Math Core

A complex number $a + bi$ is really a pair of real numbers, so you can plot it as a point. Once it's a point, you can describe it the polar way, with a distance and an angle. That polar description, called trigonometric form, turns multiplication and division of complex numbers into simple arithmetic on distances and angles.

The complex plane

To picture complex numbers, use a plane whose horizontal axis is the real axis and whose vertical axis is the imaginary axis. The number $a + bi$ is plotted at the point $(a, b)$ . For example, $3 + 4i$ sits $3$ units right and $4$ units up, and $-2i$ sits $2$ units down on the imaginary axis.

The complex number 3 + 4i is the point (3, 4). Its distance from 0 is the modulus r = 5, and θ is its argument.Open in grapher →

Modulus and argument

Just like a vector or a polar point, a complex number has a length and a direction.

Definition

Modulus and argument

For $z = a + bi$ , the modulus is its distance from $0$ :

|z| = r = \sqrt{a^2 + b^2}.

An argument of $z$ is an angle $\theta$ from the positive real axis to the point, so that $\tan\theta = \dfrac{b}{a}$ (with $\theta$ in the correct quadrant).

For $3 + 4i$ , the modulus is $\sqrt{9 + 16} = 5$ and the argument is $\tan^{-1}\tfrac{4}{3} \approx 53.1^\circ$ . As with polar coordinates, adding $360^\circ$ to an argument gives another argument, so we usually choose one between $0^\circ$ and $360^\circ$ (or between $0$ and $2\pi$ ).

Trigonometric form

Since the point is at distance $r$ and angle $\theta$ , its coordinates are $a = r\cos\theta$ and $b = r\sin\theta$ . Substituting into $a + bi$ :

Trigonometric (polar) form

A complex number $z = a + bi$ can be written

z = r(\cos\theta + i\sin\theta)

where $r = |z|$ and $\theta$ is an argument of $z$ . This is often abbreviated $z = r\operatorname{cis}\theta$ .

Worked example: Standard form to trigonometric form

Write $z = -1 + i\sqrt{3}$ in trigonometric form, with $0^\circ \le \theta < 360^\circ$ .

$r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2$ .

$\tan\theta = \dfrac{\sqrt{3}}{-1} = -\sqrt{3}$ . The reference angle is $60^\circ$ , and the point $(-1, \sqrt{3})$ is in Quadrant II, so $\theta = 120^\circ$ .

z = 2(\cos 120^\circ + i\sin 120^\circ).

Worked example: Trigonometric form to standard form

Write $z = 6\left(\cos\tfrac{7\pi}{4} + i\sin\tfrac{7\pi}{4}\right)$ in the form $a + bi$ .

Evaluate the cosine and sine, then distribute:

z = 6\left(\frac{\sqrt{2}}{2} + i\left(-\frac{\sqrt{2}}{2}\right)\right) = 3\sqrt{2} - 3\sqrt{2}\,i.

Common mistake

Finding the argument with $\tan^{-1}\left(\dfrac{b}{a}\right)$ alone ignores the quadrant. For $-1 + i\sqrt{3}$ the calculator gives $-60^\circ$ , which points into Quadrant IV, the wrong place. Plot the point first and choose the angle that actually lands there.

Multiplying and dividing

Here is where trigonometric form pays off. Multiply $z_1 = r_1(\cos\alpha + i\sin\alpha)$ by $z_2 = r_2(\cos\beta + i\sin\beta)$ and expand:

z_1 z_2 = r_1 r_2\left[(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + i(\sin\alpha\cos\beta + \cos\alpha\sin\beta)\right].

The two groups in parentheses are exactly the sum formulas for cosine and sine, so they collapse to $\cos(\alpha + \beta)$ and $\sin(\alpha + \beta)$ .

Products and quotients

z_1 z_2 = r_1 r_2\left[\cos(\alpha + \beta) + i\sin(\alpha + \beta)\right]

\frac{z_1}{z_2} = \frac{r_1}{r_2}\left[\cos(\alpha - \beta) + i\sin(\alpha - \beta)\right] \qquad (z_2 \ne 0)

To multiply, multiply the moduli and add the arguments. To divide, divide the moduli and subtract the arguments.

Geometrically, multiplying by $z_2$ stretches a number by a factor of $r_2$ and rotates it by $\beta$ . For example, multiplying by $i = 1(\cos 90^\circ + i\sin 90^\circ)$ rotates any point $90^\circ$ counterclockwise without changing its distance from $0$ .

Worked example: A product and a quotient

Let $z_1 = 8(\cos 150^\circ + i\sin 150^\circ)$ and $z_2 = 2(\cos 60^\circ + i\sin 60^\circ)$ . Find $z_1 z_2$ and $\dfrac{z_1}{z_2}$ , and write each in the form $a + bi$ .

Product. $r = 8 \cdot 2 = 16$ and $\theta = 150^\circ + 60^\circ = 210^\circ$ :

z_1 z_2 = 16(\cos 210^\circ + i\sin 210^\circ) = 16\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) = -8\sqrt{3} - 8i.

Quotient. $r = \dfrac{8}{2} = 4$ and $\theta = 150^\circ - 60^\circ = 90^\circ$ :

\frac{z_1}{z_2} = 4(\cos 90^\circ + i\sin 90^\circ) = 4(0 + i) = 4i.

Tip

Check a product with moduli alone: $|z_1 z_2|$ should equal $|z_1| \cdot |z_2|$ . For the example, $|-8\sqrt{3} - 8i| = \sqrt{192 + 64} = \sqrt{256} = 16 = 8 \cdot 2$ .