Complex Numbers

Lessons

Senior 6 · Mathematics

Complex Numbers

5 lessons

Lesson 4 of 5

nth roots, roots of unity and the exponential form

25 min

Three answers to $z^{3} = 1$

In $\mathbb{R}$ , the equation $z^{3} = 1$ has one solution, $z = 1$ . The fundamental theorem of algebra promises three in $\mathbb{C}$ . Factorise: $z^{3} - 1 = (z - 1) (z^{2} + z + 1)$ , and the quadratic gives $z = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}i$ . Each of these has modulus $\sqrt{\frac{1}{4} + \frac{3}{4}} = 1$ , so all three roots lie on the unit circle, and their arguments are $0$ , $\frac{2\pi}{3}$ and $-\frac{2\pi}{3}$ : they split the circle into three equal arcs. De Moivre's theorem explains this pattern and extends it to every $n$ .

Important

Theorem (the $n$ th roots of a complex number). Let $w = R(\cos\varphi + i\sin\varphi) \ne 0$ and let $n \ge 2$ be an integer. The equation $z^{n} = w$ has exactly $n$ distinct roots,

z_k = \sqrt[n]{R}\left[\cos\left(\frac{\varphi + 2k\pi}{n}\right) + i\sin\left(\frac{\varphi + 2k\pi}{n}\right)\right],

for $k = 0, 1, \ldots, n - 1$ .

Derivation: put $z = r(\cos\theta + i\sin\theta)$ . By De Moivre, $z^n = r^n(\cos n\theta + i\sin n\theta)$ . Two complex numbers in polar form are equal when their moduli are equal and their arguments differ by a multiple of $2\pi$ , so $r^{n} = R$ and $n\theta = \varphi + 2k\pi$ . Hence $r = \sqrt[n]{R}$ (the positive real root) and $\theta = \frac{\varphi + 2k\pi}{n}$ . The values $k = 0, 1, \ldots, n - 1$ give $n$ different angles in one turn; $k = n$ gives the angle of $k = 0$ plus $2\pi$ , the same point again.

Definition

$n$ th roots of unity: the $n$ solutions of $z^{n} = 1$ ,

\omega_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \qquad k = 0, 1, \ldots, n - 1 .

Notation: with $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$ , De Moivre gives $\omega_k = \omega^k$ , so the roots are $1, \omega, \omega^2, \ldots, \omega^{n-1}$ .

Example: the fourth roots of unity are $1, i, - 1, - i$ .

The cube roots of unity are equally spaced on the unit circle, $120^\circ$ apart, and are the vertices of an equilateral triangle.

Worked example 1: the cube roots of unity

Find the cube roots of unity, and show that $1 + \omega + \omega^2 = 0$ and $\bar{\omega} = \omega^2$ .

Solution With $n = 3$ : $\omega_0 = 1$ , $\omega_1 = \cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$ and $\omega_2 = \cos\frac{4\pi}{3} + i\sin\frac{4\pi}{3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i$ , the same as from the factorisation above.

Adding: $1 + \left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = 0$ . Also $\omega_2$ has the opposite imaginary part to $\omega_1$ , so $\omega^2 = \bar{\omega}$ . Finally $\omega^3 = 1$ , which is often used to simplify powers: $\omega^{100} = \omega^{99}\omega = (\omega^3)^{33}\omega = \omega$ .

Important

Properties of the $n$ th roots of unity ( $n \ge 2$ ):

They lie on the unit circle and are the vertices of a regular $n$ -gon with one vertex at $1$ : consecutive roots are $\frac{2\pi}{n}$ apart.
Their sum is 0. Proof: $1 + \omega + \cdots + \omega^{n-1}$ is a geometric series with ratio $\omega \ne 1$ , so it equals $\frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0$ .
The non-real roots come in conjugate pairs, $\overline{\omega^k} = \omega^{n-k}$ , because $z^{n} - 1$ has real coefficients.

The fifth roots of unity, $\omega^k = e^{\frac{2k\pi i}{5}}$ for $k = 0, 1, 2, 3, 4$ , are the vertices of a regular pentagon inscribed in the unit circle.

Worked example 2: the cube roots of $8 i$

Solve $z^{3} = 8 i$ , giving the roots in the form $a + b i$ .

Solution $8i = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$ , so $R = 8$ , $\varphi = \frac{\pi}{2}$ and

z_k = 2\left[\cos\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right) + i\sin\left(\frac{\pi}{6} + \frac{2k\pi}{3}\right)\right], \quad k = 0, 1, 2 .

$k = 0$ : $2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \sqrt{3} + i$ .
$k = 1$ : $2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right) = -\sqrt{3} + i$ .
$k = 2$ : $2\left(\cos\frac{3\pi}{2} + i\sin\frac{3\pi}{2}\right) = -2i$ .

Check: $(- 2 i)^{3} = - 8 i^{3} = 8 i$ ✓. The three roots also add to 0, as the roots of $z^{3} - 8 i = 0$ must, since that cubic has no $z^{2}$ term.

The cube roots of $8 i$ have modulus $\sqrt[3]{8} = 2$ . The first has argument $\frac{\pi}{6} = \frac{1}{3} \times \frac{\pi}{2}$ ; the others follow at intervals of $\frac{2\pi}{3}$ .

Constructing regular polygons

Multiplying by $\omega = \cos\frac{2\pi}{n} + i\sin\frac{2\pi}{n}$ is a rotation through $\frac{2\pi}{n}$ about O. So if a regular $n$ -gon has centre $c$ and one vertex $v$ , its vertices are

c + (v - c)\,\omega^k, \qquad k = 0, 1, \ldots, n - 1 .

Example: a square has centre $1 + i$ and one vertex $3 + i$ . Here $v - c = 2$ and $\omega = i$ , so the vertices are $1 + i + 2 i^{k}$ : $3 + i$ , $1 + 3 i$ , $- 1 + i$ and $1 - i$ . A draughtsman can use exactly these coordinates to set out a square or hexagonal plot on a grid.

Definition

Exponential form. For real $\theta$ , define

e^{i\theta} = \cos\theta + i\sin\theta \qquad \text{(Euler's formula)}.

Every non-zero complex number can then be written $z = re^{i\theta}$ , with $r = |z| \gt 0$ and $\theta = \arg z$ . For a complex exponent, $e^{a + bi} = e^a e^{bi}$ .

Example: $2i = 2e^{\frac{i\pi}{2}}$ , and $e^{2 + \frac{i\pi}{3}} = e^2\left(\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) \approx 3.69 + 6.40i$ .

Why this definition is sensible. The product rule of Lesson 3 says $(\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$ , which in the new notation reads $e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)}$ : the law of indices holds. De Moivre's theorem becomes $\left(e^{i\theta}\right)^n = e^{in\theta}$ . In Unit 3 (Series) you will see that the Maclaurin series of $e^{x}$ , $\cos x$ and $\sin x$ prove Euler's formula outright. With $\theta = \pi$ it gives Euler's identity, $e^{i\pi} + 1 = 0$ , which links five of the most important numbers in mathematics.

For real $\theta$ , $e^{i\theta}$ is the point of the unit circle at angle $\theta$ . Half a turn gives $e^{i\pi} = -1$ .

Important

Euler's formulae for cosine and sine. Adding and subtracting $e^{i\theta} = \cos\theta + i\sin\theta$ and $e^{-i\theta} = \cos\theta - i\sin\theta$ :

\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}, \qquad \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} .

Lesson 5 uses these to turn powers of $\cos\theta$ and $\sin\theta$ into sums of multiple angles.

Worked example 3: working in exponential form

(a) Write $- 1 - i$ in exponential form. (b) Hence find $(- 1 - i)^{8}$ . (c) Write the sixth roots of unity in exponential form.

Solution

$|-1 - i| = \sqrt{2}$ ; the point is in the third quadrant with $\alpha = \frac{\pi}{4}$ , so $\operatorname{Arg} = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$ and $-1 - i = \sqrt{2}\,e^{-\frac{3i\pi}{4}}$ .
$(-1 - i)^8 = (\sqrt{2})^8 e^{-6i\pi} = 16(\cos 6\pi - i\sin 6\pi) = 16$ . Check: $(- 1 - i)^{2} = 2 i$ and $(2 i)^{4} = 16$ ✓.
$e^{\frac{2k\pi i}{6}} = e^{\frac{k\pi i}{3}}$ for $k = 0, 1, \ldots, 5$ : $1, e^{\frac{i\pi}{3}}, e^{\frac{2i\pi}{3}}, -1, e^{\frac{4i\pi}{3}}, e^{\frac{5i\pi}{3}}$ , the vertices of a regular hexagon.

Warning

In $re^{i\theta}$ , $r$ must be positive. $-3e^{\frac{i\pi}{4}}$ is not in exponential form. Since $-1 = e^{i\pi}$ , it equals $3e^{\frac{5i\pi}{4}} = 3e^{-\frac{3i\pi}{4}}$ . The same rule holds for the polar form $r(\cos\theta + i\sin\theta)$ .

Tip

Check a set of $n$ th roots by adding them: the $n$ roots of $z^{n} = w$ always add up to 0 when $n \ge 2$ , because $z^{n} - w$ has no $z^{n-1}$ term. If your roots do not add to 0, one of them is wrong.

Eddie Woo: Roots of Unity, part 2 of 2, Insights from polar and exponential forms. Watch how the polar and exponential forms explain where the roots of unity lie on the unit circle.

✦

Practice

Draw an Argand diagram for every root-finding question.

Find the fourth roots of unity and plot them. What shape do they form?
Solve $z^{3} = - 27$ , giving the roots in the form $a + b i$ .
Solve $z^4 = -8 + 8\sqrt{3}i$ .
Let $\omega = e^{\frac{2i\pi}{5}}$ . Explain why $1 + \omega + \omega^2 + \omega^3 + \omega^4 = 0$ , and deduce that $\cos\frac{2\pi}{5} + \cos\frac{4\pi}{5} = -\frac{1}{2}$ .
Write in exponential form: (a) $- 2$ (b) $1 + i$ (c) $-\sqrt{3} - i$ .
Write in the form $a + b i$ : (a) $4e^{\frac{i\pi}{6}}$ (b) $e^{1 - \frac{i\pi}{2}}$ .
A regular hexagon has centre O and one vertex at $2 + 2 i$ . Find the two vertices next to it.
If $\omega$ is a non-real cube root of unity, simplify $(1 + \omega)(1 + \omega^2)$ and $\omega^{50} + \omega^{25}$ .

Show answersHide answers

$1, i, - 1, - i$ : the vertices of a square inscribed in the unit circle.
$-27 = 27e^{i\pi}$ , so $z = 3e^{i\left(\frac{\pi}{3} + \frac{2k\pi}{3}\right)}$ : $\frac{3}{2} + \frac{3\sqrt{3}}{2}i$ , $- 3$ , $\frac{3}{2} - \frac{3\sqrt{3}}{2}i$ .
$|-8 + 8\sqrt{3}i| = 16$ and the argument is $\frac{2\pi}{3}$ , so $z = 2e^{i\left(\frac{\pi}{6} + \frac{k\pi}{2}\right)}$ : $\sqrt{3} + i$ , $-1 + \sqrt{3}i$ , $-\sqrt{3} - i$ , $1 - \sqrt{3}i$ .
These are the five fifth roots of unity, whose sum is 0. Their real parts are $1, \cos\frac{2\pi}{5}, \cos\frac{4\pi}{5}, \cos\frac{6\pi}{5}, \cos\frac{8\pi}{5}$ , and $\cos\frac{6\pi}{5} = \cos\frac{4\pi}{5}$ , $\cos\frac{8\pi}{5} = \cos\frac{2\pi}{5}$ . So $1 + 2\cos\frac{2\pi}{5} + 2\cos\frac{4\pi}{5} = 0$ , which gives the result.
(a) $2e^{i\pi}$ (b) $\sqrt{2}\,e^{\frac{i\pi}{4}}$ (c) $2e^{-\frac{5i\pi}{6}}$ .
(a) $4\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = 2\sqrt{3} + 2i$ (b) $e \times e^{-\frac{i\pi}{2}} = -ei \approx -2.72i$ .
Multiply by $e^{\pm\frac{i\pi}{3}} = \frac{1}{2} \pm \frac{\sqrt{3}}{2}i$ : $(1 - \sqrt{3}) + (1 + \sqrt{3})i$ and $(1 + \sqrt{3}) + (1 - \sqrt{3})i$ .
$1 + \omega + \omega^2 = 0$ , so $1 + \omega = -\omega^2$ and $1 + \omega^2 = -\omega$ ; the product is $\omega^3 = 1$ . Also $\omega^{50} = \omega^{48}\omega^2 = \omega^2$ and $\omega^{25} = \omega$ , so the sum is $\omega^2 + \omega = -1$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the diagrams, the worked examples, the practice items and the quiz questions.
Image: Leonhard Euler, pastel by Jakob Emanuel Handmann (1753), Kunstmuseum Basel, public domain, via Wikimedia Commons.
Video: Roots of Unity (2 of 2: Insights from polar & exponential forms), Eddie Woo, embedded from YouTube.

1Three answers to z3=1 z^3 = 1 z3=1

2Constructing regular polygons

Three answers to $z^{3} = 1$

Constructing regular polygons