Products and Quotients in Polar Form, precalculus lesson

Multiplying $(2 + 3i)(1 - i)$ in rectangular form takes a few lines and gives $5 + i$ , but nothing in the work tells you what happened geometrically. In polar form the answer is visible at once, because lengths multiply and angles add. That one fact makes rotations of the plane, powers and roots of complex numbers easy to handle.

Key terms

Complex multiplication

Definition For real numbers $a$ , $b$ , $c$ and $d$ , the product of $a + bi$ and $c + di$ is

\begin{aligned} (a + bi)(c + di) &= (ac - bd) \\[0.4em] &\quad + (ad + bc)i, \end{aligned}

which is what you get by expanding and using $i^2 = -1$ . In polar form, if $z_1 = r_1\operatorname{cis}\theta_1$ and $z_2 = r_2\operatorname{cis}\theta_2$ with $r_1, r_2 > 0$ , then

z_1 z_2 = r_1 r_2 \operatorname{cis}(\theta_1 + \theta_2).

The proof uses the sum formulas for cosine and sine,

\begin{aligned} &\operatorname{cis}\theta_1 \operatorname{cis}\theta_2 \\[0.4em] &= (\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) \\[0.4em] &\quad + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2) \\[0.4em] &= \cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2). \end{aligned}

In plain words To multiply in polar form, multiply the lengths and add the angles.

Example In polar form the product is quick,

2\operatorname{cis}\frac{\pi}{6} \cdot 3\operatorname{cis}\frac{\pi}{3} = 6\operatorname{cis}\frac{\pi}{2} = 6i.

The rectangular route agrees. The factors are $\sqrt{3} + i$ and $\frac{3}{2} + \frac{3\sqrt{3}}{2}i$ , and their product is

\begin{aligned} &\left(\frac{3\sqrt{3}}{2} - \frac{3\sqrt{3}}{2}\right) + \left(\frac{9}{2} + \frac{3}{2}\right)i \\[0.8em] &\quad = 6i. \end{aligned}

Complex division by a nonzero number

Definition Let $w \ne 0$ . The quotient $z/w$ is the unique complex number $q$ with $wq = z$ . In rectangular form you find it by multiplying the numerator and the denominator by the conjugate $\overline{w}$ , which makes the denominator the positive real number $w\,\overline{w} = \lvert w \rvert^2$ . In polar form, if $z_1 = r_1\operatorname{cis}\theta_1$ and $z_2 = r_2\operatorname{cis}\theta_2$ with $r_1 > 0$ and $r_2 > 0$ , then

\frac{z_1}{z_2} = \frac{r_1}{r_2}\operatorname{cis}(\theta_1 - \theta_2).

This follows from the product rule, since multiplying the right side by $z_2 = r_2\operatorname{cis}\theta_2$ gives back $r_1\operatorname{cis}\theta_1$ . Division by $0$ is undefined, because $0 \cdot q = 0$ for every $q$ , so no $q$ solves $0 \cdot q = z$ when $z \ne 0$ , and every $q$ solves it when $z = 0$ .

In plain words To divide, divide the lengths and subtract the angles. The divisor must not be zero. You cannot divide by a length of zero, and zero has no angle to subtract.

Example In polar form,

\frac{8\operatorname{cis}\frac{5\pi}{6}}{2\operatorname{cis}\frac{\pi}{3}} = 4\operatorname{cis}\frac{\pi}{2} = 4i.

The same quotient in rectangular form, with the conjugate $1 - \sqrt{3}\,i$ of the denominator, is

\frac{-4\sqrt{3} + 4i}{1 + \sqrt{3}\,i} \cdot \frac{1 - \sqrt{3}\,i}{1 - \sqrt{3}\,i} = \frac{16i}{4} = 4i.

Multiplying and dividing moduli

Definition For all complex numbers $z$ and $w$ ,

\lvert zw \rvert = \lvert z \rvert\,\lvert w \rvert,

and if $w \ne 0$ , also

\left\lvert \frac{z}{w} \right\rvert = \frac{\lvert z \rvert}{\lvert w \rvert}.

For nonzero numbers these are the modulus parts of the polar product and quotient formulas, and they hold trivially when $z = 0$ or, in the product, when $w = 0$ .

In plain words The length of a product is the product of the lengths, and the length of a quotient is the quotient of the lengths. You can find $\lvert zw \rvert$ without computing $zw$ .

Example With $z = 3 + 4i$ and $w = 5 - 12i$ , the moduli are $5$ and $13$ , so $\lvert zw \rvert = 5 \cdot 13 = 65$ . Multiplying out confirms it. The product is $zw = 63 - 16i$ , and

\begin{aligned} \sqrt{63^2 + 16^2} &= \sqrt{3969 + 256} \\[0.4em] &= \sqrt{4225} \\[0.4em] &= 65. \end{aligned}

In the same way $\lvert z/w \rvert = \frac{5}{13}$ .

Adding and subtracting arguments

Definition Let $z$ and $w$ be nonzero, with arguments $\alpha$ and $\beta$ . Then $\alpha + \beta$ is an argument of $zw$ , and $\alpha - \beta$ is an argument of $z/w$ . For principal arguments the equations hold only up to a full turn,

\begin{gathered} \operatorname{Arg}(zw) = \operatorname{Arg} z + \operatorname{Arg} w + 2k\pi, \\[0.4em] k \in \{-1, 0, 1\}, \end{gathered}

where $k$ is the integer that brings the result into $(-\pi, \pi]$ . The same holds for $\operatorname{Arg}(z/w)$ with $\operatorname{Arg} z - \operatorname{Arg} w$ in place of the sum.

In plain words Angles add when you multiply and subtract when you divide. If the result leaves the principal range, add or take away one full turn.

Example Take $z = w = -1 + i$ , so $\operatorname{Arg} z = \operatorname{Arg} w = \frac{3\pi}{4}$ . The sum $\frac{3\pi}{2}$ is outside $(-\pi, \pi]$ . The product is

(-1 + i)^2 = 1 - 2i + i^2 = -2i,

and its principal argument is $-\frac{\pi}{2} = \frac{3\pi}{2} - 2\pi$ , so here $k = -1$ . The modulus is $\sqrt{2} \cdot \sqrt{2} = 2$ , as the product rule predicts.

Rotation and scaling

Definition Let $w = \rho\operatorname{cis}\varphi$ with $\rho > 0$ . Multiplying by $w$ sends each complex number $z$ to $wz$ . This multiplies the distance of every point from $0$ by $\rho$ and turns every point about $0$ through the angle $\varphi$ , counterclockwise when $\varphi > 0$ . So multiplication by $w$ is a rotation about the origin together with a scaling by the factor $\rho$ , and the two can be done in either order. Dividing by $w$ undoes it, turning by $-\varphi$ and scaling by $1/\rho$ . In particular, multiplying by $i = \operatorname{cis}\frac{\pi}{2}$ is a quarter turn counterclockwise.

In plain words Multiplying by a nonzero complex number turns the whole plane around the origin and stretches or shrinks it at the same time. A number of modulus $1$ only turns, and a positive real number only stretches or shrinks.

Example Multiply the corners $0$ , $1$ , $1 + i$ and $i$ of a unit square by

w = 1 + i = \sqrt{2}\operatorname{cis}\frac{\pi}{4}.

The results are $0$ , $1 + i$ , $2i$ and $-1 + i$ . These are the corners of a square with side $\sqrt{2}$ , turned $45^\circ$ counterclockwise about the origin. In the same way the product $i(4 + 2i) = -2 + 4i$ is the point $4 + 2i$ turned a quarter turn counterclockwise.

Check yourself

Compute $2\operatorname{cis}\frac{2\pi}{5} \cdot 5\operatorname{cis}\frac{3\pi}{5}$ and write the result in rectangular form.

Show the answer

Multiplying the moduli and adding the arguments gives

2\operatorname{cis}\frac{2\pi}{5} \cdot 5\operatorname{cis}\frac{3\pi}{5} = 10\operatorname{cis}\pi = -10.

Divide $6\operatorname{cis}\frac{\pi}{4}$ by $3\operatorname{cis}\frac{3\pi}{4}$ .

Show the answer

Dividing the moduli and subtracting the arguments gives

\frac{6\operatorname{cis}\frac{\pi}{4}}{3\operatorname{cis}\frac{3\pi}{4}} = 2\operatorname{cis}\left(-\frac{\pi}{2}\right) = -2i.

Find a complex number $w$ such that multiplying by $w$ turns every point a quarter turn clockwise about $0$ and halves its distance from $0$ . Check your answer on $z = 2$ .

Show the answer

The number needs modulus $\frac{1}{2}$ and argument $-\frac{\pi}{2}$ , so

w = \frac{1}{2}\operatorname{cis}\left(-\frac{\pi}{2}\right) = -\frac{1}{2}i.

For $z = 2$ the product is $wz = -i$ , which is $1$ unit from the origin, straight down, a quarter turn clockwise from the direction of $2$ .

Products and Quotients in Polar Form

Key terms

Complex multiplication

Complex division by a nonzero number

Multiplying and dividing moduli

Adding and subtracting arguments

Rotation and scaling

Check yourself

How this lesson connects

Builds on

Used later in