Math Core

Adding complex numbers is easy in the form $a + bi$ , but multiplying them, raising them to powers, and taking roots are much easier in polar form. Once you see a complex number as a length and a direction, multiplication turns into "multiply the lengths, add the angles," and a problem like $(1 + i)^{10}$ takes one line.

The complex plane

Plot the complex number $z = a + bi$ as the point $(a, b)$ : the horizontal axis is the real axis and the vertical axis is the imaginary axis. For example, $3 + 4i$ sits at $(3, 4)$ and $-2i$ sits at $(0, -2)$ .

Complex numbers as points. The dashed segment has length |3 + 4i| = 5.Open in grapher →

Definition

Modulus and argument

For $z = a + bi$ :

The modulus $|z| = r = \sqrt{a^2 + b^2}$ is the distance from $z$ to the origin.
An argument of $z$ is an angle $\theta$ from the positive real axis to $z$ , so that $\tan\theta = \dfrac{b}{a}$ with $\theta$ in the correct quadrant.

This is exactly the rectangular-to-polar conversion from earlier in the unit. As before, the argument is not unique: adding any multiple of $2\pi$ gives another argument. In this lesson we use $0 \le \theta < 2\pi$ unless told otherwise.

Polar (trigonometric) form

Since $a = r\cos\theta$ and $b = r\sin\theta$ ,

z = a + bi = r\cos\theta + (r\sin\theta)i = r(\cos\theta + i\sin\theta).

This is the polar form or trigonometric form of $z$ . Many books abbreviate $\cos\theta + i\sin\theta$ as $\operatorname{cis}\theta$ , so $z = r\operatorname{cis}\theta$ .

Worked example: Writing a complex number in polar form

Write $z = -1 + \sqrt{3}\,i$ in polar form.

Solution. $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2$ . The point $(-1, \sqrt{3})$ is in Quadrant II with reference angle $\tan^{-1}\sqrt{3} = \tfrac{\pi}{3}$ , so $\theta = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$ .

z = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right).

Going back to rectangular form is just evaluation: $4\left(\cos\tfrac{7\pi}{4} + i\sin\tfrac{7\pi}{4}\right) = 4\left(\tfrac{\sqrt{2}}{2} - \tfrac{\sqrt{2}}{2}i\right) = 2\sqrt{2} - 2\sqrt{2}\,i$ .

Multiplying and dividing

Multiply $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ by $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ and expand:

z_1 z_2 = r_1 r_2\big[(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2)\big].

The brackets are the sum formulas for cosine and sine. So the product collapses beautifully.

Products and quotients in polar form

z_1 z_2 = r_1 r_2\big[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big]

\dfrac{z_1}{z_2} = \dfrac{r_1}{r_2}\big[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big], \quad z_2 \ne 0

Multiply the moduli and add the arguments. Divide the moduli and subtract the arguments.

Geometrically, multiplying by $z_2$ stretches by the factor $r_2$ and rotates by the angle $\theta_2$ . That is why multiplying by $i = 1(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2})$ rotates any point $90^\circ$ counterclockwise.

Worked example: A product and a quotient

Let $z_1 = 6\left(\cos\tfrac{5\pi}{12} + i\sin\tfrac{5\pi}{12}\right)$ and $z_2 = 2\left(\cos\tfrac{\pi}{12} + i\sin\tfrac{\pi}{12}\right)$ . Find $z_1 z_2$ and $\dfrac{z_1}{z_2}$ in rectangular form.

Solution.

$z_1 z_2 = 12\left(\cos\tfrac{6\pi}{12} + i\sin\tfrac{6\pi}{12}\right) = 12\left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right) = 12i$ .

$\dfrac{z_1}{z_2} = 3\left(\cos\tfrac{4\pi}{12} + i\sin\tfrac{4\pi}{12}\right) = 3\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = \tfrac{3}{2} + \tfrac{3\sqrt{3}}{2}i$ .

Powers: De Moivre's Theorem

Multiplying $z$ by itself $n$ times multiplies the modulus by itself $n$ times and adds the argument $n$ times.

De Moivre's Theorem

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

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

Worked example: A high power

Find $(1 + i)^{10}$ in the form $a + bi$ .

Solution. $1 + i$ has $r = \sqrt{2}$ and $\theta = \tfrac{\pi}{4}$ . By De Moivre,

(1 + i)^{10} = (\sqrt{2})^{10}\left(\cos\tfrac{10\pi}{4} + i\sin\tfrac{10\pi}{4}\right) = 32\left(\cos\tfrac{5\pi}{2} + i\sin\tfrac{5\pi}{2}\right).

Since $\tfrac{5\pi}{2}$ is coterminal with $\tfrac{\pi}{2}$ , the result is $32(0 + i) = 32i$ .

Common mistake

Raise the modulus to the power, but multiply the argument by it. A frequent slip is writing $(\sqrt{2})^{10}$ as $10\sqrt{2}$ or writing the angle as $\left(\tfrac{\pi}{4}\right)^{10}$ . It's $r^n$ and $n\theta$ .

Roots of complex numbers

Running De Moivre backward finds roots. If $w^n = z$ , then the modulus of $w$ must be $\sqrt[n]{r}$ and $n$ times the argument of $w$ must be coterminal with $\theta$ . Because $\theta$ , $\theta + 2\pi$ , $\theta + 4\pi, \ldots$ are all arguments of $z$ , you get $n$ different roots.

The n-th roots

The nonzero complex number $z = r(\cos\theta + i\sin\theta)$ has exactly $n$ distinct $n$ -th roots:

w_k = \sqrt[n]{r}\left[\cos\left(\dfrac{\theta + 2\pi k}{n}\right) + i\sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right], \quad k = 0, 1, \ldots, n - 1.

They all lie on the circle of radius $\sqrt[n]{r}$ , equally spaced $\tfrac{2\pi}{n}$ apart.

Worked example: Cube roots of 8

Find all cube roots of $8$ .

Solution. $8 = 8(\cos 0 + i\sin 0)$ , so each root has modulus $\sqrt[3]{8} = 2$ and argument $\dfrac{0 + 2\pi k}{3}$ for $k = 0, 1, 2$ : that is, $0$ , $\tfrac{2\pi}{3}$ and $\tfrac{4\pi}{3}$ .

$w_0 = 2(\cos 0 + i\sin 0) = 2$
$w_1 = 2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right) = -1 + \sqrt{3}\,i$
$w_2 = 2\left(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\right) = -1 - \sqrt{3}\,i$

The three cube roots of 8 form an equilateral triangle on the circle of radius 2.Open in grapher →

Tip

Check a root by cubing it with De Moivre: $\left(-1 + \sqrt{3}\,i\right)^3 = 2^3\left(\cos 2\pi + i\sin 2\pi\right) = 8$ .

Practice

Practice 1

Find the modulus of $z = 3 - 4i$ .

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

Practice 2

Find the argument $\theta$ of $z = -4 + 4i$ , with $0 \le \theta < 2\pi$ .

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

Practice 3

Write $z = 2\left(\cos\tfrac{5\pi}{6} + i\sin\tfrac{5\pi}{6}\right)$ in the form $a + bi$ . Enter the real and imaginary parts as $(a, b)$ .

Enter a point like (2, -3)

Practice 4

Let $z_1 = 3\left(\cos\tfrac{\pi}{5} + i\sin\tfrac{\pi}{5}\right)$ and $z_2 = 2\left(\cos\tfrac{3\pi}{10} + i\sin\tfrac{3\pi}{10}\right)$ . Write $z_1 z_2$ as $a + bi$ and enter $(a, b)$ .

Enter a point like (2, -3)

Practice 5

Let $z_1 = 10\left(\cos\tfrac{4\pi}{3} + i\sin\tfrac{4\pi}{3}\right)$ and $z_2 = 5\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right)$ . Write $\dfrac{z_1}{z_2}$ as $a + bi$ and enter $(a, b)$ .

Enter a point like (2, -3)

Practice 6

Use De Moivre's Theorem to evaluate $(1 + i)^8$ .

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

Practice 7

Write $\left(1 - \sqrt{3}\,i\right)^5$ in the form $a + bi$ and enter $(a, b)$ .

Enter a point like (2, -3)

Practice 8

The three cube roots of $8i$ all have modulus $2$ . Find their arguments in $[0, 2\pi)$ .

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