Chapter 6.6, Problem 61E

Solution

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

The value of $z$ is $-8$ and cube roots are $1+\sqrt{3}i$ , $-2$ , and $1-\sqrt{3}i$ .

Given information:

The one cube root is $1+\sqrt{3}i$ .

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 $1+\sqrt{3}i$ . Then,

  $z={\left(1+\sqrt{3}i\right)}^3$

From the given complex number,

  $\begin{array}{c}a=1\\ b=\sqrt{3}\end{array}$

Substitute $1$ for $a$ and $\sqrt{3}$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(1\right)}^2+{\left(\sqrt{3}\right)}^2}\\ =\sqrt{1+3}\\ =\sqrt{4}\\ =2\end{array}$

Substitute $1$ for $a$ and $\sqrt{3}$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

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

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

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

From the above complex number,

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

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

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

  $\begin{array}{c}z=2^3\left(\cos \left(3\cdot \frac{\pi }{3}\right)+i\sin \left(3\cdot \frac{\pi }{3}\right)\right)\\ =8\left(\cos \pi +i\sin \pi \right)\\ =8\left(-1+0\right)\\ =-8\end{array}$

Substitute $8$ for $r$ , $\pi$ 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 second root.

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

Substitute $8$ for $r$ , $\pi$ 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.

  $\begin{array}{c}{z}_2=\sqrt[3]{8}\left(\cos \frac{\pi +2\pi \cdot 1}{3}+i\sin \frac{\pi +2\pi \cdot 1}{3}\right)\\ =2\left(\cos \pi +i\sin \pi \right)\\ =2\left(-1+0\right)\\ =-2\end{array}$

Substitute $8$ for $r$ , $\pi$ 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 fourth root.

  $\begin{array}{c}{z}_3=\sqrt[3]{8}\left(\cos \frac{\pi +2\pi \cdot 2}{3}+i\sin \frac{\pi +2\pi \cdot 2}{3}\right)\\ =2\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\\ =2\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\\ =1-\sqrt{3}i\end{array}$

Therefore, the value of $z$ is $-8$ and cube roots are $1+\sqrt{3}i$ , $-2$ , and $1-\sqrt{3}i$ .