Lesson 10.5 : Polar Form of Complex Numbers

New Concept

Unlock a new perspective on complex numbers by expressing them in polar form, $z = r(\operatorname{cos}\theta + i\operatorname{sin}\theta)$. This powerful representation simplifies multiplication, division, finding powers, and extracting roots, turning complex calculations into straightforward steps.

What’s next

This is your foundation. Soon, you'll master converting between forms and apply these new skills in worked examples, practice cards, and challenge problems.

Property

Given $z = x + yi$, a complex number, the absolute value of $z$ is defined as

It is the distance from the origin to the point $(x, y)$. Notice that the absolute value of a real number gives the distance of the number from $0$, while the absolute value of a complex number gives the distance of the number from the origin, $(0, 0)$.

Examples

• Given $z = 4 - 3i$, its absolute value is $|z| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

• For $z = \sqrt{7} + 3i$, the absolute value is $|z| = \sqrt{(\sqrt{7})^2 + 3^2} = \sqrt{7 + 9} = \sqrt{16} = 4$.

Property

Writing a complex number in polar form involves the following conversion formulas:

Making a direct substitution, we have

where $r$ is the modulus and $\theta$ is the argument. We often use the abbreviation $r \operatorname{cis} \theta$ to represent $r (\operatorname{cos} \theta + i \operatorname{sin} \theta)$.

Examples

• To write $z = 5 + 5i$ in polar form, find $r = \sqrt{5^2+5^2} = 5\sqrt{2}$ and $\theta = \frac{\pi}{4}$. The polar form is $5\sqrt{2} \operatorname{cis}(\frac{\pi}{4})$.

• For $z = -6i$, the radius is $r=6$. The point is on the negative imaginary axis, so the angle is $\theta = \frac{3\pi}{2}$. The polar form is $6 \operatorname{cis}(\frac{3\pi}{2})$.

Property

Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Given $z = r (\operatorname{cos} \theta + i \operatorname{sin} \theta)$, first evaluate the trigonometric functions $\operatorname{cos} \theta$ and $\operatorname{sin} \theta$. Then, multiply through by $r$ to get $z = x + yi$.

Examples

• Convert $z = 8(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3}))$. This becomes $z = 8(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = 4 + 4i\sqrt{3}$.

• Convert $z = 10 \operatorname{cis}(210^\circ)$. This becomes $z = 10(-\frac{\sqrt{3}}{2} + i(-\frac{1}{2})) = -5\sqrt{3} - 5i$.

Property

If $z_1 = r_1 (\operatorname{cos} \theta_1 + i \operatorname{sin} \theta_1)$ and $z_2 = r_2 (\operatorname{cos} \theta_2 + i \operatorname{sin} \theta_2)$, then the product of these numbers is given as:

Notice that the product calls for multiplying the moduli and adding the angles.

Examples

• For $z_1 = 3 \operatorname{cis}(40^\circ)$ and $z_2 = 5 \operatorname{cis}(20^\circ)$, the product is $z_1 z_2 = (3 \cdot 5) \operatorname{cis}(40^\circ + 20^\circ) = 15 \operatorname{cis}(60^\circ)$.

• For $z_1 = 2 \operatorname{cis}(\frac{\pi}{2})$ and $z_2 = 4 \operatorname{cis}(\frac{3\pi}{4})$, the product is $z_1 z_2 = (2 \cdot 4) \operatorname{cis}(\frac{\pi}{2} + \frac{3\pi}{4}) = 8 \operatorname{cis}(\frac{5\pi}{4})$.

Property

If $z_1 = r_1 (\operatorname{cos} \theta_1 + i \operatorname{sin} \theta_1)$ and $z_2 = r_2 (\operatorname{cos} \theta_2 + i \operatorname{sin} \theta_2)$, then the quotient of these numbers is

Notice that the moduli are divided, and the angles are subtracted.

Examples

• For $z_1 = 12 \operatorname{cis}(150^\circ)$ and $z_2 = 4 \operatorname{cis}(60^\circ)$, the quotient is $\frac{z_1}{z_2} = \frac{12}{4} \operatorname{cis}(150^\circ - 60^\circ) = 3 \operatorname{cis}(90^\circ)$.

• For $z_1 = 10 \operatorname{cis}(\frac{5\pi}{6})$ and $z_2 = 2 \operatorname{cis}(\frac{\pi}{3})$, the quotient is $\frac{z_1}{z_2} = \frac{10}{2} \operatorname{cis}(\frac{5\pi}{6} - \frac{\pi}{3}) = 5 \operatorname{cis}(\frac{\pi}{2})$.

Property

If $z = r (\operatorname{cos} \theta + i \operatorname{sin} \theta)$ is a complex number, then

where $n$ is a positive integer.

Examples

• To find $(2 \operatorname{cis}(30^\circ))^3$, calculate $2^3 \operatorname{cis}(3 \cdot 30^\circ) = 8 \operatorname{cis}(90^\circ)$.

• To find $(\sqrt{3}(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})))^4$, calculate $(\sqrt{3})^4 (\cos(4\frac{\pi}{6}) + i\sin(4\frac{\pi}{6})) = 9(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))$.

Property

To find the $n$th root of a complex number in polar form, use the formula given as

where $k = 0, 1, 2, 3, \ldots, n-1$. We add $\frac{2k\pi}{n}$ to $\frac{\theta}{n}$ in order to obtain the periodic roots.

Examples

• The two square roots of $9 \operatorname{cis}(60^\circ)$ are $3 \operatorname{cis}(\frac{60^\circ}{2} + \frac{360^\circ k}{2})$. For $k=0$, we get $3 \operatorname{cis}(30^\circ)$. For $k=1$, we get $3 \operatorname{cis}(210^\circ)$.

• The three cube roots of $27 \operatorname{cis}(180^\circ)$ are $3 \operatorname{cis}(\frac{180^\circ}{3} + \frac{360^\circ k}{3})$. The roots are $3 \operatorname{cis}(60^\circ)$, $3 \operatorname{cis}(180^\circ)$, and $3 \operatorname{cis}(300^\circ)$ for $k=0,1,2$.