Chapter 6.6, Problem 62E

Solution

To find: The $z$ and the three cube roots of $z$ .

The value of $z$ is $-64$ and cube roots are $2\pm 2i$ , and $-2\pm 2i$ .

Given information:

The one cube root 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 $\tan \theta =\frac{b}{a}$ .

By the De Moivre’s Theorem,

If $z=r\left(\cos \theta +i\sin \theta \right)$ and $n$ is a positive integer then,

  $\begin{array}{c}{z}^n={\left[r\left(\cos \theta +i\sin \theta \right)\right]}^n\\ =r^n\left(\cos n\theta +i\sin n\theta \right)\end{array}$

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:

The cube root is $-2-2i$ . Then,

  $z={\left(-2-2i\right)}^4$

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 $r$ .

  $\begin{array}{c}r=\sqrt{{\left(-2\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{4+4}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

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

  $\begin{array}{c}\tan \theta =\frac{-2}{-2}\\ \theta ={\tan }^{-1}\left(1\right)\\ =\frac{5\pi }{4}\end{array}$

Substitute $2\sqrt{2}$ for $r$ and $\frac{5\pi }{4}$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=2\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$

From the above complex number,

  $\begin{array}{c}r=2\sqrt{2}\\ \theta =\frac{5\pi }{4}\\ n=4\end{array}$

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

Substitute $2\sqrt{2}$ for $r$ , $\frac{5\pi }{4}$ for $\theta$ , and $4$ for $n$ in expression $z=r^n\left(\cos n\theta +i\sin n\theta \right)$ to find first root.

  $\begin{array}{c}z={\left(2\sqrt{2}\right)}^5\left(\cos \left(4\cdot \frac{5\pi }{4}\right)+i\sin \left(4\cdot \frac{5\pi }{4}\right)\right)\\ =64\left(\cos 5\pi +i\sin 5\pi \right)\\ =8\left(-1+0\right)\\ =-64\end{array}$

Substitute $2\sqrt{2}$ for $r$ , $\pi$ for $\theta$ , and $4$ 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 second root.

  $\begin{array}{c}{z}_1=\sqrt[4]{64}\left(\cos \frac{\pi +2\pi \cdot 0}{4}+i\sin \frac{\pi +2\pi \cdot 0}{4}\right)\\ =2\sqrt{2}\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ =2\sqrt{2}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\\ =2+2i\end{array}$

Substitute $2\sqrt{2}$ for $r$ , $\pi$ for $\theta$ , and $4$ 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.

  $\begin{array}{c}{z}_2=\sqrt[4]{64}\left(\cos \frac{\pi +2\pi \cdot 1}{4}+i\sin \frac{\pi +2\pi \cdot 1}{4}\right)\\ =2\sqrt{2}\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ =2\sqrt{2}\left(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\right)\\ =-2+2i\end{array}$

Substitute $2\sqrt{2}$ for $r$ , $\pi$ for $\theta$ , and $4$ 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 fourth root.

  $\begin{array}{c}{z}_3=\sqrt[3]{64}\left(\cos \frac{\pi +2\pi \cdot 2}{4}+i\sin \frac{\pi +2\pi \cdot 2}{4}\right)\\ =2\sqrt{2}\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ =2\sqrt{2}\left(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\\ =-2-2i\end{array}$

Substitute $2\sqrt{2}$ for $r$ , $\pi$ for $\theta$ , and $4$ 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.

  $\begin{array}{c}{z}_4=\sqrt[3]{64}\left(\cos \frac{\pi +2\pi \cdot 3}{4}+i\sin \frac{\pi +2\pi \cdot 3}{4}\right)\\ =2\sqrt{2}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\\ =2\sqrt{2}\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\right)\\ =2-2i\end{array}$

Therefore, the value of $z$ is $-64$ and cube roots are $2\pm 2i$ , and $-2\pm 2i$ .