Complex Numbers

Lessons

Senior 6 · Mathematics

Complex Numbers

5 lessons

Lesson 3 of 5

Argument, polar form and De Moivre's theorem

25 min

What does multiplying do to a point?

In Lesson 2 you saw that multiplying by $i$ turns a point through a right angle. Try another multiplier: $(1 + i) (2 + i) = 2 + i + 2 i + i^{2} = 1 + 3 i$ . Measure the result. The modulus has grown from $|2 + i| = \sqrt{5}$ to $|1 + 3i| = \sqrt{10}$ , a factor of $\sqrt{2} = |1 + i|$ , and the direction has turned through exactly $45^\circ$ . Lengths multiply and angles add. To see why, and to use it, you need to describe a complex number by its length and its angle.

Definition

Argument of a non-zero complex number $z$ : an angle $\theta$ from the positive real axis to the vector $\overrightarrow{OZ}$ , measured anticlockwise. It is written $\arg z$ .

Adding a whole turn does not move the point, so $\arg z$ is defined only up to a multiple of $2\pi$ . The principal argument, $\operatorname{Arg} z$ , is the one value in $(-\pi, \pi]$ . The number $0$ has no argument.

Example: $\arg i = \frac{\pi}{2} + 2k\pi$ for $k \in \mathbb{Z}$ , and $\operatorname{Arg}(-1) = \pi$ .

The argument is measured from the positive real axis: anticlockwise gives a positive angle, clockwise a negative one. The principal argument lies in $(-\pi, \pi]$ .

Finding the argument of $a + b i$

In the right triangle drawn from O, $\cos\theta = \frac{a}{r}$ , $\sin\theta = \frac{b}{r}$ and $\tan\theta = \frac{b}{a}$ , where $r = ∣ z ∣$ . A calculator's $\tan^{-1}$ only returns angles in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so the quadrant must be fixed by hand. With $\alpha = \tan^{-1}\left|\frac{b}{a}\right|$ , the acute angle to the real axis:

Position of $z$	Principal argument	Example
first quadrant ( $a \gt 0, b \ge 0$ )	$\alpha$	$\operatorname{Arg}(\sqrt{3} + i) = \frac{\pi}{6}$
second quadrant ( $a \lt 0, b \ge 0$ )	$\pi - \alpha$	$\operatorname{Arg}(-2 + 2i) = \frac{3\pi}{4}$
third quadrant ( $a \lt 0, b \lt 0$ )	$-\pi + \alpha$	$\operatorname{Arg}(-1 - i) = -\frac{3\pi}{4}$
fourth quadrant ( $a \gt 0, b \lt 0$ )	$-\alpha$	$\operatorname{Arg}(1 - \sqrt{3}i) = -\frac{\pi}{3}$
imaginary axis ( $a = 0$ )	$\frac{\pi}{2}$ if $b \gt 0$ ; $-\frac{\pi}{2}$ if $b \lt 0$	$\operatorname{Arg}(-3i) = -\frac{\pi}{2}$

Warning

$\tan^{-1}\frac{b}{a}$ is not the argument when $a \lt 0$ . For $- 2 + 2 i$ , the calculator gives $\tan^{-1}(-1) = -\frac{\pi}{4}$ , which points into the fourth quadrant; the point is in the second. Always sketch the point first, then choose the angle.

Definition

Polar (trigonometric) form of $z \ne 0$ : since $a = r\cos\theta$ and $b = r\sin\theta$ ,

z = r(\cos\theta + i\sin\theta), \qquad r = |z| \gt 0, \quad \theta = \arg z .

Notation: this is often shortened to $z = r\operatorname{cis}\theta$ . Two numbers in polar form are equal when their moduli are equal and their arguments differ by a multiple of $2\pi$ .

Example: $2i = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$ .

Worked example 1: converting between forms

(a) Write $z_{1} = - 2 + 2 i$ and $z_2 = 1 - \sqrt{3}i$ in polar form. (b) Write $4\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)$ in the form $a + b i$ .

Solution

$|z_1| = \sqrt{4 + 4} = 2\sqrt{2}$ . The point is in the second quadrant with $\alpha = \tan^{-1} 1 = \frac{\pi}{4}$ , so $\operatorname{Arg} z_1 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$ and $z_1 = 2\sqrt{2}\left(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}\right)$ .
$|z_2| = \sqrt{1 + 3} = 2$ . Fourth quadrant, $\alpha = \tan^{-1}\sqrt{3} = \frac{\pi}{3}$ , so $z_2 = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$ .
$\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ and $\sin\frac{5\pi}{6} = \frac{1}{2}$ , so the number is $4\left(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = -2\sqrt{3} + 2i$ .

Loci defined by angles

The condition $\arg(z - z_0) = \theta$ says that the vector from $z_{0}$ to $z$ points in the direction $\theta$ . The locus is a half-line starting at $z_{0}$ ; the point $z_{0}$ itself is excluded, because $0$ has no argument. For $\arg(z - 1 - i) = \frac{\pi}{4}$ : with $z = x + y i$ , the vector $(x - 1) + (y - 1) i$ points at $45^\circ$ , so $y - 1 = x - 1$ with $x \gt 1$ , that is, $y = x$ for $x \gt 1$ .

$\arg(z - z_0) = \theta$ is a half-line that starts at $z_{0}$ (not included) and makes the angle $\theta$ with the direction of the positive real axis.

Multiplying and dividing in polar form

Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ . Expanding, and using the addition formulae from S5 (Trigonometric Formulae),

\begin{aligned} z_1 z_2 &= r_1 r_2\big[(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) \\ &\qquad + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2)\big] \\ &= r_1 r_2\left[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\right]. \end{aligned}

In the same way (multiply the top and bottom by $\cos\theta_2 - i\sin\theta_2$ ),

\frac{z_1}{z_2} = \frac{r_1}{r_2}\left[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right].

Important

$|z_1 z_2| = |z_1|\,|z_2|$ and $\arg(z_1 z_2) = \arg z_1 + \arg z_2$ (up to a multiple of $2\pi$ ).
$\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ and $\arg\frac{z_1}{z_2} = \arg z_1 - \arg z_2$ .

Geometric meaning: multiplying by $w = \rho(\cos\alpha + i\sin\alpha)$ is a rotation about O through $\alpha$ combined with an enlargement of scale factor $\rho$ . Multiplying by $\cos\alpha + i\sin\alpha$ (modulus 1) is a pure rotation.

$1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$ . Multiplying by it turns $z$ through $45^\circ$ and multiplies its modulus by $\sqrt{2}$ : $\sqrt{5} \times \sqrt{2} = \sqrt{10}$ .

Worked example 2: product and quotient

$z_1 = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$ and $z_2 = 3\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$ . Find $z_{1} z_{2}$ and $\frac{z_1}{z_2}$ in the form $a + b i$ .

Solution

z_1 z_2 = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6i,

\begin{aligned} \frac{z_1}{z_2} &= \frac{2}{3}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) \\ &= \frac{2}{3}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{3}}{3} + \frac{1}{3}i . \end{aligned}

Important

De Moivre's theorem. For every real $\theta$ and every integer $n$ ,

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

Proof by induction for $n \ge 1$ .

Base case $n = 1$ : both sides equal $\cos\theta + i\sin\theta$ .
Inductive step: assume $(\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta$ for some $k \ge 1$ . Then $\begin{aligned} (\cos\theta + i\sin\theta)^{k+1} &= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) \\ &= \cos(k + 1)\theta + i\sin(k + 1)\theta, \end{aligned}$ by the product rule above (moduli 1 and 1, arguments added). So the statement is true for $k + 1$ .
By the principle of mathematical induction, it is true for every integer $n \ge 1$ .

For $n = 0$ both sides are 1. For $n = - m$ with $m \ge 1$ : $(\cos\theta + i\sin\theta)^{-m} = \frac{1}{(\cos\theta + i\sin\theta)^m} = \frac{1}{\cos m\theta + i\sin m\theta} = \cos m\theta - i\sin m\theta = \cos(-m\theta) + i\sin(-m\theta)$ , using the case $n = m$ , then multiplying the top and bottom by $\cos m\theta - i\sin m\theta$ , which makes the denominator $\cos^2 m\theta + \sin^2 m\theta = 1$ . ∎

For $z = r(\cos\theta + i\sin\theta)$ it follows that $z^n = r^n(\cos n\theta + i\sin n\theta)$ .

Abraham de Moivre (1667–1754), painted by Joseph Highmore in 1736. Born in France, de Moivre worked in London; the theorem carries his name, and later mathematicians, Euler among them, gave it the form used today. Painting: public domain, via Wikimedia Commons.

Worked example 3: large powers without expanding

Calculate (a) $(1 + i)^{10}$ (b) $(1 - \sqrt{3}i)^{-6}$ .

Solution

$1 + i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$ , so $(1 + i)^{10} = (\sqrt{2})^{10}\left(\cos\frac{10\pi}{4} + i\sin\frac{10\pi}{4}\right) = 32\left(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right) = 32i$ , since $\frac{5\pi}{2} = 2\pi + \frac{\pi}{2}$ .
$1 - \sqrt{3}i = 2\left(\cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\right)$ , so $(1 - \sqrt{3}i)^{-6} = 2^{-6}\left(\cos 2\pi + i\sin 2\pi\right) = \frac{1}{64}$ .

Moduli $\sqrt{2}, 2, 2\sqrt{2}, 4$ ; arguments $\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi$ . De Moivre's theorem predicts every point without multiplying out.

Worked example 4: multiple-angle formulae

Use De Moivre's theorem to express $\cos 3\theta$ in terms of $\cos\theta$ , and $\sin 3\theta$ in terms of $\sin\theta$ .

Solution Write $c = \cos\theta$ and $s = \sin\theta$ . By De Moivre and the binomial expansion (S4, Combinatorics),

\begin{aligned} \cos 3\theta + i\sin 3\theta &= (c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3 \\ &= (c^3 - 3cs^2) + i(3c^2 s - s^3). \end{aligned}

Equating real parts and using $s^{2} = 1 - c^{2}$ : $\cos 3\theta = c^3 - 3c(1 - c^2) = 4\cos^3\theta - 3\cos\theta$ .

Equating imaginary parts and using $c^{2} = 1 - s^{2}$ : $\sin 3\theta = 3(1 - s^2)s - s^3 = 3\sin\theta - 4\sin^3\theta$ .

Check with $\theta = \frac{\pi}{3}$ : $\cos\pi = -1$ and $4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{3}{2} = -1$ ✓.

Tip

For a power, convert to polar form, apply De Moivre, then convert back. For a sum or difference, stay in algebraic form. Choosing the right form is half of every complex-number question.

Eddie Woo: Investigating de Moivre's Theorem, part 3 of 3, Proof by mathematical induction. Watch the proof by induction, and compare each step with the proof above.

✦

Practice

Sketch each number before you find its argument.

Find the modulus and the principal argument of (a) $-\sqrt{3} + i$ (b) $- 4 - 4 i$ (c) $- 5$ (d) $3 i$ .
Write $6\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)$ in the form $a + b i$ .
$z_1 = 4\left(\cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12}\right)$ and $z_2 = 2\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)$ . Find $z_{1} z_{2}$ and $\frac{z_1}{z_2}$ in the form $a + b i$ .
Sketch the locus $\arg(z + 2i) = \frac{3\pi}{4}$ and describe it with a Cartesian equation.
Use De Moivre's theorem to calculate (a) $(\sqrt{3} + i)^6$ (b) $(1 - i)^{-8}$ .
Prove that $\cos 4\theta = 8\cos^4\theta - 8\cos^2\theta + 1$ .
Find the smallest positive integer $n$ for which $(1 + \sqrt{3}i)^n$ is real, and find that real number.
Prove by induction that $(\cos\theta - i\sin\theta)^n = \cos n\theta - i\sin n\theta$ for every integer $n \ge 1$ .

Show answersHide answers

(a) $2$ , $\frac{5\pi}{6}$ (b) $4\sqrt{2}$ , $-\frac{3\pi}{4}$ (c) $5$ , $π \pi$ (d) $3$ , $\frac{\pi}{2}$ .
$6\left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = -3 - 3\sqrt{3}i$ .
$z_1 z_2 = 8\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 8i$ ; $\frac{z_1}{z_2} = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 1 + \sqrt{3}i$ .
A half-line from $- 2 i$ (excluded) in the direction $135^\circ$ : $y + 2 = - x$ with $x \lt 0$ , that is, $y = - x - 2$ for $x \lt 0$ .
(a) $\sqrt{3} + i = 2\operatorname{cis}\frac{\pi}{6}$ , so the power is $64\operatorname{cis}\pi = -64$ . (b) $1 - i = \sqrt{2}\operatorname{cis}\left(-\frac{\pi}{4}\right)$ , so the power is $(\sqrt{2})^{-8}\operatorname{cis} 2\pi = \frac{1}{16}$ .
The real part of $(c + i s)^{4}$ is $c^{4} - 6 c^{2} s^{2} + s^{4}$ . With $s^{2} = 1 - c^{2}$ : $c^{4} - 6 c^{2} + 6 c^{4} + 1 - 2 c^{2} + c^{4} = 8 c^{4} - 8 c^{2} + 1$ .
$1 + \sqrt{3}i = 2\operatorname{cis}\frac{\pi}{3}$ , so $(1 + \sqrt{3}i)^n = 2^n\operatorname{cis}\frac{n\pi}{3}$ , which is real when $\frac{n\pi}{3}$ is a multiple of $π \pi$ . The smallest is $n = 3$ , giving $8\operatorname{cis}\pi = -8$ .
Base case $n = 1$ is clear. If $(\cos\theta - i\sin\theta)^k = \cos k\theta - i\sin k\theta$ , then multiplying by $\cos\theta - i\sin\theta$ gives $\cos k\theta\cos\theta - \sin k\theta\sin\theta - i(\sin k\theta\cos\theta + \cos k\theta\sin\theta) = \cos(k + 1)\theta - i\sin(k + 1)\theta$ . So the statement holds for all $n \ge 1$ .

Sources and credits

Written by Ganzaa: the explanations, the proofs, the diagrams, the worked examples, the practice items and the quiz questions.
Image: Portrait of Abraham de Moivre by Joseph Highmore (1736), Royal Society, public domain, via Wikimedia Commons.
Video: Investigating de Moivre's Theorem (3 of 3: Proof by mathematical induction), Eddie Woo, embedded from YouTube.

1What does multiplying do to a point?

2Finding the argument of a+bi a + bi a+bi

3Loci defined by angles

4Multiplying and dividing in polar form

What does multiplying do to a point?

Finding the argument of $a + b i$

Loci defined by angles

Multiplying and dividing in polar form