Chapter 6.6, Problem 55E

Solution

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

The sixth roots of the given complex number are $\frac{1+i}{\sqrt[6]{4}}$ , $\sqrt[6]{2}\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ , $\sqrt[6]{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ , $\sqrt[6]{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ , $\sqrt[6]{2}\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ and $\sqrt[6]{2}\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)$ .

Given information:

The complex number is $-2i$ .

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 $\tan \theta =\frac{b}{a}$ .

The $n\text{th}$ roots of a complex number $z=r\left(\cos \theta +i\sin \theta \right)$ which are $n$ distinct complex numbers are given by,

  $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

Here, $k=0,1,2,\mathrm{...},n-1$

Calculation:

From the given complex number,

  $\begin{array}{c}a=0\\ b=-2\end{array}$

Substitute $0$ for $a$ and $-2$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(0\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{4}\\ =2\end{array}$

Substitute $0$ for $a$ and $-2$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{-2}{0}\\ \theta ={\tan }^{-1}\left(\text{undefined}\right)\\ =\frac{3\pi }{2}\end{array}$

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

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

From the above complex number,

  $\begin{array}{c}r=2\\ \theta =\frac{3\pi }{2}\\ n=6\end{array}$

Use $k=0,1,2,3,4\text{and}5$ to determine the cube roots.

Substitute $2$ for $r$ , $\frac{3\pi }{2}$ for $\theta$ , $6$ for $n$ , and $0$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find first root.

  $\begin{array}{c}{z}_1=\sqrt[6]{2}\left(\cos \frac{\frac{3\pi }{2}+2\pi \cdot 0}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi \cdot 0}{6}\right)\\ =\sqrt[6]{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ =\sqrt[6]{2}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\\ =\frac{1+i}{\sqrt[6]{4}}\end{array}$

Substitute $2$ for $r$ , $\frac{3\pi }{2}$ for $\theta$ , $6$ for $n$ , and $1$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find second root.

  $\begin{array}{c}{z}_2=\sqrt[6]{2}\left(\cos \frac{\frac{3\pi }{2}+2\pi \cdot 1}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi \cdot 1}{6}\right)\\ =\sqrt[6]{2}\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{3\pi }{2}$ for $\theta$ , $6$ for $n$ , and $2$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find third root.

  $\begin{array}{c}{z}_3=\sqrt[6]{2}\left(\cos \frac{\frac{3\pi }{2}+2\pi \cdot 2}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi \cdot 2}{6}\right)\\ =\sqrt[6]{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{3\pi }{2}$ for $\theta$ , $6$ for $n$ , and $3$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find fourth root.

  $\begin{array}{c}{z}_4=\sqrt[6]{2}\left(\cos \frac{\frac{3\pi }{2}+2\pi \cdot 3}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi \cdot 3}{6}\right)\\ =\sqrt[6]{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{3\pi }{2}$ for $\theta$ , $6$ for $n$ , and $4$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find fifth root.

  $\begin{array}{c}{z}_5=\sqrt[6]{2}\left(\cos \frac{\frac{3\pi }{2}+2\pi \cdot 4}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi \cdot 4}{6}\right)\\ =\sqrt[6]{2}\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

Substitute $2$ for $r$ , $\frac{3\pi }{2}$ for $\theta$ , $6$ for $n$ , and $5$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find sixth root.

  $\begin{array}{c}{z}_6=\sqrt[6]{2}\left(\cos \frac{\frac{3\pi }{2}+2\pi \cdot 5}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi \cdot 5}{6}\right)\\ =\sqrt[6]{2}\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)\end{array}$

Therefore, the sixth roots of the given complex number are $\frac{1+i}{\sqrt[6]{4}}$ , $\sqrt[6]{2}\left(\cos \frac{7\pi }{12}+i\sin \frac{7\pi }{12}\right)$ , $\sqrt[6]{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ , $\sqrt[6]{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ , $\sqrt[6]{2}\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ and $\sqrt[6]{2}\left(\cos \frac{23\pi }{12}+i\sin \frac{23\pi }{12}\right)$ .