Chapter 8.3, Problem 39E

To determine

To calculate: To write the complex number in polar form with argument $\theta$ between $0$ and $2\pi$

Answer to Problem 39E

Complex number in polar form is $z=8\left(\cos \frac{11\pi }{6}+isin\frac{11\pi }{6}\right)$

Explanation of Solution

Given information: Complex numbers is $z=4\sqrt{3}-4i$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

Polar form of the complex number is given as

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

Where,

  $r=\left|z\right|=\sqrt{a^2+b^2}$

  $\tan \theta =\frac{b}{a}$

Calculation:

Complex number is given as

  $z=4\sqrt{3}-4i$

Polar form of the complex number is given as

  $z=r\left(\cos \theta +isin\theta \right)----(1)$

Calculating the value of $r$ as follows:

  $r=\left|z\right|=\sqrt{a^2+b^2}$

Substituting the values,

  $\begin{array}{l}r=\sqrt{{\left(4\sqrt{3}\right)}^2+{\left(-4\right)}^2}\\ r=\sqrt{48+16}\\ r=\sqrt{64}\\ r=8\end{array}$

Calculating the value of $\theta$ as follows:

  $\tan \theta =\frac{b}{a}$

Substituting the values,

  $\begin{array}{l}\tan \theta =\frac{-4}{4\sqrt{3}}\\ \tan \theta =\frac{-1}{\sqrt{3}}\\ \theta ={\tan }^{-1}\frac{-1}{\sqrt{3}}\end{array}$

Here, solution lies in fourth quadrant

  $\begin{array}{l}\theta =2\pi -\frac{\pi }{6}\\ \theta =\frac{11\pi }{6}\end{array}$

Now, substituting the values of $r=8$ and $\theta =\frac{11\pi }{6}$ in equation (1) to get the complex number in polar form

  $z=8\left(\cos \frac{11\pi }{6}+isin\frac{11\pi }{6}\right)$

Conclusion:

Hence, complex number in polar form is $z=8\left(\cos \frac{11\pi }{6}+isin\frac{11\pi }{6}\right)$