Exercise 1.4 — Question 5

Polar Representation of Complex Numbers

This question involves converting complex numbers to polar form and applying operations in polar representation.

Polar Coordinate System

In the polar coordinate system, a point $P$ in the plane is located by:

• $r$ — the radial distance from the origin (pole) to the point $P$, where $r\ge 0$
• $\theta$ — the polar angle (argument) measured counterclockwise from the positive $x$-axis

A point is written as $P(r,\theta )$.

Polar Form of a Complex Number

For a complex number $z=a+ib$, the polar (trigonometric) form is:

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

where:

• $r=|z|=\sqrt{a^2+b^2}$ is the modulus
• $\theta =\arg (z)={\tan }^{-1}\left(\frac{b}{a}\right)$ is the argument
• $a=r\cos \theta$, $b=r\sin \theta$

Operations in Polar Form

Multiplication: $z_1{z}_2=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]$

Division: $\frac{z_1}{z_2}=\frac{r_1}{r_2}[\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)],z_2\ne 0$

De Moivre's Theorem: $z^n=r^n(\cos n\theta +i\sin n\theta )$

Worked Example

Express $z_1=1+i$ and $z_2=\sqrt{3}-i$ in polar form, then find $z_1{z}_2$ and $\frac{z_1}{z_2}$.

Step 1 — Polar form of $z_1=1+i$: $r_1=\sqrt{1^2+1^2}=\sqrt{2},{\theta }_1={\tan }^{-1}\left(\frac{1}{1}\right)=45°$ $z_1=\sqrt{2}(\cos 45°+i\sin 45°)$

Step 2 — Polar form of $z_2=\sqrt{3}-i$: $r_2=\sqrt{(\sqrt{3}{)}^2+(-1{)}^2}=\sqrt{4}=2,{\theta }_2=-30°\text{ (4th quadrant)}$ $z_2=2(\cos (-30°)+i\sin (-30°))$

Step 3 — Product $z_1{z}_2$: $z_1{z}_2=\sqrt{2}\cdot 2[\cos (45°+(-30°))+i\sin (45°+(-30°))]$ $=2\sqrt{2}(\cos 15°+i\sin 15°)$

Step 4 — Quotient $\frac{z_1}{z_2}$: $\frac{z_1}{z_2}=\frac{\sqrt{2}}{2}[\cos (45°-(-30°))+i\sin (45°-(-30°))]$ $=\frac{\sqrt{2}}{2}(\cos 75°+i\sin 75°)$