Chapter 6.6, Problem 44E

Solution

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

The cube roots of the given complex number are $1+i$ , $\sqrt{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ and $\sqrt{2}\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ .

Given information:

The complex number is $-2+2i$ .

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,

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

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

  $\begin{array}{c}r=\sqrt{{\left(-2\right)}^2+{\left(2\right)}^2}\\ =\sqrt{8}\end{array}$

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

  $\begin{array}{c}\tan \theta =\frac{-2}{2}\\ \theta ={\tan }^{-1}\left(-1\right)\\ =\frac{3\pi }{4}\end{array}$

Substitute $\sqrt{8}$ for $r$ and $\frac{3\pi }{4}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=\sqrt{8}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

From the above complex number,

  $\begin{array}{c}r=\sqrt{8}\\ \theta =\frac{3\pi }{4}\\ n=3\end{array}$

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

Substitute $\sqrt{8}$ for $r$ , $\frac{3\pi }{4}$ for $\theta$ , $3$ 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.

  $\begin{array}{c}{z}_1=\sqrt[3]{\sqrt{8}}\left(\cos \frac{\frac{3\pi }{4}+2\pi \cdot 0}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi \cdot 0}{3}\right)\\ =\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ =\sqrt{2}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\\ =1+i\end{array}$

Substitute $\sqrt{8}$ for $r$ , $\frac{3\pi }{4}$ for $\theta$ , $3$ 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.

  $\begin{array}{c}{z}_2=\sqrt[3]{\sqrt{8}}\left(\cos \frac{\frac{3\pi }{4}+2\pi \cdot 1}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi \cdot 1}{3}\right)\\ =\sqrt{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\end{array}$

Substitute $\sqrt{8}$ for $r$ , $\frac{3\pi }{4}$ for $\theta$ , $3$ 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.

  $\begin{array}{c}{z}_3=\sqrt[3]{\sqrt{8}}\left(\cos \frac{\frac{3\pi }{4}+2\pi \cdot 2}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi \cdot 2}{3}\right)\\ =\sqrt{2}\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

Therefore, the cube roots of the given complex number are $1+i$ , $\sqrt{2}\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ and $\sqrt{2}\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ .