Complex polar/exponential form: $z = re^{i\theta}$

The formula

If $z$ has modulus $r$ and argument $\theta$:

$z = r \cdot e^{i\theta}$

with $e^{i\theta} = \cos \theta + i \sin \theta$ (Euler’s formula).

From trig to polar

$z = r(\cos \theta + i \sin \theta) = r \cdot e^{i\theta}$.

From algebraic to polar

$z = a + ib$:

• Compute $r = \sqrt{a^2 + b^2}$
• Compute $\theta$ (argument)
• Write $z = r \cdot e^{i\theta}$

Examples

• $1 + i$: $r = \sqrt{2}$, $\theta = \pi/4$. So $1 + i = \sqrt{2} \cdot e^{i\pi/4}$.
• $-1 = e^{i\pi}$. Euler’s identity: $e^{i\pi} + 1 = 0$.
• $i = e^{i\pi/2}$.

Properties (Euler)

$e^{i\alpha} \cdot e^{i\beta} = e^{i(\alpha + \beta)}$

$\dfrac{e^{i\alpha}}{e^{i\beta}} = e^{i(\alpha - \beta)}$

$(e^{i\theta})^n = e^{in\theta} \quad (\text{de Moivre})$

$\overline{e^{i\theta}} = e^{-i\theta}$

Why is this great?

Multiplying two complex numbers = adding arguments:

$z_1 z_2 = r_1 r_2 \cdot e^{i(\theta_1 + \theta_2)}$

Application: powers

Compute $(1 + i)^8$.

$1 + i = \sqrt{2} \cdot e^{i\pi/4}$ $(1 + i)^8 = (\sqrt{2})^8 \cdot e^{i \cdot 2\pi} = 16$.

Slick, one line!

Inverse Euler

$\cos \theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2}, \quad \sin \theta = \dfrac{e^{i\theta} - e^{-i\theta}}{2i}$

Very useful for linearizing trig expressions.

Common mistakes

1. Forgetting $r$: $z = e^{i\theta}$ implies $r = 1$.
2. Wrong argument: watch the quadrant if $a < 0$.
3. Confusing Euler with Taylor.

?À toi de jouer

In polar form, i equals:

A.e^{iπ/2}B.e^{iπ}C.e^{i}D.e^{iπ/4}

Valider