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Roots of unity

Numbers · level 8, lesson 3 of 6

A question to start

Which numbers, cubed, give 1?

Among the real numbers there is only one: \(1\) itself. (\(2^3 = 8\) is too big, \(0.5^3\) too small, and any negative number cubed is negative.) But try this complex number:

\[\omega = -\frac12 + \frac{\sqrt3}{2}i.\]

Square it: \(\omega^2 = \tfrac14 - \tfrac{\sqrt3}{2}i + \tfrac34 i^2 = -\tfrac12 - \tfrac{\sqrt3}{2}i\). Multiply by \(\omega\) once more: \(\omega^3 = \left(-\tfrac12 - \tfrac{\sqrt3}{2}i\right)\left(-\tfrac12 + \tfrac{\sqrt3}{2}i\right) = \tfrac14 + \tfrac34 = 1\). So 1 has at least three cube roots: \(1\), \(\omega\) and \(\omega^2\). Plot them and they are the corners of an equilateral triangle. Is that luck? This lesson shows that 1 has exactly \(n\) different \(n\)th roots for every \(n\), and that they always sit at the corners of a regular polygon.

The idea

Solving \(z^n = 1\)

A complex number \(z\) with \(z^n = 1\) is called an \(n\)th root of unity. ("Unity" is an old word for the number 1.) Polar form and De Moivre's theorem find all of them.

Write \(z = r\operatorname{cis}\theta\). Then \(z^n = r^n\operatorname{cis}(n\theta)\), and we want this to equal \(1 = 1\operatorname{cis}0\).

When are two numbers in polar form equal? When they have the same length, and angles that point the same way, that is, angles that differ by a whole number of full turns. So we need

\[r^n = 1 \qquad\text{and}\qquad n\theta = 2\pi k \text{ for some integer } k.\]

Since \(r\) is a positive real number, \(r^n = 1\) forces \(r = 1\): every root of unity lies on the unit circle. And \(\theta = \dfrac{2\pi k}{n}\). Taking \(k = 0, 1, 2, \ldots, n - 1\) gives \(n\) different angles between 0 and \(2\pi\). Taking \(k = n\) gives angle \(2\pi\), the same point as \(k = 0\), and from then on the points repeat. So:

The equation \(z^n = 1\) has exactly \(n\) solutions: \(\;z = \operatorname{cis}\dfrac{2\pi k}{n}\) for \(k = 0, 1, \ldots, n - 1\).

They form a regular polygon

The roots all lie on the unit circle, and neighbouring roots are separated by the same angle \(\dfrac{2\pi}{n}\). Points equally spaced round a circle are the corners of a regular polygon. One corner is always at \(1\).

1 ω = cis 120° ω² = cis 240° cis 60° cis 300° −1

The six sixth roots of unity make a regular hexagon. Every other one of them is a cube root of unity (an angle of \(120°\) is two steps of \(60°\)), and those three make the equilateral triangle from the opening question.

The roots are powers of one root. Let \(\omega = \operatorname{cis}\dfrac{2\pi}{n}\), the first root after 1 going anticlockwise. By De Moivre, \(\omega^k = \operatorname{cis}\dfrac{2\pi k}{n}\), so the \(n\) roots are simply

\[1, \ \omega, \ \omega^2, \ \ldots, \ \omega^{n-1}.\]

Each multiplication by \(\omega\) turns one step round the polygon. (\(\omega\) is the Greek letter omega.)

The cube roots of unity

For \(n = 3\), \(\omega = \operatorname{cis}\dfrac{2\pi}{3} = -\tfrac12 + \tfrac{\sqrt3}{2}i\) and \(\omega^2 = \operatorname{cis}\dfrac{4\pi}{3} = -\tfrac12 - \tfrac{\sqrt3}{2}i\). Some facts worth knowing:

  • \(\omega^2 = \bar\omega\): the two non-real roots are conjugates (as they must be, since \(z^3 - 1\) has real coefficients; level 7, lesson 5).
  • \(\omega^2 = \dfrac1\omega\), since \(\omega \cdot \omega^2 = \omega^3 = 1\).
  • \(1 + \omega + \omega^2 = 0\).

The last fact also comes from algebra. Factor: \(z^3 - 1 = (z - 1)(z^2 + z + 1)\). The root \(z = 1\) comes from the first bracket; \(\omega\) and \(\omega^2\) are the roots of \(z^2 + z + 1 = 0\), which gives \(z = \dfrac{-1 \pm \sqrt{-3}}{2}\), the same two numbers. And \(\omega^2 + \omega + 1 = 0\) is exactly that equation.

The roots of unity add to zero

Theorem. For \(n \ge 2\), the \(n\)th roots of unity add to 0:

\[1 + \omega + \omega^2 + \cdots + \omega^{n-1} = 0.\]

Proof 1 (a geometric series). The sum is a geometric series with ratio \(\omega\). Since \(\omega \ne 1\) (because \(n \ge 2\)), the sum formula from Patterns applies, and it works for complex numbers because its proof uses only algebra:

\[1 + \omega + \cdots + \omega^{n-1} = \frac{\omega^n - 1}{\omega - 1} = \frac{1 - 1}{\omega - 1} = 0.\]

Proof 2 (a turn of the polygon). Call the sum \(S\). Multiplying every root by \(\omega\) moves each corner of the polygon one step round, so the list \(\omega, \omega^2, \ldots, \omega^n = 1\) is the same set of numbers in a different order. So \(\omega S = S\), which means \((\omega - 1)S = 0\). Since \(\omega \ne 1\), \(S = 0\). ∎

Proof 2 has a physical picture: \(n\) equal forces pulling outwards from the centre of a regular polygon towards its corners balance perfectly. Nothing moves.

Solving \(z^n = w\) for any \(w\)

The same method solves \(z^n = w\) for any non-zero complex \(w = R\operatorname{cis}\phi\). We need \(r^n = R\) and \(n\theta = \phi + 2\pi k\), so

\[z = R^{1/n}\operatorname{cis}\frac{\phi + 2\pi k}{n}, \qquad k = 0, 1, \ldots, n - 1.\]

Every non-zero complex number has exactly \(n\) different \(n\)th roots. They sit on a circle of radius \(R^{1/n}\), at the corners of a regular \(n\)-gon, one corner at angle \(\dfrac\phi n\). Equivalently: take any one root and multiply it by each of the \(n\)th roots of unity.

This is why the symbol \(\sqrt[n]{w}\) is slippery for complex \(w\): there is no single obvious choice among the \(n\) roots. It also explains the warning in level 7 about \(\sqrt a\sqrt b\).

A famous application

In 1796, the 18-year-old Carl Friedrich Gauss showed that a regular 17-sided polygon can be constructed with only a straight edge and compasses, something nobody had known for two thousand years. His method studied the 17th roots of unity and how they can be grouped and added. Today roots of unity also drive the fast Fourier transform, the algorithm that splits sound and signals into frequencies inside phones and computers.

Worked examples

1. Fourth roots of unity. Solve \(z^4 = 1\).

\(z = \operatorname{cis}\dfrac{2\pi k}{4}\) for \(k = 0, 1, 2, 3\): angles \(0\), \(\tfrac\pi2\), \(\pi\), \(\tfrac{3\pi}2\). So \(z = 1, i, -1, -i\): a square. These are the four values of the cycle of powers of \(i\) from level 7. Check: \(1 + i - 1 - i = 0\). ✓

2. Sixth roots of unity. List them in the form \(a + bi\).

Angles \(0°, 60°, 120°, 180°, 240°, 300°\):

\[1, \quad \tfrac12 + \tfrac{\sqrt3}{2}i, \quad -\tfrac12 + \tfrac{\sqrt3}{2}i, \quad -1, \quad -\tfrac12 - \tfrac{\sqrt3}{2}i, \quad \tfrac12 - \tfrac{\sqrt3}{2}i.\]

They pair off into opposites (\(z\) and \(-z\)), so they clearly add to 0.

3. Cube roots of \(8i\). Solve \(z^3 = 8i\).

  1. \(8i = 8\operatorname{cis}\tfrac\pi2\), so \(R = 8\), \(\phi = 90°\).
  2. Length of each root: \(8^{1/3} = 2\). Angles: \(\dfrac{90° + 360°k}{3} = 30°, 150°, 270°\).
  3. Roots: \(2\operatorname{cis}30° = \sqrt3 + i\), \(2\operatorname{cis}150° = -\sqrt3 + i\), \(2\operatorname{cis}270° = -2i\).
  4. Check the easiest: \((-2i)^3 = -8i^3 = -8(-i) = 8i\). ✓

4. Fourth roots of \(-16\). Solve \(z^4 = -16\).

\(-16 = 16\operatorname{cis}180°\). Length \(16^{1/4} = 2\); angles \(\dfrac{180° + 360°k}{4} = 45°, 135°, 225°, 315°\). So \(z = \sqrt2 + \sqrt2 i\), \(-\sqrt2 + \sqrt2 i\), \(-\sqrt2 - \sqrt2 i\), \(\sqrt2 - \sqrt2 i\): a square tilted by 45°.

5. An exact value from a pentagon. Use the fifth roots of unity to find \(\cos72°\) exactly.

  1. The fifth roots are at \(0°, 72°, 144°, 216°, 288°\), and they add to 0.
  2. Real parts: \(1 + \cos72° + \cos144° + \cos216° + \cos288° = 0\). Since \(\cos288° = \cos72°\) and \(\cos216° = \cos144°\), this is \(1 + 2\cos72° + 2\cos144° = 0\).
  3. Let \(c = \cos72°\). By the double-angle formula (lesson 2), \(\cos144° = 2c^2 - 1\). So \(1 + 2c + 4c^2 - 2 = 0\), that is \(4c^2 + 2c - 1 = 0\).
  4. \(c = \dfrac{-2 \pm \sqrt{4 + 16}}{8} = \dfrac{-1 \pm \sqrt5}{4}\). Since \(\cos72° > 0\), \(\cos72° = \dfrac{\sqrt5 - 1}{4} \approx 0.309\). (A calculator agrees.) The golden ratio is hiding here: \(\dfrac{\sqrt5 - 1}{2}\) is \(\dfrac{1}{\varphi}\).

Try it

  1. Check that \(\omega^2 = -\tfrac12 - \tfrac{\sqrt3}{2}i\) also cubes to 1, by computing \((\omega^2)^3\) two ways.
  2. Find all eighth roots of unity in the form \(a + bi\).
  3. Solve \(z^3 = -27\).
  4. Solve \(z^2 = 2i\) using polar form, and check one answer by squaring.
  5. Solve \(z^6 = -64\).
  6. If \(\omega\) is a non-real cube root of unity, simplify \(\omega^{100} + \omega^{200}\).
  7. Show that \((1 + \omega)^3 = -1\) for a non-real cube root of unity \(\omega\). (Hint: \(1 + \omega = -\omega^2\).)
  8. Find the product of all the fourth roots of unity, and of all the cube roots of unity. Can you guess the rule?
Answers
  1. \((\omega^2)^3 = (\omega^3)^2 = 1^2 = 1\). Directly: \(\omega^2 = \operatorname{cis}240°\), cubed is \(\operatorname{cis}720° = 1\). ✓
  2. Angles \(0°, 45°, \ldots, 315°\): \(1\), \(\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2}i\), \(i\), \(-\tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2}i\), \(-1\), \(-\tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2}i\), \(-i\), \(\tfrac{\sqrt2}{2} - \tfrac{\sqrt2}{2}i\).
  3. \(-27 = 27\operatorname{cis}180°\). Length 3, angles \(60°, 180°, 300°\): \(\tfrac32 + \tfrac{3\sqrt3}{2}i\), \(-3\), \(\tfrac32 - \tfrac{3\sqrt3}{2}i\).
  4. \(2i = 2\operatorname{cis}90°\). Length \(\sqrt2\), angles \(45°\) and \(225°\): \(z = 1 + i\) or \(z = -1 - i\). Check: \((1 + i)^2 = 1 + 2i - 1 = 2i\). ✓
  5. \(-64 = 64\operatorname{cis}180°\). Length \(64^{1/6} = 2\), angles \(30° + 60°k\): \(\sqrt3 + i\), \(2i\), \(-\sqrt3 + i\), \(-\sqrt3 - i\), \(-2i\), \(\sqrt3 - i\).
  6. \(\omega^3 = 1\), so only the remainder of the exponent after dividing by 3 matters. \(100 = 99 + 1\) and \(200 = 198 + 2\), so the sum is \(\omega + \omega^2 = -1\) (because \(1 + \omega + \omega^2 = 0\)).
  7. From \(1 + \omega + \omega^2 = 0\), \(1 + \omega = -\omega^2\). So \((1 + \omega)^3 = -\omega^6 = -(\omega^3)^2 = -1\).
  8. Fourth roots: \(1 \times i \times (-1) \times (-i) = -1\) (since \(i \times (-i) = 1\)). Cube roots: \(1 \times \omega \times \omega^2 = \omega^3 = 1\). The rule: the product of the \(n\)th roots of unity is \((-1)^{n+1}\), that is \(1\) for odd \(n\) and \(-1\) for even \(n\). (For even \(n\), the roots pair off into conjugates with product 1, except \(1\) and \(-1\).)

Common mistake

Stopping after one root. \(z^3 = 8\) does not just mean \(z = 2\). There are three cube roots, \(2\), \(2\omega\) and \(2\omega^2\), and in the complex numbers all three count. Always look for \(n\) roots of an \(n\)th power equation.

Dividing only the first angle by \(n\). In \(z^3 = 8i\), it is tempting to find \(30°\) and then add \(360°\) to get more roots. But \(30° + 360°\) is the same point. Add the full turn before dividing: \(\dfrac{90° + 360°k}{3}\), so the roots are \(120°\) apart.

Where it leads