Chapter 14, Problem 6CT

To determine

To find:The five fifth roots in polar form.

Answer to Problem 6CT

The five fifth roots in polar form is calculated.

Explanation of Solution

Given:

  $-i$

Calculation:

Calculate.

  $\begin{array}{l}z=-i\\ r=|-i|\\ r=1\\ \theta =270\end{array}$

Now,

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right)\\ z=1\left(\cos 270+i\sin 270\right)\end{array}$

The roots have the form of,

  $\sqrt[5]{1}\left(\cos \frac{270+2\pi k}{5}+i\sin \frac{270+2\pi k}{5}\right)$

For k=0,

  $\begin{array}{c}\sqrt[5]{1}\left(\cos \frac{270+2\pi k}{5}+i\sin \frac{270+2\pi k}{5}\right)=\sqrt[5]{1}\left(\cos \frac{270+2\pi 0}{5}+i\sin \frac{270+2\pi 0}{5}\right)\\ =\cos \frac{3\pi }{10}+i\sin \frac{3\pi }{10}\\ =0.5878+i0.8090\end{array}$

For $k=1$ ,

  $\begin{array}{c}\sqrt[5]{1}\left(\cos \frac{270+2\pi k}{5}+i\sin \frac{270+2\pi k}{5}\right)=\sqrt[5]{1}\left(\cos \frac{270+2\pi 1}{5}+i\sin \frac{270+2\pi 1}{5}\right)\\ =\cos \frac{7\pi }{10}+i\sin \frac{7\pi }{10}\\ =-0.5878+i0.8090\end{array}$

For $k=2$ ,

  $\begin{array}{c}\sqrt[5]{1}\left(\cos \frac{270+2\pi k}{5}+i\sin \frac{270+2\pi k}{5}\right)=\sqrt[5]{1}\left(\cos \frac{270+2\pi 2}{5}+i\sin \frac{270+2\pi 2}{5}\right)\\ =\cos \frac{11\pi }{10}+i\sin \frac{11\pi }{10}\\ =-0.9511+i0.3090\end{array}$

For $k=3$ ,

  $\begin{array}{c}\sqrt[5]{1}\left(\cos \frac{270+2\pi k}{5}+i\sin \frac{270+2\pi k}{5}\right)=\sqrt[5]{1}\left(\cos \frac{270+2\pi 3}{5}+i\sin \frac{270+2\pi 3}{5}\right)\\ =\cos \frac{3\pi }{10}+i\sin \frac{3\pi }{10}\\ =-i\end{array}$

For $k=4$ ,

  $\begin{array}{c}\sqrt[5]{1}\left(\cos \frac{270+2\pi k}{5}+i\sin \frac{270+2\pi k}{5}\right)=\sqrt[5]{1}\left(\cos \frac{270+2\pi 4}{5}+i\sin \frac{270+2\pi 4}{5}\right)\\ =\cos \frac{19\pi }{10}+i\sin \frac{19\pi }{10}\\ =0.9511-i0.3090\end{array}$

Thus, the five fifth roots in polar form is calculated.