Chapter 6, Problem 115RE

(a)

To determine

To find: The sixth root of the given complex number.

Answer to Problem 115RE

The sixth roots are $3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ , $3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ , $3\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ , $3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ and $3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)$ .

Explanation of Solution

Given:

The complex number is $-729i$ .

Calculation:

For any complex number $z=r\left(\cos \theta +i\sin \theta \right)$ , the nth root is given by,

  $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right),k=0,1,2,\dots ,n-1$

So, to find the sixth root of $-729i$ , let $z=-729i$ and convert it into the form $r\left(\cos \theta +i\sin \theta \right)$ .

  $\begin{array}{c}r=\sqrt{a^2+b^2}\\ =\sqrt{{\left(0\right)}^2+{\left(-729\right)}^2}\\ =\sqrt{{\left(729\right)}^2}\\ =729\end{array}$

  $\begin{array}{c}\tan \theta =\frac{b}{a}\\ =-\frac{729}{0}\\ =\frac{3\pi }{2}\end{array}$

So, $-729i=729\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)$ and the sixth root is calculated as shown below:

  $\sqrt[6]{729}\left(\cos \frac{\frac{3\pi }{2}+2\pi k}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi k}{6}\right)=3\left(\cos \frac{3\pi +4\pi k}{12}+i\sin \frac{3\pi +4\pi k}{12}\right)$

For $k=0$ ,

  $3\left(\cos \frac{3\pi +4\pi \left(0\right)}{12}+i\sin \frac{3\pi +4\pi \left(0\right)}{12}\right)=3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

For $k=1$ ,

  $3\left(\cos \frac{3\pi +4\pi \left(1\right)}{12}+i\sin \frac{3\pi +4\pi \left(1\right)}{12}\right)=3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$

For $k=2$ ,

  $3\left(\cos \frac{3\pi +4\pi \left(2\right)}{12}+i\sin \frac{3\pi +4\pi \left(2\right)}{12}\right)=3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$

For $k=3$ ,

  $3\left(\cos \frac{3\pi +4\pi \left(3\right)}{12}+i\sin \frac{3\pi +4\pi \left(3\right)}{12}\right)=3\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$

For $k=4$ ,

  $3\left(\cos \frac{3\pi +4\pi \left(4\right)}{12}+i\sin \frac{3\pi +4\pi \left(4\right)}{12}\right)=3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$

For $k=5$ ,

  $3\left(\cos \frac{3\pi +4\pi \left(5\right)}{12}+i\sin \frac{3\pi +4\pi \left(5\right)}{12}\right)=3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)$

So, the roots are $3\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $3\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ , $3\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ , $3\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ , $3\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ and $3\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)$ .