Polar Form

A2 · P3 · 3 min

Writing a complex number by its modulus and argument, $z = r(\cos\theta + i\sin\theta)$ , makes multiplication and division simple: multiply the moduli and add the arguments. Powers follow at once, and the geometry of rotation and enlargement becomes visible.

Modulus-argument form

Key result

z = r(\cos\theta + i\sin\theta), \qquad r = |z| \geq 0,\quad \theta = \arg z.

Converting from $x + iy$ : $r = \sqrt{x^2 + y^2}$ , $\theta$ from the quadrant. Converting back: $x = r\cos\theta$ , $y = r\sin\theta$ .

Converting to polar form

Write $-1 + i\sqrt{3}$ and $-2i$ in modulus-argument form.

Solution

$|-1 + i\sqrt{3}| = 2$ , second quadrant with $\alpha = \tfrac{\pi}{3}$ : $\arg = \tfrac{2\pi}{3}$ . So $2\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right)$ .

$-2i$ : modulus $2$ , argument $-\tfrac{\pi}{2}$ : $2\left(\cos\left(-\tfrac{\pi}{2}\right) + i\sin\left(-\tfrac{\pi}{2}\right)\right)$ .

Multiplication and division

Key result

If $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$ :

z_1 z_2 = r_1 r_2\big(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\big), \qquad \frac{z_1}{z_2} = \frac{r_1}{r_2}\big(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\big).

So $|z_1 z_2| = |z_1||z_2|$ , $\arg(z_1 z_2) = \arg z_1 + \arg z_2$ , and for division the modulus divides and the arguments subtract. Adjust the argument by $\pm 2\pi$ if it leaves $(-\pi, \pi]$ .

Proof

$(\cos\theta_1 + i\sin\theta_1)(\cos\theta_2 + i\sin\theta_2) = (\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2) = \cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)$ by the compound-angle formulae.

Multiplying in polar form

$z = 2\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)$ and $w = 3\left(\cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3}\right)$ . Find $zw$ and $\dfrac{z}{w}$ in polar form, and $zw$ in Cartesian form.

Solution

$zw = 6\left(\cos\tfrac{7\pi}{12} + i\sin\tfrac{7\pi}{12}\right)$ . $\dfrac{z}{w} = \tfrac{2}{3}\left(\cos\left(-\tfrac{\pi}{12}\right) + i\sin\left(-\tfrac{\pi}{12}\right)\right)$ .

Cartesian: $zw = 6\cos\tfrac{7\pi}{12} + 6i\sin\tfrac{7\pi}{12} \approx -1.55 + 5.80i$ .

Argument outside the principal range

$z$ has argument $\tfrac{3\pi}{4}$ and $w$ has argument $\tfrac{2\pi}{3}$ . Find the principal argument of $zw$ and of $\dfrac{w}{z}$ .

Solution

$\arg(zw) = \tfrac{3\pi}{4} + \tfrac{2\pi}{3} = \tfrac{17\pi}{12}$ , which exceeds $\pi$ ; subtract $2\pi$ : $-\tfrac{7\pi}{12}$ .

$\arg\dfrac{w}{z} = \tfrac{2\pi}{3} - \tfrac{3\pi}{4} = -\tfrac{\pi}{12}$ .

Powers

Repeated multiplication gives $z^n = r^n(\cos n\theta + i\sin n\theta)$ . This is de Moivre's theorem; the syllabus only needs the case that follows from multiplying a few times.

A power via polar form

Find $(1 + i)^8$ .

Solution

$1 + i = \sqrt{2}\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right)$ , so $(1 + i)^8 = (\sqrt{2})^8(\cos 2\pi + i\sin 2\pi) = 16$ .

Finding a complex number from conditions

$z$ satisfies $|z| = 2$ , $\arg(z^2) = \tfrac{\pi}{2}$ and $0 < \arg z < \pi$ . Find $z$ in Cartesian form.

Solution

$\arg z^2 = 2\arg z$ , so $\arg z = \tfrac{\pi}{4}$ (the other option, $\tfrac{\pi}{4} + \pi$ , is outside the range). $z = 2\left(\cos\tfrac{\pi}{4} + i\sin\tfrac{\pi}{4}\right) = \sqrt{2} + i\sqrt{2}$ .

Watch out

$r$ must be positive. $-2(\cos\theta + i\sin\theta)$ is not in polar form; rewrite it as $2\big(\cos(\theta + \pi) + i\sin(\theta + \pi)\big)$ .

Exam tip

Questions often give $z$ and $w$ in Cartesian form, ask for $zw$ or $z/w$ by direct calculation, then ask for the modulus and argument of the result. Compute the modulus and argument of $z$ and $w$ separately and combine; it is quicker and provides a check on the direct calculation.

Practice

Question

Write $-3 - 3i$ and $\sqrt{3} - i$ in polar form.
$z = 4\left(\cos\tfrac{\pi}{6} + i\sin\tfrac{\pi}{6}\right)$ , $w = 2\left(\cos\tfrac{\pi}{2} + i\sin\tfrac{\pi}{2}\right)$ . Find $zw$ and $\dfrac{z}{w}$ in Cartesian form.
Find $\left(\sqrt{3} + i\right)^6$ .
$|z| = 3$ and $\arg z = -\tfrac{2\pi}{3}$ . Find $\arg(z^2)$ and $\arg\left(\dfrac{1}{z}\right)$ as principal arguments.

Answers

$3\sqrt{2}\left(\cos\left(-\tfrac{3\pi}{4}\right) + i\sin\left(-\tfrac{3\pi}{4}\right)\right)$ ; $2\left(\cos\left(-\tfrac{\pi}{6}\right) + i\sin\left(-\tfrac{\pi}{6}\right)\right)$ .
$zw = 8\left(\cos\tfrac{2\pi}{3} + i\sin\tfrac{2\pi}{3}\right) = -4 + 4\sqrt{3}i$ ; $\dfrac{z}{w} = 2\left(\cos\left(-\tfrac{\pi}{3}\right) + i\sin\left(-\tfrac{\pi}{3}\right)\right) = 1 - \sqrt{3}i$ .
$2^6(\cos\pi + i\sin\pi) = -64$ .
$\arg z^2 = -\tfrac{4\pi}{3} + 2\pi = \tfrac{2\pi}{3}$ ; $\arg\dfrac{1}{z} = \tfrac{2\pi}{3}$ .

Modulus-argument form

Multiplication and division

Powers

Practice

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