Writing Complex Numbers in Polar Form Revision Notes for Leaving Cert Mathematics

Writing Complex Numbers in Polar Form

Introduction

The polar form of a complex number represents it in terms of its magnitude (modulus) and angle (argument) with respect to the origin in the complex plane. This form provides a geometric interpretation of complex numbers and is especially useful for multiplication, division, and exponentiation.

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

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

where

$r=|z|=\sqrt{a^2+b^2}$
$\theta=\tan^{-1}\left( \frac{|b|}{|a|}\right)$

Example

infoNote

Express $z=2+3i$ in Polar form

First, it's always best to make a rough sketch of the complex number on the Argand diagram.

The modulus is the length of the red line and $\theta$ is the angle that the complex number makes with the real axis.

First, let's derive the modulus :

\sqrt{(2)^2+(3)^2}=\sqrt{13}

Next, find the value of $\theta$ , which can be found using basic trigonometry. Notice that the plotted complex number outlines a right-angled triangle.

\begin{align*} \tan{\theta}&=\frac{3}{2} \\\\ \theta &= \tan^{-1}\frac{3}{2} \\\\ & \approx 0.983 \end{align*}

Write in polar form :

z=\sqrt{13} \left(\cos{0.983+i\sin{0.983}} \right)

The value of $\theta$ will depend on the quadrant that the complex number is in :

For quadrant 1 : $\theta=A$
For quadrant 2 : $\theta= 180\degree-A$
For quadrant 3 : $\theta=180\degree +A$
For quadrant 4 : $\theta=360\degree -A$

For radians :

For quadrant 1 : $\theta=A$
For quadrant 2 : $\theta= \pi-A$
For quadrant 3 : $\theta=\pi +A$
For quadrant 4 : $\theta=2\pi -A$

Example

infoNote

Express $z=-3+\sqrt{3}i$ in Polar form

Derive the modulus :

\sqrt{(-3)^2+(\sqrt{3})^2}=2\sqrt{3}

Next, find the value of $\theta$ , which can be found using basic trigonometry. Notice that the plotted complex number outlines a right-angled triangle.

\begin{align*} \tan{\theta}&=\frac{\sqrt{3}}{3} \\\\ \theta &= \tan^{-1}\frac{\sqrt{3}} {3} \\\\ &=\frac{\pi}{6} \end{align*}

Since $z$ lies in the second quadrant, $\theta$ is derived as :

\pi-\frac{\pi}{6}=\frac{5\pi}{6}

z=2\sqrt{3} \left(\cos{\frac{5\pi}{6}+i\sin{\frac{5\pi}{6}}} \right)

Writing Complex Numbers in Polar Form (Leaving Cert Mathematics): Revision Notes