Regular Polygons from the Roots of Unity

Six roots, six corners

The six roots of $z^6=1$z⁶ = 1 lie on the unit circle, 360° ÷ 6 = 60° apart, starting at 1. Points spaced equally round a circle are the vertices of a regular polygon. Neighbors are joined by equal arcs, so the chords between them, the sides, are equal; and each vertex sees the same picture of its two neighbors, so the angles are equal too.

So the six roots are the corners of a regular hexagon. In general the n roots of $z^n=1$zⁿ = 1 are the corners of a regular n-gon inscribed in the unit circle, with one corner at 1.

The six roots of $z^6=1$z⁶ = 1, joined in order into a regular hexagon. With $\omega =\cos 60^{\circ }+i\sin 60^{\circ }$ω = cos 60° + i sin 60°, the corners going counterclockwise from 1 are 1, $\omega$ω, ${\omega }^2$ω², ${\omega }^3$ω³, ${\omega }^4$ω⁴ and ${\omega }^5$ω⁵.

Coordinates without measuring

Corner k is a turn of 60k° from 1 and is 1 from the origin, so it is cos 60k° + i sin 60k°. The first past 1 is $\omega =\cos 60^{\circ }+i\sin 60^{\circ }=\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i$ω = cos 60° + i sin 60° = 1/2 + (√3/2)i, with $\frac{\sqrt{3}}{2}\approx 0.866$√3/2 ≈ 0.866.

All six: 1, $\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i$1/2 + (√3/2)i, $-\frac{1}{2}+\left(\frac{\sqrt{3}}{2}\right)i$−1/2 + (√3/2)i, −1, $-\frac{1}{2}-\left(\frac{\sqrt{3}}{2}\right)i$−1/2 − (√3/2)i and $\frac{1}{2}-\left(\frac{\sqrt{3}}{2}\right)i$1/2 − (√3/2)i. Their arguments are 0°, 60°, 120°, 180°, −120° and −60°. The real and imaginary parts are the coordinates of the corners.

The corners add to 0

Add the six roots: $S=1+\omega +{\omega }^2+{\omega }^3+{\omega }^4+{\omega }^5$S = 1 + ω + ω² + ω³ + ω⁴ + ω⁵. This is a geometric series with first term 1 and ratio $\omega$ω, so $S=\frac{{\omega }^6-1}{\omega -1}$S = (ω⁶ − 1)/(ω − 1). Since ${\omega }^6=1$ω⁶ = 1, the top is 0. The bottom is not 0, because $\omega \ne 1$ω ≠ 1. So S = 0.

There is a second way to see it. Multiplying by $\omega$ω turns every corner 60° on to the next one, so $\omega S$ωS is the same six numbers added in a different order, and $\omega S=S$ωS = S. Then $\left(\omega -1\right)S=0$(ω − 1)S = 0, and since $\omega \ne 1$ω ≠ 1, S = 0.

Both arguments work for every n from 2 up, with $\omega =\cos \left(\frac{360^{\circ }}{n}\right)+i\sin \left(\frac{360^{\circ }}{n}\right)$ω = cos(360°/n) + i sin(360°/n). For n = 1 the only root is 1, so the sum is 1: there $\omega =1$ω = 1, and both arguments would divide by $\omega -1=0$ω − 1 = 0. In the hexagon the corners also cancel in opposite pairs: 1 and −1, $\omega$ω and $-\omega$−ω, ${\omega }^2$ω² and $-{\omega }^2$−ω².

The length of a side

One side joins 1 to $\omega$ω, so its length is $|\omega -1|$|ω − 1|. The triangle with corners 0, 1 and $\omega$ω has two sides of length 1, the radii, and the angle between them at the origin is $\frac{360^{\circ }}{n}$360°/n.

The line from 0 to the middle of the side cuts that triangle into two right-angled triangles, each with hypotenuse 1 and angle $\frac{180^{\circ }}{n}$180°/n at the origin. Half the side is opposite that angle, so half the side is $\sin \left(\frac{180^{\circ }}{n}\right)$sin(180°/n), and the side is $2\sin \left(\frac{180^{\circ }}{n}\right)$2 sin(180°/n).

Hexagon, n = 6: 2 sin 30° = 1, the same as the radius. Square, n = 4: $2\sin 45^{\circ }=\sqrt{2}$2 sin 45° = √2, and $|i-1|=\sqrt{1+1}=\sqrt{2}$|i − 1| = √(1 + 1) = √2. Triangle, n = 3: $2\sin 60^{\circ }=\sqrt{3}$2 sin 60° = √3. Octagon, n = 8: $2\sin 22.5^{\circ }\approx 0.765$2 sin 22.5° ≈ 0.765.

The eighth roots of 1 make a regular octagon. The radii to 1 and to $\omega =\cos 45^{\circ }+i\sin 45^{\circ }$ω = cos 45° + i sin 45° are 1 long and 45° apart, and the side between their tips is 2 sin 22.5°, about 0.765.

A bigger circle

The roots of $z^n=R^n$zⁿ = Rⁿ, for a positive real R, lie on the circle of radius R. Write z = Rw: then $z^n=R^n{w}^n$zⁿ = Rⁿwⁿ, which equals $R^n$Rⁿ exactly when $w^n=1$wⁿ = 1. So the roots are R times the roots of unity, the same polygon scaled by R, and every length scales by R: the side is $R\times 2\sin \left(\frac{180^{\circ }}{n}\right)$R × 2 sin(180°/n).

$z^4=16=2^4$z⁴ = 16 = 2⁴, so R = 2, and the roots are 2, 2i, −2 and −2i. Each to the fourth power is 16; for example ${\left(2i\right)}^4=16i^4=16$(2i)⁴ = 16i⁴ = 16. They make a square with side $2\times 2\sin 45^{\circ }=2\sqrt{2}$2 × 2 sin 45° = 2√2, which is $|2i-2|=\sqrt{4+4}=2\sqrt{2}$|2i − 2| = √(4 + 4) = 2√2. The radius is 2 and the diagonal, from 2 to −2, is 4.

The roots of $z^4=16$z⁴ = 16 on the circle of radius 2. The square they make has side $2\sqrt{2}$2√2, about 2.83, and diagonal 4.

The usual mistakes

Miscounting the corners. $z^6=1$z⁶ = 1 has six roots, so a hexagon; a pentagon comes from $z^5=1$z⁵ = 1 and a square from $z^4=1$z⁴ = 1.

Giving the sum as 1 or 6. Only the first root is 1, and the six arrows point six different ways, so they cancel: the sum is 0.

Taking the number of roots as the spacing. The roots of $z^8=1$z⁸ = 1 are 360° ÷ 8 = 45° apart, not 8°.

Taking the radius or the diagonal for the side. For $z^4=16$z⁴ = 16 the radius is 2 and the diagonal 4; the side is $2\sqrt{2}$2√2.

Six bolts round a pipe flange

In the application below, the bolt circle has radius 60 mm, so each root of $z^6=1$z⁶ = 1 is scaled by 60, and the gap between neighboring holes is a side of the hexagon: 60 × 2 sin 30° = 60 mm. The application writes the angles in radians; $\frac{k\pi }{3}$kπ/3 is 60k°.