Roots of Unity

Over the real numbers, x^n = 1 has at most two solutions (1, and -1 when n is even). Over the complex numbers it always has exactly n of them — the nth roots of unity — and they sit in a beautiful pattern: equally spaced points around the unit circle. Finding them is a direct application of De Moivre's theorem.

Solving zⁿ = 1

Step 1 — write z in polar form. Any solution lies on the unit circle (its modulus must be 1, since |z|^n = 1), so write z = e^{i\theta}:

z^n = \left(e^{i\theta}\right)^n = e^{in\theta}.

Step 2 — set it equal to 1. The number 1 is e^{i\cdot 0}, but it is also e^{i2\pi k} for every integer k — one full turn lands you back at 1. That extra freedom is where the many roots come from:

e^{in\theta} = e^{i 2\pi k}.

Step 3 — equate the angles and solve for \theta. Matching exponents gives n\theta = 2\pi k, so

\theta = \frac{2\pi k}{n}.

Step 4 — list the distinct roots. Taking k = 0, 1, 2, \dots, n - 1 gives n different angles (after that they repeat). The nth roots of unity are

\omega_k = e^{2\pi i k / n}, \qquad k = 0, 1, \dots, n - 1.

These are n equally spaced points on the unit circle, separated by \tfrac{2\pi}{n} each — the vertices of a regular n-gon, with \omega_0 = 1 always one of them.

For each integer n \ge 1:

z^n = 1 has exactly n complex solutions, \omega_k = e^{2\pi i k/n} for k = 0, \dots, n - 1;
they are the vertices of a regular n-gon inscribed in the unit circle, one vertex at 1;
for n \ge 2 they sum to zero: \sum_{k=0}^{n-1}\omega_k = 0;
the nth roots of a general w = \rho e^{i\varphi} are z = \rho^{1/n}\,e^{i(\varphi + 2\pi k)/n}.

Roots of any complex number

The same idea finds the nth roots of any w = \rho e^{i\varphi}, not just 1. Solve z^n = w: the modulus must satisfy |z|^n = \rho so |z| = \rho^{1/n}, and the angles spread out exactly as before:

z = \rho^{1/n}\,e^{i(\varphi + 2\pi k)/n}, \qquad k = 0, 1, \dots, n - 1.

Worked example: the cube roots of unity

Find all solutions of z^3 = 1. Here n = 3, so the angles are \theta = \tfrac{2\pi k}{3} for k = 0, 1, 2.

Step 1 — k = 0: angle 0:

\omega_0 = e^{0} = 1.

Step 2 — k = 1: angle \tfrac{2\pi}{3} = 120°:

\omega_1 = e^{2\pi i/3} = \cos 120° + i\sin 120° = -\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i.

Step 3 — k = 2: angle \tfrac{4\pi}{3} = 240°:

\omega_2 = e^{4\pi i/3} = \cos 240° + i\sin 240° = -\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}i.

Step 4 — check they sum to zero. The imaginary parts cancel and the real parts give 1 - \tfrac12 - \tfrac12 = 0:

\omega_0 + \omega_1 + \omega_2 = 1 + \left(-\tfrac12 + \tfrac{\sqrt3}{2}i\right) + \left(-\tfrac12 - \tfrac{\sqrt3}{2}i\right) = 0.

Three points 120° apart — the corners of an equilateral triangle on the unit circle.

For n \ge 2 the nth roots of unity always add up to 0. One slick reason: they are the roots of z^n - 1 = 0, a polynomial whose z^{n-1} coefficient is 0. By the relationship between roots and coefficients, the sum of the roots equals minus that coefficient — which is 0. Geometrically: the vectors to n equally spaced points on a circle cancel by symmetry, like the spokes of a wheel pulling equally in every direction.

Build the n-gon

Set n with the slider. The diagram plots all n roots \omega_k = e^{2\pi i k/n} on the unit circle and joins them into a regular polygon. One vertex always sits at 1; the rest are evenly spaced by \tfrac{360°}{n}.