Polar Form & De Moivre's Theorem
Polar Form & De Moivre's Theorem
A complex number $z = a+bi$ can be written as:
$$z = r(\cos\theta + i\sin\theta) = r\operatorname{cis}\theta$$
where $r = |z| = \sqrt{a^2+b^2}$ and $\theta = \arg(z) = \arctan\frac{b}{a}$ (adjusted for quadrant).
$$z_1 z_2 = r_1 r_2 \operatorname{cis}(\theta_1 + \theta_2)$$
$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \operatorname{cis}(\theta_1 - \theta_2)$$
$$[r\operatorname{cis}\theta]^n = r^n \operatorname{cis}(n\theta)$$
Convert $z = 1 + i$ to polar form.
$r = \sqrt{1+1} = \sqrt{2}$, $\theta = \arctan(1/1) = \pi/4$.
$$z = \sqrt{2}\operatorname{cis}\frac{\pi}{4}$$
Use De Moivre's Theorem to compute $(1+i)^8$.
$(\sqrt{2})^8 \operatorname{cis}(8 \cdot \pi/4) = 16\operatorname{cis}(2\pi) = 16(1+0i) = 16$
Convert $z = 3\operatorname{cis}(\pi/3)$ to rectangular form.
$a = 3\cos(\pi/3) = 3/2$, $b = 3\sin(\pi/3) = 3\sqrt{3}/2$.
$z = \frac{3}{2} + \frac{3\sqrt{3}}{2}i$
Practice Problems
Show Answer Key
1. $r = \sqrt{2}$, $\theta = 3\pi/4$; $z = \sqrt{2}\operatorname{cis}(3\pi/4)$
2. $2\cos(\pi/6)+2i\sin(\pi/6) = \sqrt{3}+i$
3. $(\sqrt{2})^4\operatorname{cis}(\pi) = 4(-1) = -4$
4. $r = 3$, $\theta = \pi$
5. $6\operatorname{cis}(7\pi/12)$
6. $r = 4$, $\theta = \pi/2$; $4\operatorname{cis}(\pi/2)$
7. $r = 2$, $\theta = \pi/6$; $2^3\operatorname{cis}(\pi/2) = 8i$
8. $3\operatorname{cis}(\pi/2) = 3i$
9. $|z| = 2\sqrt{2}$, $\arg = 5\pi/4$ (QIII)
10. $i = \operatorname{cis}(\pi/2)$; $i^6 = \operatorname{cis}(3\pi) = -1$
11. $-5$
12. $1$