The Roots of Unity

Three cube roots of 1

Solve $z^{3} = 1$. The real number 1 is one answer. There are two more, off the real axis: $- \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)i$ and $- \frac{1}{2} - \left(\frac{\sqrt{3}}{2}\right)i$. All three are 1 from the origin, so they lie on the unit circle, and they are a third of a turn apart.

Factoring confirms the count. $z^{3} - 1 = \left(z - 1\right)\left(z^{2} + z + 1\right)$. The quadratic $z^{2} + z + 1 = 0$ has discriminant $1^{2} - 4 \times 1 \times 1 = -3$, so its roots are $\frac{{-1 \pm \sqrt{3} i}}{2}$, the two roots off the real axis.

The three cube roots of 1 on the unit circle: 1, $\omega$ at 120° and $\omega^{2}$ at −120°, a third of a turn apart. Joined, they make an equilateral triangle.

Why cubing lands on 1

Call the root at $120^\circ \omega$, the Greek letter omega: $\omega = \cos 120^\circ + i \sin 120^\circ = - \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)i$. By De Moivre’s theorem, $\omega^{3} = \cos 360^\circ + i \sin 360^\circ$. 360° is a full turn, which points along the positive real axis, so $\omega^{3} = 1$.

$\omega^{2} = \cos 240^\circ + i \sin 240^\circ$. The turn of 240° ends below the real axis, and in the range $-180^\circ < \theta \le 180^\circ$ that direction is 240° − 360° = −120°. So $\omega^{2} = - \frac{1}{2} - \left(\frac{\sqrt{3}}{2}\right)i$, the third root, and it is the conjugate of $\omega$. Cubed, $\omega^{2}$ turns through 3 × 240° = 720°, two full turns, so $\left(\omega^{2}\right)^{3} = 1$ as well.

The powers of $\omega$ repeat every three: $\omega^{3} = 1$, so $\omega^{4} = \omega$ and $\omega^{5} = \omega^{2}$.

Finding every root

Take any cube root of 1, z = r at $\theta$. By De Moivre, $z^{3} = r^{3}$ at $3\theta$, and this must be 1, which is 1 at 0°. The lengths must agree, so $r^{3} = 1$, and r = 1 because a length is a positive real number. The directions must agree, so $3\theta$ is a whole number of full turns: $3\theta = 0^\circ$, 360° or 720°, and $\theta = 0^\circ$, 120° or 240°.

The next whole number of turns gives nothing new: $3\theta = 1080^\circ$ makes $\theta = 360^\circ$, which is 1 again. Negative turns repeat the same three directions too. So $z^{3} = 1$ has exactly three roots.

The same argument works for $z^{n} = 1$. The length gives r = 1, and the direction gives $n\theta = 360^\circ k$, so $\theta = \frac{360^\circ k}{n}$ for k = 0, 1, …, n − 1. That is n roots, all on the unit circle, spaced $\frac{360^\circ}{n}$ apart and starting at 1.

Fourth and fifth roots

For $z^{4} = 1$ the spacing is 360° ÷ 4 = 90°, and the roots are 1, i, −1 and −i, the four powers of i. Each one to the fourth power is 1: $i^{4} = 1$, $\left(-1\right)^{4} = 1$ and $\left(-i\right)^{4} = 1$.

For $z^{5} = 1$ the spacing is 360° ÷ 5 = 72°. Going round from 1, the roots are at turns of 0°, 72°, 144°, 216° and 288°. As arguments in the range they are 0°, 72°, 144°, −144° and −72°. The first past 1 is cos 72° + i sin 72°, about 0.309 + 0.951i, and the one at 144° is about −0.809 + 0.588i.

The five fifth roots of 1, each labeled with its argument. They sit 72° apart round the unit circle, starting at 1.

The roots add to 0

For the cube roots, $1 + \omega + \omega^{2} = 1 + \left(- \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)i\right) + \left(- \frac{1}{2} - \left(\frac{\sqrt{3}}{2}\right)i\right) = 0$. That is the quadratic $z^{2} + z + 1 = 0$ at $z = \omega$, read the other way.

For the fourth roots, 1 + i + (−1) + (−i) = 0. The n roots of $z^{n} = 1$ add to 0 for every n from 2 up; Regular Polygons from the Roots of Unity proves it.

The usual mistakes

Counting only the real root. 1 is a cube root of 1, but so are $\omega$ and $\omega^{2}$, off the real axis: $z^{3} = 1$ has three roots.

Giving $a \pm$ pair. That is what a square root gives; $\left(-1\right)^{3} = -1$, so −1 is not a cube root of 1.

Spacing the cube roots 90° or 180° apart. Three roots share the full turn, so they are 360° ÷ 3 = 120° apart.

Putting $\omega$ at 60°. 3 × 60° = 180°, so (1 at $60^\circ)^{3} = -1$, not 1.

Listing 360° as another root. It is the same direction as 0°, the root 1; $z^{3} = 1$ has three roots, not four.