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Writing Complex Numbers in Polar Form Simplified Revision Notes

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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+biz=a+bi can be written as

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

where

  • r=z=a2+b2r=|z|=\sqrt{a^2+b^2}
  • θ=tan1(ba)\theta=\tan^{-1}\left( \frac{|b|}{|a|}\right)

Example

infoNote

Express z=2+3iz=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 θ\theta is the angle that the complex number makes with the real axis.

First, let's derive the modulus :

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

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

tanθ=32θ=tan1320.983\begin{align*} \tan{\theta}&=\frac{3}{2} \\\\ \theta &= \tan^{-1}\frac{3}{2} \\\\ & \approx 0.983 \end{align*}

Write in polar form :

z=13(cos0.983+isin0.983)z=\sqrt{13} \left(\cos{0.983+i\sin{0.983}} \right)

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


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

For radians :


  • For quadrant 1 : θ=A\theta=A
  • For quadrant 2 : θ=πA\theta= \pi-A
  • For quadrant 3 : θ=π+A\theta=\pi +A
  • For quadrant 4 : θ=2πA\theta=2\pi -A

Example

infoNote

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

Derive the modulus :

(3)2+(3)2=23\sqrt{(-3)^2+(\sqrt{3})^2}=2\sqrt{3}

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

tanθ=33θ=tan133=π6\begin{align*} \tan{\theta}&=\frac{\sqrt{3}}{3} \\\\ \theta &= \tan^{-1}\frac{\sqrt{3}} {3} \\\\ &=\frac{\pi}{6} \end{align*}

Since zz lies in the second quadrant, θ\theta is derived as :

ππ6=5π6\pi-\frac{\pi}{6}=\frac{5\pi}{6}z=23(cos5π6+isin5π6)z=2\sqrt{3} \left(\cos{\frac{5\pi}{6}+i\sin{\frac{5\pi}{6}}} \right)
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