Every complex number $z = a + bi$ can be written in polar form:
$$z = r(\cos\theta + i\sin\theta) = r\,\text{cis}\,\theta$$
where $r = |z| = \sqrt{a^2+b^2}$ and $\theta = \arg(z) \in (-\pi, \pi]$.
The shorthand $\text{cis}\,\theta$ means $\cos\theta + i\sin\theta$.
| Direction | Formulas |
|---|---|
| Cartesian $\to$ Polar | $r = \sqrt{a^2+b^2}$; $\theta$ from quadrant check |
| Polar $\to$ Cartesian | $a = r\cos\theta$, $b = r\sin\theta$ |
Example 1: Convert $z = -\sqrt{3} + i$ to polar form.
$r = \sqrt{3+1} = 2$. Since $z$ is in Q2: $\theta = \pi - \arctan!\left(\dfrac{1}{\sqrt{3}}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.
$$z = 2\,\text{cis}\,\dfrac{5\pi}{6}$$
Example 2: Convert $z = 4\,\text{cis}\,\dfrac{\pi}{3}$ to Cartesian form.
$$z = 4\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right) = 4\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = 2 + 2\sqrt{3}\,i$$
Let $z_1 = r_1\,\text{cis}\,\theta_1$ and $z_2 = r_2\,\text{cis}\,\theta_2$.
$$z_1 z_2 = r_1 r_2\,\text{cis}(\theta_1 + \theta_2)$$
$$\frac{z_1}{z_2} = \frac{r_1}{r_2}\,\text{cis}(\theta_1 - \theta_2)$$
Proof of multiplication:
$$z_1 z_2 = r_1 r_2 (\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2)$$
$$= r_1 r_2 [(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2)]$$
$$= r_1 r_2 [\cos(\theta_1+\theta_2) + i\sin(\theta_1+\theta_2)] = r_1 r_2\,\text{cis}(\theta_1+\theta_2). \quad\square$$
Geometric meaning:
- Multiplying by $z_2$: scale by $r_2$ and rotate by $\theta_2$.
- Dividing by $z_2$: scale by $1/r_2$ and rotate by $-\theta_2$.
Let $z_1 = 3\,\text{cis}\,\dfrac{\pi}{4}$ and $z_2 = 2\,\text{cis}\,\dfrac{\pi}{6}$.
$$z_1 z_2 = 6\,\text{cis}\,\frac{5\pi}{12}, \qquad \frac{z_1}{z_2} = \frac{3}{2}\,\text{cis}\,\frac{\pi}{12}$$
To verify $z_1 z_2$ in Cartesian: $6\cos\dfrac{5\pi}{12} + 6i\sin\dfrac{5\pi}{12}$.
Using $\cos\dfrac{5\pi}{12} = \dfrac{\sqrt{6}-\sqrt{2}}{4}$ gives the exact Cartesian form.
From repeated multiplication:
$$(\text{cis}\,\theta)^n = \text{cis}(n\theta) \quad \text{for } n \in \mathbb{Z}$$
This is the foundation for De Moivre’s theorem.
KEY TAKEAWAY: In polar form, multiplication rotates and scales; division un-rotates and scales. This makes polar form far more efficient than Cartesian for products and powers.
EXAM TIP: Always reduce the resulting argument to $(-\pi, \pi]$ after multiplying or dividing. Add or subtract $2\pi$ as needed.
COMMON MISTAKE: Adding moduli instead of multiplying them when multiplying two polar-form complex numbers.
