Math Core

Computing $(1 + i)^{8}$ by multiplying $1 + i$ by itself eight times is slow and easy to get wrong. In trigonometric form, you already know that multiplying means multiply the moduli and add the arguments. Repeating that $n$ times gives a shortcut called De Moivre's theorem, and running it backward lets you find every $n$ th root of a complex number.

Powers of a complex number

Take $z = r(\cos\theta + i\sin\theta)$ and square it. Multiplying $z$ by itself multiplies $r$ by $r$ and adds $\theta$ to $\theta$ :

z^2 = r^2(\cos 2\theta + i\sin 2\theta).

Multiply by $z$ once more and you get $z^3 = r^3(\cos 3\theta + i\sin 3\theta)$ . The pattern continues for every power.

De Moivre's theorem

If $z = r(\cos\theta + i\sin\theta)$ and $n$ is a positive integer, then

z^n = r^n(\cos n\theta + i\sin n\theta).

Raise the modulus to the $n$ th power and multiply the argument by $n$ .

Worked example: A power of 1 + i

Find $(1 + i)^8$ .

First write $1 + i$ in trigonometric form. $r = \sqrt{1 + 1} = \sqrt{2}$ and $\theta = 45^\circ$ , so $1 + i = \sqrt{2}(\cos 45^\circ + i\sin 45^\circ)$ .

By De Moivre's theorem:

(1 + i)^8 = (\sqrt{2})^8(\cos 360^\circ + i\sin 360^\circ) = 16(1 + 0i) = 16.

Eight multiplications collapse into one line, and the answer is a plain real number.

Worked example: A power with a Quadrant IV base

Find $(\sqrt{3} - i)^5$ in the form $a + bi$ .

$r = \sqrt{3 + 1} = 2$ . The point $(\sqrt{3}, -1)$ is in Quadrant IV with reference angle $30^\circ$ , so $\theta = 330^\circ$ .

(\sqrt{3} - i)^5 = 2^5(\cos 1650^\circ + i\sin 1650^\circ).

Reduce the angle: $1650^\circ - 4(360^\circ) = 210^\circ$ . So

(\sqrt{3} - i)^5 = 32(\cos 210^\circ + i\sin 210^\circ) = 32\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = -16\sqrt{3} - 16i.

Common mistake

The modulus gets raised to the power, but the argument gets multiplied, not raised. $(2\operatorname{cis} 15^\circ)^3$ is $8\operatorname{cis} 45^\circ$ , not $8\operatorname{cis} 3375^\circ$ and not $6\operatorname{cis} 45^\circ$ .

Roots of a complex number

A number $w$ is an $n$ th root of $z$ if $w^n = z$ . Every nonzero complex number has exactly $n$ different $n$ th roots. That may be surprising: the real number $8$ has only one real cube root, but it has three complex cube roots.

Here is how to find them. Suppose $w = s(\cos\phi + i\sin\phi)$ is an $n$ th root of $z = r(\cos\theta + i\sin\theta)$ . By De Moivre's theorem, $w^n = s^n(\cos n\phi + i\sin n\phi)$ . For this to equal $z$ :

the moduli must match, so $s^n = r$ and $s = \sqrt[n]{r}$ ;
the angles must point the same way, so $n\phi = \theta + 360^\circ k$ for some integer $k$ .

Dividing by $n$ gives $\phi = \dfrac{\theta + 360^\circ k}{n}$ . The values $k = 0, 1, \dots, n - 1$ give different angles; after that, they repeat.

The nth roots of a complex number

The $n$ distinct $n$ th roots of $z = r(\cos\theta + i\sin\theta)$ are

w_k = \sqrt[n]{r}\left[\cos\left(\frac{\theta + 360^\circ k}{n}\right) + i\sin\left(\frac{\theta + 360^\circ k}{n}\right)\right], \qquad k = 0, 1, \dots, n - 1.

In radians, replace $360^\circ$ with $2\pi$ .

All $n$ roots have the same modulus $\sqrt[n]{r}$ , so they lie on one circle. Their arguments are spaced $\dfrac{360^\circ}{n}$ apart, so they form the vertices of a regular $n$ -gon.

Worked example: The cube roots of 8

Find all three cube roots of $8$ .

Write $8 = 8(\cos 0^\circ + i\sin 0^\circ)$ . The roots have modulus $\sqrt[3]{8} = 2$ and arguments $\dfrac{0^\circ + 360^\circ k}{3}$ , which are $0^\circ$ , $120^\circ$ and $240^\circ$ .

\begin{aligned} w_0 &= 2(\cos 0^\circ + i\sin 0^\circ) = 2 \\ w_1 &= 2(\cos 120^\circ + i\sin 120^\circ) = -1 + i\sqrt{3} \\ w_2 &= 2(\cos 240^\circ + i\sin 240^\circ) = -1 - i\sqrt{3} \end{aligned}

The three cube roots of 8 lie on a circle of radius 2, spaced 120° apart. They form an equilateral triangle.Open in grapher →

You can check $w_1$ directly: $(-1 + i\sqrt{3})^3 = 2^3(\cos 360^\circ + i\sin 360^\circ) = 8$ .

Worked example: Fourth roots of a non-real number

Find the fourth roots of $z = 16(\cos 120^\circ + i\sin 120^\circ)$ . Give each in trigonometric form.

The modulus of each root is $\sqrt[4]{16} = 2$ . The arguments are $\dfrac{120^\circ + 360^\circ k}{4} = 30^\circ + 90^\circ k$ :

2\operatorname{cis} 30^\circ, \quad 2\operatorname{cis} 120^\circ, \quad 2\operatorname{cis} 210^\circ, \quad 2\operatorname{cis} 300^\circ.

They are the corners of a square inscribed in the circle of radius $2$ .

Tip

Once you have the first root, you can get the rest by adding $\dfrac{360^\circ}{n}$ to the argument again and again. Stop when you have $n$ roots.

Practice

Practice 1

Find $\left[2(\cos 15^\circ + i\sin 15^\circ)\right]^4$ . Write it as $r(\cos\theta + i\sin\theta)$ and enter $(r, \theta)$ with $\theta$ in degrees.

Enter a point like (2, -3)

Practice 2

How many distinct fifth roots does the complex number $3 - 7i$ have?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Find $(\sqrt{3} + i)^6$ . The answer is a real number.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Write $(1 + i\sqrt{3})^5$ in the form $a + bi$ . What is the real part $a$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

The cube roots of $27$ are written in the form $3(\cos\theta + i\sin\theta)$ . List their arguments $\theta$ in degrees, with $0^\circ \le \theta < 360^\circ$ , separated by commas.

Separate answers with commas, e.g. 2, -5

Practice 6

Find the arguments, in degrees from $0^\circ$ to $360^\circ$ , of the four fourth roots of $81(\cos 80^\circ + i\sin 80^\circ)$ . Separate them with commas.

Separate answers with commas, e.g. 2, -5

Practice 7

What is the modulus of each sixth root of $64i$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

Which of these is one of the square roots of $-4i$ ?