Question

Convert the following complex number into its polar representation:2-2√3i

Answer 1

Given:

`=2-2√(3)i`

Find-:

Convert complex numbers to a polar representation

Explanation-:

Polar from of the complex number

`z=a+ib=r(\cos\theta+i\sin\theta)`

Where,

`\begin{gathered} r=√(a^2+b^2) \\ \\ \theta=\tan^(-1)((b)/(a)) \end{gathered}`

Given complex form is:

`\begin{gathered} z=a+ib \\ \\ z=2-i2√(3) \\ \\ a=2 \\ \\ b=-2√(3) \end{gathered}`
`\begin{gathered} r=|z|=√(a^2+b^2) \\ \\ r=|z|=\sqrt{2^2+(2√(3))^2} \\ \\ =√(4+12) \\ \\ =√(16) \\ \\ =4 \end{gathered}`

For the angle value is:

`\begin{gathered} \theta=\tan^(-1)((b)/(a)) \\ \\ \theta=\tan^(-1)((-2√(3))/(2)) \\ \\ =\tan^(-1)(-√(3)) \\ \\ =-60 \\ \\ =-(\pi)/(3) \end{gathered}`

So, the polar form is:

`\begin{gathered} z=r(\cos\theta+i\sin\theta) \\ \\ z=4(\cos(-(\pi)/(3))+i\sin(-(\pi)/(3))) \end{gathered}`

Use the formula:

`\begin{gathered} \sin(-\theta)=-\sin\theta \\ \\ \cos(-\theta)=+\cos\theta \end{gathered}`

Then value is:

`\begin{gathered} z=4(\cos(-(\pi)/(3))+i\sin(-(\pi)/(3))) \\ \\ z=4(\cos(\pi)/(3)-i\sin(\pi)/(3)) \end{gathered}`