Euler's Formula and Exponential Form

In Precalculus you wrote $z = r(\cos\theta + i\sin\theta)$ and found that multiplying complex numbers multiplies their moduli and adds their angles. Calculus gives that rule a short name. The Maclaurin series from known series work for complex inputs too, and they explain why $\cos\theta + i\sin\theta$ deserves to be written as a power of $e$.

Key idea

Euler's formula: $e^{i\theta} = \cos\theta + i\sin\theta$. Every nonzero complex number has an exponential form $z = re^{i\theta}$ with $r = |z|$ and $\theta$ an argument (angle) of $z$. Then $$re^{i\alpha} \cdot se^{i\beta} = rs\,e^{i(\alpha + \beta)}, \qquad \frac{re^{i\alpha}}{se^{i\beta}} = \frac{r}{s}e^{i(\alpha - \beta)}, \qquad \left(re^{i\theta}\right)^n = r^ne^{in\theta}.$$

Why it works

Put $i\theta$ into $e^{x} = \sum \frac{x^n}{n!}$. The powers of $i$ cycle $1, i, -1, -i$, so the even terms are real and the odd terms carry an $i$: $$e^{i\theta} = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) = \cos\theta + i\sin\theta.$$ The product rule is the angle-sum formulas in disguise: $(\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$, which is exactly what $e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)}$ says.

Exponential form is the natural way to multiply, divide and take powers; the form $a + bi$ is the natural way to add. Most problems switch between the two. Going from $re^{i\theta}$ to $a + bi$ is direct: $a = r\cos\theta$ and $b = r\sin\theta$. Going the other way needs care with the angle, as the first example shows. Two special values are worth knowing by heart: $e^{i\pi} = -1$ and $e^{2\pi i} = 1$. The second says that a full turn brings you back, so the angle of a complex number is only defined up to adding $2\pi k$; the next lesson deals with that.

A few numbers in both forms.
$z$	$1$	$i$	$-1$	$-i$	$1 + i$	$1 + \sqrt{3}\,i$
$re^{i\theta}$	$e^{0}$	$e^{\pi i/2}$	$e^{\pi i}$	$e^{-\pi i/2}$	$\sqrt{2}e^{\pi i/4}$	$2e^{\pi i/3}$

Multiplying: the lengths multiply and the angles add. Try $w = i$ (length 1, angle $\frac{\pi}{2}$): multiplying by $i$ turns $z$ a quarter turn.

Worked examples

Worked example: Rectangular to exponential

Write $z = -1 - i$ in the form $re^{i\theta}$ with $-\pi < \theta \le \pi$.

$r = |z| = \sqrt{1 + 1} = \sqrt{2}$.
The point $(-1, -1)$ is in quadrant III. The reference angle is $\tan^{-1}\left|\frac{-1}{-1}\right| = \frac{\pi}{4}$.
In quadrant III, measured the short way from the positive real axis, $\theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4}$.
$z = \sqrt{2}\,e^{-3\pi i/4}$. Check: $\sqrt{2}\left(\cos\frac{3\pi}{4} - i\sin\frac{3\pi}{4}\right) = \sqrt{2}\left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i\right) = -1 - i$.

Answer: $\sqrt{2}\,e^{-3\pi i/4}$

Worked example: A product

Compute $\left(2e^{\pi i/6}\right)\left(3e^{\pi i/3}\right)$ in the form $a + bi$.

Multiply lengths: $2 \cdot 3 = 6$. Add angles: $\frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$.
$6e^{\pi i/2} = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6i$.

Answer: $6i$

Worked example: A power

Compute $(1 + i)^{8}$.

$1 + i = \sqrt{2}\,e^{\pi i/4}$.
$(1 + i)^8 = (\sqrt{2})^8 e^{8\pi i/4} = 16e^{2\pi i}$.
$e^{2\pi i} = 1$, so the answer is $16$. Multiplying out eight factors would take much longer.

Answer: $16$

Common mistake

Taking the angle straight from $\tan^{-1}\frac{b}{a}$. A calculator's $\tan^{-1}$ only returns angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, so for points on the left half of the plane it points the wrong way: for $-1 - i$ it gives $\frac{\pi}{4}$, the angle of $1 + i$. Always locate the quadrant first. The correct angle is $-\frac{3\pi}{4}$ (or $\frac{5\pi}{4}$, the same direction). Check by converting back: $\sqrt{2}e^{i\pi/4} = 1 + i$, not $-1 - i$.

Euler's Formula and Exponential Form

Worked examples

Practice

Builds on

Leads to

Sources and further reading