}, and -\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 + \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 x3 – 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: x3 – (1 + α + β)x2 + (α + β + αβ)x – αβ = 0. But we already know what cubic polynomial we should get when we multiply this out: x3 – 1. So x3 – (1 + α + β)x2 + (α + β + αβ)x – αβ is really x3 – 1, which means that the coefficient of x2 on the left side, 1 + α + β, has to equal the coefficient of x2 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 xn, the sum of the roots of the polynomial will always be the negation of the coefficient of xn-1. Since roots of unity come from polynomials of the form xn– 1, where the coefficient of xn-1 is always 0, Vieta’s formula tells us that the sum of the nth 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 nth roots of unity, then α × β is also an nth root of unity! And if α and β are both nth roots of unity, then αn = 1 and βn = 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 nth roots of unity is assumed to always be an integer, the basic rules of exponents still apply, like this one: (α × β)n = αn × βn . So (α × β)n = αn × βn = 1 × 1 = 1. This means α × β satisfies the equation xn = 1 and so is an nth 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: \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 nth roots of unity: a “group” structure. A group is a set of elements (here, the nth 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 nth roots of unity is always another nth root of unity. Another important property of groups is that inverses always exist. This means that for any nth root of unity there is another nth 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) = −i2 = −(−1) = 1, and among the third roots of unity the inverse of -\frac{1}{2}+i \frac{\sqrt{3}}{2} happens to be -\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 nth 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 xn − 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: \sqrt{i} is one of them, but you need to write it in a + bi form.) 3. When is an nth root of unity also an mth root of unity? 4. A “primitive nth root of unity” is an nth root of unity whose powers include all the nth 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 nth roots of unity? (Hint: Take a closer look at Vieta’s formulas.) Answers Click for Answer 1: Let α be a fourth root of unity. Then α4 = 1. Now square both sides of the equation to get α8 = 1. Since α satisfies x8 = 1, it is an eighth root of unity. Click for Answer 2: Notice that \sqrt{i} is an eighth root of unity, since if α = \sqrt{i}, then α2 = i, and so α8 = i4 = 1. To find \sqrt{i}, set \sqrt{i} = a + bi and square both sides to get i = a2 – b2 + 2abi. This tells you that a2 – b2 = 0 and 2ab = 1, and some algebra establishes that one square root of i is \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}. The other three eighth roots of unity are \frac{\sqrt{2}}{2}i \frac{\sqrt{2}}{2}, −\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}, and −\frac{\sqrt{2}}{2}i \frac{\sqrt{2}}{2}. Click for Answer 3: This is true when m is a multiple of n, and it follows from simple rules of exponents. Suppose αn = 1 and m = kn. Then αm = αkn = (αn)k = (1)k = 1, so α is an mth root of unity as well. Click for Answer 4: Let α = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}. Then α is a primitive eighth root of unity, since α2 = i, α3 = -\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}, α4 = -1, α5 = -\frac{\sqrt{2}}{2}i \frac{\sqrt{2}}{2}, α6 = –i, α7 = \frac{\sqrt{2}}{2} – $latexi \frac{\sqrt{2}}{2}$, and α8 = 1. Some multiplication, or some clever use of laws of exponents, will show you that the other primitive eighth roots of unity are α3, α5 and α7. In fact, there’s a very beautiful relationship between n and the primitive nth roots of unity, which you can find just using laws of exponents. See if you can find it!