Chapter 6.6, Problem 43E

Solution

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

The cube roots of the given complex number are $\sqrt[3]{5}\left(\cos 0.3091-i\sin 0.3091\right)$ , $\sqrt[3]{5}\left(\cos 1.79+i\sin 1.79\right)$ and $\sqrt[3]{5}\left(\cos 3.88+i\sin 3.88\right)$ .

Given information:

The complex number is $3-4i$ .

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,

  $\begin{array}{c}a=3\\ b=-4\end{array}$

Substitute $3$ for $a$ and $-4$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(3\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{25}\\ =5\end{array}$

Substitute $3$ for $a$ and $-4$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

  $\begin{array}{c}\tan \theta =\frac{-4}{3}\\ \theta ={\tan }^{-1}\left(\frac{-4}{3}\right)\\ =-0.9273\end{array}$

Substitute $5$ for $r$ and $-0.9273$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=5\left(\cos \left(-0.9273\right)+i\sin \left(-0.9273\right)\right)$

From the above complex number,

  $\begin{array}{c}r=5\\ \theta =-0.9273\\ n=3\end{array}$

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

Substitute $5$ for $r$ , $-0.9273$ 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.

  $\begin{array}{c}{z}_1=\sqrt[3]{5}\left(\cos \frac{-0.9273+2\pi \cdot 0}{3}+i\sin \frac{-0.9273+2\pi \cdot 0}{3}\right)\\ =\sqrt[3]{5}\left(\cos 0.3091-i\sin 0.3091\right)\end{array}$

Substitute $5$ for $r$ , $-0.9273$ 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.

  $\begin{array}{c}{z}_2=\sqrt[3]{5}\left(\cos \frac{-0.9273+2\pi \cdot 1}{3}+i\sin \frac{-0.9273+2\pi \cdot 1}{3}\right)\\ =\sqrt[3]{5}\left(\cos 1.79+i\sin 1.79\right)\end{array}$

Substitute $5$ for $r$ , $-0.9273$ 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.

  $\begin{array}{c}{z}_3=\sqrt[3]{5}\left(\cos \frac{-0.9273+2\pi \cdot 2}{3}+i\sin \frac{-0.9273+2\pi \cdot 2}{3}\right)\\ =\sqrt[3]{5}\left(\cos 3.88+i\sin 3.88\right)\end{array}$

Therefore, the cube roots of the given complex number are $\sqrt[3]{5}\left(\cos 0.3091-i\sin 0.3091\right)$ , $\sqrt[3]{5}\left(\cos 1.79+i\sin 1.79\right)$ and $\sqrt[3]{5}\left(\cos 3.88+i\sin 3.88\right)$ .