Sums of roots of unity

Figure 1. Ninth roots of unity on the complex plane.

This note, which I’ll probably expand upon later, is about roots of unity. I was reading Shafarevich’s excellent Discourses on Algebra, and wondered what the identity at the end of section 2.4 meant for the polynomial xⁿ − 1, the defining polynomial for the nth roots of unity (where n > 1).

The reason I wondered was that the roots of unity are very symmetric: they lie evenly along the unit circle in the complex plane. As an example, the ninth roots of unity (i.e., the roots of the polynomial when n is 9) are represented in figure 1.

Really, I started thinking about the identity (as discussed below), but then I wondered what other ways there are to show that the sum of all of the roots of unity is 0. This note has a number of ways of showing this fact.

1. The intuitive way

The nth roots of unity lie evenly on the unit circle, so their center of mass better be at the origin. So, the sum of the complex numbers as vectors is zero.

2. The direct way

The most direct way to find the sum of the nth roots of unity is as follows. Let x = ω₀ + ⋯ + ω_{n − 1} be the sum of all n of the roots of unity. Since roots of unity have unit length, since ω_iⁿ = 1 for all i, and since 1/ω_i is also an nth root of unity, ω_ix must be exactly x (ω_i just permutes the components of the sum). That is, (1 − ω_i)x = 0, which means either x = 0 or ω_i = 1, and because we can choose an i such that ω_i ≠ 1, we conclude that x = 0.

3. By representation theory

The direct way reminded me of one-dimensional representations of the cyclic group C_n. Take an irreducible complex one-dimensional representation ρ of C_n = ⟨α⟩, and first note that the value of ρ_α completely characterizes ρ since ρ_α^k = (ρ_α)^k.

One can show that all of the irreducible representations of C_n are one-dimensional from the fact that C_n is abelian, and furthermore, using the fact that the sum of the squares of the dimensions of the irreducible representations is equal to |C_n|, we can conclude that each root of unity corresponds to a different irreducible representation of C_n.

Next, we compute x = 1 + α + α² + ⋯ + α^{n − 1}. Since ρ_x is an endomorphism of complex representations from irreducible ρ, it must be some constant in C. We can see, as before, that 1 − α is in the kernel of ρ_x, so either ρ is the trivial representation or x = 0, and since we can choose ρ not to be trivial, this completes the proof.

4. By polynomial theory

This kind of method should be well known to anyone who did high-school competition math. We take the polynomials f(x) = xⁿ − 1 and g(x) = (x − ω₀)(x − ω₁)⋯(x − ω_{n − 1}) and note that they are the same because they are both degree n, have the same n roots, and are monic. We use the fact they are monic by seeing that dividing each polynomial through by x − ω_i results in a monic polynomial, and so dividing through by all such monomials will result in 1 for each of them, proving f = g. Now, the x^{n − 1} term has 0 as a coefficient according to f and −ω₀ − ω₁ − ⋯ − ω_{n − 1} according to g, which completes the proof.

5. By Shafarevich’s identity

This method is the reason I thought about any of this to begin with. It’s not very direct, but I think it is kind of interesting. First, we need the identity from the end of section 2.4 of Discourses on Algebra, which I will derive here. To do this, we will develop the theory of the interpolating polynomial. The idea is that, given a set of n + 1 values v_i at distinct values x_i, what is the nth degree polynomial f with f(x_i) = v_i for all i?

Let F(x) = (x − x₁)(x − x₂)⋯(x − x_{n + 1}), and let F_i(x) be the polynomial obtained by dividing F through by x − x_i. We define

so that f_i(x_i) = 1 and f_i(x_j) = 0 when i ≠ j. This provides a convenient basis for polynomials: observe that

is the interpolating polynomial as discussed.

If k is a natural number not exceeding n, if we have v_i = x_i^k, then since f(x) must be x^k, we have

and since each F_i is a monic degree-n polynomial, this means that if k < n,

and if k = n, then

These are the identities at the end of section 2.4.

We won’t be using these identities directly, but rather in spirit^[1]. We let x_i = ω_i and v_i = 1, where ω_i is an nth root of unity (with ω₁ = 1). Notice that each F_i is an (n − 1)-degree polynomial, so since both f(x) and g(x) = 1 are at-most-(n − 1)-degree polynomials which agree at n points, f = g. By taking the leading x^{n − 1} coefficient, this means

Notice that F_i(ω_i) is the product of all ω_i − ω_j for j ≠ i, and ω_i − ω_j = ω_i(1 − ω_i⁻¹ω_j). Then, we can see F_i(ω_i) = ω_i^{n − 1}F₁(ω₁) since ω_i⁻¹ is just permuting the roots of unity. Because (ω_i^{n − 1})⁻¹ = ω_i, we have

and so ω₁ + ⋯ + ω_n = 0.