9709 P3 3.9 Complex Numbers II

💡 Lead In

Quick Recall

Pick the correct option in each question before moving on.

Review 1. In the polar form $z = r(\cos\theta + \mathrm{i}\sin\theta) \equiv re^{\mathrm{i}\theta}$, the quantity $r$ is:

Correct! $r = |z| = \sqrt{x^2 + y^2}$ is the modulus (distance from the origin), while $\theta$ is the argument.

Review 2. In the polar form $z = r(\cos\theta + \mathrm{i}\sin\theta) \equiv re^{\mathrm{i}\theta}$, the quantity $\theta$ is:

Correct! $\theta = \arg z$ is the argument — the angle the position vector of $z$ makes with the positive real axis.

The Wall

If we multiply two complex numbers in Cartesian form, the algebra can be messy. But what happens when we multiply two complex numbers given in polar form? Does the product have a simple geometric interpretation?

Multiplying two complex numbers: the product's modulus is the product of the moduli, and its argument is the sum of the arguments.

The Need

We need a systematic way to multiply and divide complex numbers in polar form. Understanding the geometric meaning of these operations will allow us to solve problems involving rotations and scaling in the complex plane.

📖 Key Concepts — Multiplication and Division in Polar Form

Multiplication in Polar Form

If $z_1 = r_1(\cos\theta_1 + \mathrm{i}\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + \mathrm{i}\sin\theta_2)$, then:

Multiplication: $$z_1 z_2 = r_1 r_2\left(\cos(\theta_1 + \theta_2) + \mathrm{i}\sin(\theta_1 + \theta_2)\right)$$

$|z_1 z_2| = |z_1||z_2|$, $\arg(z_1 z_2) = \arg z_1 + \arg z_2$

Division: $$\frac{z_1}{z_2} = \frac{r_1}{r_2}\left(\cos(\theta_1 - \theta_2) + \mathrm{i}\sin(\theta_1 - \theta_2)\right)$$

$\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$, $\arg\left(\dfrac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$

Exponential Form: $$z_1 z_2 = r_1 r_2 e^{\mathrm{i}(\theta_1 + \theta_2)} \qquad \frac{z_1}{z_2} = \frac{r_1}{r_2} e^{\mathrm{i}(\theta_1 - \theta_2)}$$

The exponential form makes multiplication and division especially elegant.

Group Activity Polar Operation Race

Task: In groups of 3–4, compete to compute products and quotients in polar form. You have 8 minutes.

Instructions:

Each group receives a set of 4 problems involving multiplication and division of complex numbers in polar form.
Work together to solve each problem using the polar form rules.
Write your answers in both $r(\cos\theta + \mathrm{i}\sin\theta)$ and $re^{\mathrm{i}\theta}$ forms.
The first group to correctly solve all 4 problems wins.

Race Problems:

Given $z_1 = 3\left(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\right)$ and $z_2 = 2\left(\cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}\right)$, find $z_1 z_2$.
Given $z_1 = 8\left(\cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3}\right)$ and $z_2 = 4\left(\cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}\right)$, find $\dfrac{z_1}{z_2}$.
Given $z_1 = 5e^{\mathrm{i}\pi/3}$ and $z_2 = 2e^{-\mathrm{i}\pi/6}$, find $z_1 z_2$ in exponential form.
Given $z_1 = 6e^{\mathrm{i}3\pi/4}$ and $z_2 = 3e^{\mathrm{i}\pi/4}$, find $\dfrac{z_1}{z_2}$ in exponential form.

Answer 1

$z_1 z_2 = 6\left(\cos\frac{5\pi}{12} + \mathrm{i}\sin\frac{5\pi}{12}\right) = 6e^{\mathrm{i}5\pi/12}$

Answer 2

$\dfrac{z_1}{z_2} = 2\left(\cos\frac{\pi}{2} + \mathrm{i}\sin\frac{\pi}{2}\right) = 2e^{\mathrm{i}\pi/2} = 2\mathrm{i}$

Answer 3

$z_1 z_2 = 10e^{\mathrm{i}\pi/6}$

Answer 4

$\dfrac{z_1}{z_2} = 2e^{\mathrm{i}\pi/2} = 2\mathrm{i}$

Individual Activity Products and Quotients Practice

Task: Work independently to compute each product or quotient. Show your working clearly. You have 5 minutes.

$4\left(\cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}\right) \times 3\left(\cos\frac{\pi}{4} + \mathrm{i}\sin\frac{\pi}{4}\right)$
$10\left(\cos\frac{5\pi}{6} + \mathrm{i}\sin\frac{5\pi}{6}\right) \div 2\left(\cos\frac{\pi}{3} + \mathrm{i}\sin\frac{\pi}{3}\right)$
$6e^{\mathrm{i}2\pi/3} \times 2e^{\mathrm{i}\pi/4}$
$8e^{\mathrm{i}\pi/2} \div 4e^{\mathrm{i}\pi/6}$

Solution 1

$4 \times 3(\cos(\frac{\pi}{3}+\frac{\pi}{4}) + \mathrm{i}\sin(\frac{\pi}{3}+\frac{\pi}{4})) = 12(\cos\frac{7\pi}{12} + \mathrm{i}\sin\frac{7\pi}{12})$.

Solution 2

$\frac{10}{2}(\cos(\frac{5\pi}{6}-\frac{\pi}{3}) + \mathrm{i}\sin(\frac{5\pi}{6}-\frac{\pi}{3})) = 5(\cos\frac{\pi}{2} + \mathrm{i}\sin\frac{\pi}{2}) = 5\mathrm{i}$.

Solution 3

$6 \times 2 e^{\mathrm{i}(2\pi/3 + \pi/4)} = 12e^{\mathrm{i}11\pi/12}$.

Solution 4

$\frac{8}{4} e^{\mathrm{i}(\pi/2 - \pi/6)} = 2e^{\mathrm{i}\pi/3}$.

✏️ Worked Example

Worked Example — Ex 11G Q2 Medium

Given that $z = 2\left(\cos\dfrac{\pi}{4} + \mathrm{i}\sin\dfrac{\pi}{4}\right)$ and $w = 3\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right)$, find the following in polar form.

(i) $wz$

(ii) $\dfrac{w}{z}$

Step 1: Identify $|z|$, $\arg z$, $|w|$, $\arg w$

$|z| = 2$, $\arg z = \dfrac{\pi}{4}$; $|w| = 3$, $\arg w = \dfrac{\pi}{3}$.

Step 2: Find $wz$

$|wz| = |w||z| = 3 \times 2 = 6$.

$\arg(wz) = \arg w + \arg z = \dfrac{\pi}{3} + \dfrac{\pi}{4} = \dfrac{7\pi}{12}$.

$wz = 6\left(\cos\dfrac{7\pi}{12} + \mathrm{i}\sin\dfrac{7\pi}{12}\right)$.

Step 3: Find $\dfrac{w}{z}$

$\left|\dfrac{w}{z}\right| = \dfrac{|w|}{|z|} = \dfrac{3}{2}$.

$\arg\left(\dfrac{w}{z}\right) = \arg w - \arg z = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}$.

$\dfrac{w}{z} = \dfrac{3}{2}\left(\cos\dfrac{\pi}{12} + \mathrm{i}\sin\dfrac{\pi}{12}\right)$.

📝 Concept Quiz

Q1. Given $z_1 = 3(\cos 0.5 + \mathrm{i}\sin 0.5)$ and $z_2 = 4(\cos 1.2 + \mathrm{i}\sin 1.2)$, what is $|z_1 z_2|$?

Correct! $|z_1 z_2| = |z_1||z_2| = 3 \times 4 = 12$.

Q2. For the same $z_1$ and $z_2$, what is $\arg(z_1 z_2)$?

Correct! $\arg(z_1 z_2) = \arg z_1 + \arg z_2 = 0.5 + 1.2 = 1.7$.

Q3. If $z = 5e^{\mathrm{i}\pi/4}$ and $w = 2e^{\mathrm{i}\pi/6}$, what is $\dfrac{z}{w}$ in exponential form?

Correct! $\frac{z}{w} = \frac{5}{2}e^{\mathrm{i}(\pi/4 - \pi/6)} = \frac{5}{2}e^{\mathrm{i}\pi/12}$.

Q4. What is $\left|\dfrac{z}{w}\right|$ for the same $z$ and $w$ as in Q3?

Correct! $\left|\frac{z}{w}\right| = \frac{|z|}{|w|} = \frac{5}{2}$.

✏️ More Practice

Ex 11G Q1 Standard

Find the following.

(i) $8(\cos 0.2 + \mathrm{i}\sin 0.2) \times 4(\cos 0.4 + \mathrm{i}\sin 0.4)$

(ii) $8(\cos 0.2 + \mathrm{i}\sin 0.2) \div 4(\cos 0.4 + \mathrm{i}\sin 0.4)$

(iii) $6\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right) \times 2\left(\cos\dfrac{\pi}{6} + \mathrm{i}\sin\dfrac{\pi}{6}\right)$

(iv) $6\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right) \div 2\left(\cos\dfrac{\pi}{6} + \mathrm{i}\sin\dfrac{\pi}{6}\right)$

(v) $12(\cos\pi + \mathrm{i}\sin\pi) \times 2\left(\cos\dfrac{\pi}{4} + \mathrm{i}\sin\dfrac{\pi}{4}\right)$

(vi) $12(\cos\pi + \mathrm{i}\sin\pi) \div 2\left(\cos\dfrac{\pi}{4} + \mathrm{i}\sin\dfrac{\pi}{4}\right)$

Solution (i)

$8 \times 4(\cos(0.2+0.4) + \mathrm{i}\sin(0.2+0.4)) = 32(\cos 0.6 + \mathrm{i}\sin 0.6)$.

Solution (ii)

$\frac{8}{4}(\cos(0.2-0.4) + \mathrm{i}\sin(0.2-0.4)) = 2(\cos(-0.2) + \mathrm{i}\sin(-0.2))$.

Solution (iii)

$6 \times 2(\cos(\frac{\pi}{3}+\frac{\pi}{6}) + \mathrm{i}\sin(\frac{\pi}{3}+\frac{\pi}{6})) = 12(\cos\frac{\pi}{2} + \mathrm{i}\sin\frac{\pi}{2}) = 12\mathrm{i}$.

Solution (iv)

$\frac{6}{2}(\cos(\frac{\pi}{3}-\frac{\pi}{6}) + \mathrm{i}\sin(\frac{\pi}{3}-\frac{\pi}{6})) = 3(\cos\frac{\pi}{6} + \mathrm{i}\sin\frac{\pi}{6}) = 3\left(\frac{\sqrt{3}}{2} + \frac{1}{2}\mathrm{i}\right)$.

Solution (v)

$12 \times 2(\cos(\pi+\frac{\pi}{4}) + \mathrm{i}\sin(\pi+\frac{\pi}{4})) = 24(\cos\frac{5\pi}{4} + \mathrm{i}\sin\frac{5\pi}{4})$.

Solution (vi)

$\frac{12}{2}(\cos(\pi-\frac{\pi}{4}) + \mathrm{i}\sin(\pi-\frac{\pi}{4})) = 6(\cos\frac{3\pi}{4} + \mathrm{i}\sin\frac{3\pi}{4})$.

Ex 11G Q2 Medium

Given that $z = 2\left(\cos\dfrac{\pi}{4} + \mathrm{i}\sin\dfrac{\pi}{4}\right)$ and $w = 3\left(\cos\dfrac{\pi}{3} + \mathrm{i}\sin\dfrac{\pi}{3}\right)$, find the following in polar form.

(i) $wz$

(ii) $\dfrac{w}{z}$

(iii) $\dfrac{z}{w}$

(iv) $\dfrac{1}{z}$

(v) $w^2$

(vi) $z^5$

Solution (i) $wz$

$|wz| = 3 \times 2 = 6$, $\arg(wz) = \frac{\pi}{3} + \frac{\pi}{4} = \frac{7\pi}{12}$.

$wz = 6\left(\cos\frac{7\pi}{12} + \mathrm{i}\sin\frac{7\pi}{12}\right)$.

Solution (ii) $\dfrac{w}{z}$

$\left|\frac{w}{z}\right| = \frac{3}{2}$, $\arg(\frac{w}{z}) = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$.

$\frac{w}{z} = \frac{3}{2}\left(\cos\frac{\pi}{12} + \mathrm{i}\sin\frac{\pi}{12}\right)$.

Solution (iii) $\dfrac{z}{w}$

$\left|\frac{z}{w}\right| = \frac{2}{3}$, $\arg(\frac{z}{w}) = \frac{\pi}{4} - \frac{\pi}{3} = -\frac{\pi}{12}$.

$\frac{z}{w} = \frac{2}{3}\left(\cos\left(-\frac{\pi}{12}\right) + \mathrm{i}\sin\left(-\frac{\pi}{12}\right)\right)$.

Solution (iv) $\dfrac{1}{z}$

$\left|\frac{1}{z}\right| = \frac{1}{2}$, $\arg(\frac{1}{z}) = -\arg z = -\frac{\pi}{4}$.

$\frac{1}{z} = \frac{1}{2}\left(\cos\left(-\frac{\pi}{4}\right) + \mathrm{i}\sin\left(-\frac{\pi}{4}\right)\right)$.

Solution (v) $w^2$

$|w^2| = 3^2 = 9$, $\arg(w^2) = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$.

$w^2 = 9\left(\cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3}\right)$.

Solution (vi) $z^5$

$|z^5| = 2^5 = 32$, $\arg(z^5) = 5 \times \frac{\pi}{4} = \frac{5\pi}{4}$.

$z^5 = 32\left(\cos\frac{5\pi}{4} + \mathrm{i}\sin\frac{5\pi}{4}\right)$.

Ex 11G Q7 Standard

Find the following.

(a) $2e^{3\mathrm{i}} \times 5e^{-2\mathrm{i}}$

(b) $8e^{5\mathrm{i}} \div 2e^{5\mathrm{i}}$

Solution (a)

$2 \times 5 e^{(3-2)\mathrm{i}} = 10e^{\mathrm{i}}$.

Solution (b)

$\frac{8}{2} e^{(5-5)\mathrm{i}} = 4e^{0} = 4$.

Solution (c)

$3 \times 2 e^{(7+1)\mathrm{i}} = 6e^{8\mathrm{i}}$.

🎫

Exit Ticket

Check the learning objectives you feel confident about:

I can multiply two complex numbers in polar form using $|z_1z_2| = |z_1||z_2|$ and $\arg(z_1z_2) = \arg z_1 + \arg z_2$. I can divide two complex numbers in polar form using $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$ and $\arg\left(\frac{z_1}{z_2}\right) = \arg z_1 - \arg z_2$. I can multiply and divide complex numbers in exponential form $re^{\mathrm{i}\theta}$. I can interpret multiplication by a complex number geometrically as a combined scaling and rotation.