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Multiplying and dividing in polar form

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Why polar form exists

Multiplying two complex numbers in rectangular form takes four products and a collection step. In polar form it takes one multiplication and one addition.

The multiplication rule

$r_1\,\text{cis}\,\theta_1 \times r_2\,\text{cis}\,\theta_2 = r_1r_2\,\text{cis}(\theta_1 + \theta_2)$

In words: multiply the moduli, add the arguments.
$3\,\text{cis}\,\frac{\pi}{6} \times 4\,\text{cis}\,\frac{\pi}{3} = 12\,\text{cis}\,\frac{\pi}{2}$

The division rule

$\frac{r_1\,\text{cis}\,\theta_1}{r_2\,\text{cis}\,\theta_2} = \frac{r_1}{r_2}\,\text{cis}(\theta_1 - \theta_2)$

In words: divide the moduli, subtract the arguments.
$\dfrac{10\,\text{cis}\,\frac{3\pi}{4}}{5\,\text{cis}\,\frac{\pi}{4}} = 2\,\text{cis}\,\frac{\pi}{2}$

What multiplication does geometrically

Multiplying by $r\,\text{cis}\,\theta$ $r cis θ$ does exactly two things to a point on the Argand diagram:
- stretches its distance from the origin by a factor of $r$
- rotates it anticlockwise through an angle $\theta$

Special cases that follow immediately:
- Multiplying by $i = 1\,\text{cis}\,\frac{\pi}{2}$ rotates a quarter turn anticlockwise and changes nothing else.
- Multiplying by $-1 = 1\,\text{cis}\,\pi$ rotates a half turn.
- Multiplying by a positive real number is a pure stretch, with no rotation.

Keeping the argument principal

Adding two arguments can push the total outside $(-\pi, \pi]$ .
Add or subtract $2\pi$ until the result is back in range.
$\text{cis}\,\frac{3\pi}{4} \times \text{cis}\,\frac{3\pi}{4} = \text{cis}\,\frac{6\pi}{4} = \text{cis}\,\frac{3\pi}{2}$ , and $\frac{3\pi}{2} - 2\pi = -\frac{\pi}{2}$ , so the answer is $\text{cis}\left(-\frac{\pi}{2}\right)$ .

The conjugate in polar form

Reflecting in the real axis leaves the distance unchanged and negates the angle:

$\overline{r\,\text{cis}\,\theta} = r\,\text{cis}(-\theta)$

It follows that $z\overline{z} = r^2\,\text{cis}(0) = r^2 = |z|^2$ , which is the polar-form version of the identity from the conjugate page.

The reciprocal in polar form

Setting $r_1 = 1$ and $\theta_1 = 0$ in the division rule:

$\frac{1}{r\,\text{cis}\,\theta} = \frac{1}{r}\,\text{cis}(-\theta)$

So the reciprocal inverts the modulus and negates the argument.

Choosing your route

A question like "find $\dfrac{(2 + 2i)(1 - \sqrt{3}i)}{\sqrt{3} + i}$ " can be done either way. In polar form:
- Convert each factor: $2\sqrt{2}\,\text{cis}\,\frac{\pi}{4}$ , $2\,\text{cis}\left(-\frac{\pi}{3}\right)$ , $2\,\text{cis}\,\frac{\pi}{6}$
- Moduli: $\dfrac{2\sqrt{2} \times 2}{2} = 2\sqrt{2}$
- Arguments: $\dfrac{\pi}{4} - \dfrac{\pi}{3} - \dfrac{\pi}{6} = -\dfrac{\pi}{4}$
- Answer: $2\sqrt{2}\,\text{cis}\left(-\frac{\pi}{4}\right)$ , or $2 - 2i$ in rectangular form.
Note how little arithmetic that took compared with expanding three brackets and rationalising.

Tips and Tricks

The rules are multiply/add and divide/subtract. Getting them crossed — adding the moduli, or multiplying the arguments — is the standard error, and it produces an answer that is wrong in both parts. Say the rule out loud as you apply it.

Reduce the argument to the principal range at the end, not part-way through. Doing it once, at the finish, is less error-prone than adjusting at every step.

When a question gives you numbers in mixed forms, convert everything to one form before you start. Half-polar, half-rectangular working is where sign errors breed.

Use the modulus rules as a check. If you multiply $3\,\text{cis}\,\alpha$ by $5\,\text{cis}\,\beta$ and your answer has modulus anything other than 15, you have made an arithmetic slip and you can catch it before it propagates.

Recognise multiplying by $i$ as a quarter-turn rotation. Questions about squares, rectangles and equilateral triangles on the Argand diagram are almost always asking you to rotate one vertex about another, and rotation is just multiplication by $\text{cis}\,\theta$ .

Worked ExampleCombining products and quotients in polar form

Given $z = 2\,\text{cis}\,\dfrac{5\pi}{6}$ and $w = 4\,\text{cis}\left(-\dfrac{\pi}{2}\right)$ , find $zw$ and $\dfrac{z}{w}$ , giving each answer in polar form with a principal argument.

Step 1 — The product: handle the moduli

Multiply the moduli:

$|zw| = 2 \times 4 = 8$

Step 2 — The product: handle the arguments

Add the arguments:

$\arg(zw) = \frac{5\pi}{6} + \left(-\frac{\pi}{2}\right) = \frac{5\pi}{6} - \frac{3\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3}$

Step 3 — Check the range

$\frac{\pi}{3}$ is comfortably inside $(-\pi, \pi]$ , so no adjustment is needed.

Step 4 — The quotient: handle the moduli

Divide the moduli:

$\left|\frac{z}{w}\right| = \frac{2}{4} = \frac{1}{2}$

Step 5 — The quotient: handle the arguments

Subtract the arguments. Take care with the double negative:

$\arg\left(\frac{z}{w}\right) = \frac{5\pi}{6} - \left(-\frac{\pi}{2}\right) = \frac{5\pi}{6} + \frac{3\pi}{6} = \frac{8\pi}{6} = \frac{4\pi}{3}$

Step 6 — Bring the argument into the principal range

$\frac{4\pi}{3} \approx 4.19$ , which is greater than $\pi \approx 3.14$ , so it is outside the range. Subtract $2\pi$ :

$\frac{4\pi}{3} - 2\pi = \frac{4\pi}{3} - \frac{6\pi}{3} = -\frac{2\pi}{3}$

Step 7 — Sense-check both answers

$z$ is in the second quadrant and $w$ points straight down. Rotating $z$ clockwise by a quarter turn (multiplying by $w$ 's direction) should bring it into the first quadrant — and $\frac{\pi}{3}$ is first-quadrant ✓

Dividing rotates $z$ anticlockwise by a quarter turn instead, taking it from the second quadrant into the third — and $-\frac{2\pi}{3}$ is third-quadrant ✓