Chapter 6.6, Problem 7E

Solution

To find: The polar form of the complex number.

The polar form of the complex number is $z=4\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ .

Given data:

The complex number is $-2+2i\sqrt{3}$ .

Formula used:

The polar form of a complex number $z=a+bi$ is given by,

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

Here, $r=\sqrt{a^2+b^2}$ and $\tan \theta =\frac{b}{a}$ .

Calculation:

From the given complex number,

  $\begin{array}{c}a=-2\\ b=2\sqrt{3}\end{array}$

Substitute $-2$ for $a$ and $2\sqrt{3}$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(2\right)}^2+{\left(2\sqrt{3}\right)}^2}\\ =\sqrt{4+12}\\ =\sqrt{16}\\ =4\end{array}$

Substitute $-2$ for $a$ and $2\sqrt{3}$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{2\sqrt{3}}{-2}\\ \theta ={\tan }^{-1}\left(-\sqrt{3}\right)\end{array}$

The reference angle is,

  $\begin{array}{c}{\theta }^{\prime }={\tan }^{-1}\left(\sqrt{3}\right)\\ =\frac{\pi }{3}\end{array}$

Since the complex number is in QII, then,

  $\begin{array}{c}\theta =\pi -{\theta }^{\prime }\\ =\pi -\frac{\pi }{3}\\ =\frac{2\pi }{3}\end{array}$

Substitute $4$ for $r$ and $\frac{2\pi }{3}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=4\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$

Therefore, the polar form of the complex number is $z=4\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ .