The Roots of Unity

Three cube roots of 1

Solve $z^3=1$z³ = 1. The real number 1 is one answer. There are two more, off the real axis: $-\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i$−1/2 + (√3/2)i and $-\frac{1}{2}-\left(\frac{\sqrt{3}}{2}\right)i$−1/2 − (√3/2)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. $z^3-1=\left(z-1\right)\left(z^2+z+1\right)$z³ − 1 = (z − 1)(z² + z + 1). The quadratic $z^2+z+1=0$z² + z + 1 = 0 has discriminant $1^2-4\times 1\times 1=-3$1² − 4 × 1 × 1 = −3, so its roots are $\frac{-1\pm \sqrt{3}i}{2}$(−1 ± √3 i)/2, the two roots off the real axis.

The three cube roots of 1 on the unit circle: 1, $\omega$ω at 120° and ${\omega }^2$ω² 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$120° ω, the Greek letter omega: $\omega =\cos 120^{\circ }+i\sin 120^{\circ }=-\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i$ω = cos 120° + i sin 120° = −1/2 + (√3/2)i. By De Moivre’s theorem, ${\omega }^3=\cos 360^{\circ }+i\sin 360^{\circ }$ω³ = cos 360° + i sin 360°. 360° is a full turn, which points along the positive real axis, so ${\omega }^3=1$ω³ = 1.

${\omega }^2=\cos 240^{\circ }+i\sin 240^{\circ }$ω² = cos 240° + i sin 240°. The turn of 240° ends below the real axis, and in the range $-180^{\circ }<\theta \le 180^{\circ }$−180° < θ ≤ 180° that direction is 240° − 360° = −120°. So ${\omega }^2=-\frac{1}{2}-\left(\frac{\sqrt{3}}{2}\right)i$ω² = −1/2 − (√3/2)i, the third root, and it is the conjugate of $\omega$ω. Cubed, ${\omega }^2$ω² turns through 3 × 240° = 720°, two full turns, so ${\left({\omega }^2\right)}^3=1$(ω²)³ = 1 as well.

The powers of $\omega$ω repeat every three: ${\omega }^3=1$ω³ = 1, so ${\omega }^4=\omega$ω⁴ = ω and ${\omega }^5={\omega }^2$ω⁵ = ω².

Finding every root

Take any cube root of 1, z = r at $\theta$θ. By De Moivre, $z^3=r^3$z³ = r³ at $3\theta$3θ, and this must be 1, which is 1 at 0°. The lengths must agree, so $r^3=1$r³ = 1, and r = 1 because a length is a positive real number. The directions must agree, so $3\theta$3θ is a whole number of full turns: $3\theta =0^{\circ }$3θ = 0°, 360° or 720°, and $\theta =0^{\circ }$θ = 0°, 120° or 240°.

The next whole number of turns gives nothing new: $3\theta =1080^{\circ }$3θ = 1080° makes $\theta =360^{\circ }$θ = 360°, which is 1 again. Negative turns repeat the same three directions too. So $z^3=1$z³ = 1 has exactly three roots.

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

Fourth and fifth roots

For $z^4=1$z⁴ = 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: $i^4=1$i⁴ = 1, ${\left(-1\right)}^4=1$(−1)⁴ = 1 and ${\left(-i\right)}^4=1$(−i)⁴ = 1.

For $z^5=1$z⁵ = 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$1 + ω + ω² = 1 + (−1/2 + (√3/2)i) + (−1/2 − (√3/2)i) = 0. That is the quadratic $z^2+z+1=0$z² + z + 1 = 0 at $z=\omega$z = ω, read the other way.

For the fourth roots, 1 + i + (−1) + (−i) = 0. The n roots of $z^n=1$zⁿ = 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 $\omega$ω and ${\omega }^2$ω², off the real axis: $z^3=1$z³ = 1 has three roots.

Giving $a\pm$a ± pair. That is what a square root gives; ${\left(-1\right)}^3=-1$(−1)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 $\omega$ω at 60°. 3 × 60° = 180°, so (1 at $60^{\circ }{)}^3=-1$60°)³ = −1, not 1.

Listing 360° as another root. It is the same direction as 0°, the root 1; $z^3=1$z³ = 1 has three roots, not four.