Sums and symmetry

Let’s continue our exploration of roots of unity. Recall that for any positive integer $n$ , there are $n$ complex numbers, evenly spaced around the unit circle, whose $n$ th power is equal to $1$ . These are called the $n$ th roots of unity.

Today let’s answer a simple question: what happens if you add all the $n$ th roots of unity?

For example, the third roots of unity are $1$ and $-1/2 \pm i\sqrt{3}/2$ . If we sum them, we get

$\displaystyle 1 + (-1/2 + i\sqrt{3}/2) + (-1/2 - i \sqrt{3}/2) = 1 - 1 = 0$

Notice how the two $\pm i \sqrt{3}/2$ terms cancel, and then we are left with $1 - 1/2 - 1/2 = 0$ .

What about the fourth roots of unity? That’s easy:

$\displaystyle 1 + i + (-1) + (-i) = 0.$

Hmm, zero again. Will we always get zero?

In fact, we will, because of symmetry. Here’s one way to think about it. Adding complex numbers is like adding vectors: we can think of a complex number as trying to “pull” the origin toward itself. Adding complex numbers means they “pull” the origin in multiple directions at the same time. Since the $n$ th roots of unity are evenly spaced around the unit circle, all their “pulls” exactly cancel out. It’s like having a bunch of people evenly spaced around the edge of a circular parachute, all pulling at the same time; the parachute doesn’t go anywhere since it is being pulled equally in all directions. (Of course the parachute could rip, but metaphors can only take you so far…)

Photo by dolanh on Flickr, CC BY-NC-SA 2.0

We can see this a bit more formally as follows. Consider the sum

$\displaystyle S = 1 + \omega_n^1 + \omega_n^2 + \dots + \omega_n^{n-1}.$

If we multiply both sides by $\omega_n$ , we can distribute $\omega_n$ over the sum on the right side, and we get

$\displaystyle \omega_n S = \omega_n^1 + \omega_n^2 + \omega_n^3 + \dots + \omega_n^n.$

But $\omega_n^n = 1$ , so the right-hand side is really the same as $S$ . That is, $\omega_n S = S$ . If we subtract $S$ from both sides and factor out $S$ , we get $(\omega_n - 1) S = 0$ . The only way for this to be true is if either $\omega_n - 1 = 0$ or $S = 0$ . $S = 0$ is what we were trying to show, but what’s this $\omega_n - 1 = 0$ possibility? Well, $\omega_n = 1$ only happens when $n = 1$ . When $n = 1$ , we have only a single dot:

In this case, the sum is actually $1$ , not zero. So when $n = 1$ , the sum of all $n$ th roots is $1$ , and for any larger $n$ the sum is $0$ :

$latex \displaystyle \sum_{k=0}^{n-1} \omega_n^k = \begin{cases} 1 \qquad n=1 \\ 0
\qquad n > 1 \end{cases}$

So what did this proof have to do with symmetry? Well, remember that complex multiplication corresponds to rotation. In particular, multiplying by $\omega_n$ is the same as rotating by $1/n$ of a full circle. The equation $\omega_n S = S$ then follows from the fact that the roots of unity have rotational symmetry: if we rotate all of them by $1/n$ of a turn, we end up with exactly the same set of points, so their sum must still be the same. And the only number that doesn’t change when you rotate it is—you guessed it—zero. Unless you rotate by a complete turn, that is, which is what happens when $n = 1$ .

Next time, we’ll try adding up all the primitive $n$ th roots of unity—something much more interesting happens!

About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

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