Chapter 6.6, Problem 70E

Solution

To find: The correct option.

The correct answer is option (E).

Given information:

The question asks to identify the option that is not a fourth root of $1$ .

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:

Write the unity function in $a+bi$ form.

  $1+0i$

From the given complex number,

  $\begin{array}{c}a=1\\ b=0\end{array}$

Substitute $1$ for $a$ and $0$ for $b$ in equation $r=\sqrt{a^2+b^2}$ to find modulus $r$ .

  $\begin{array}{c}r=\sqrt{{\left(1\right)}^2+{\left(0\right)}^2}\\ =1\end{array}$

Substitute $1$ for $a$ and $0$ for $b$ in equation $\tan \theta =\frac{b}{a}$ to find argument.

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

Substitute $1$ for $r$ and $0$ for $\theta$ in equation $z=r\left(\cos \theta +i\sin \theta \right)$ .

  $z=\left(\cos 0+i\sin 0\right)$

From the above complex number,

  $\begin{array}{c}r=1\\ \theta =0\\ n=4\end{array}$

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

Substitute $1$ for $r$ , $0$ for $\theta$ , $4$ 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 first root.

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

Substitute $1$ for $r$ , $0$ for $\theta$ , $4$ 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.

  $\begin{array}{c}{z}_2=\sqrt[4]{1}\left(\cos \frac{0+2\pi \cdot 1}{4}+i\sin \frac{0+2\pi \cdot 1}{4}\right)\\ =1\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)\\ =0+i\\ =i\end{array}$

Substitute $1$ for $r$ , $0$ for $\theta$ , $4$ 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 third root.

  $\begin{array}{c}{z}_3=\sqrt[4]{1}\left(\cos \frac{0+2\pi \cdot 2}{4}+i\sin \frac{0+2\pi \cdot 2}{4}\right)\\ =1\left(\cos \pi +i\sin \pi \right)\\ =-1+0\\ =-1\end{array}$

Substitute $1$ for $r$ , $0$ for $\theta$ , $4$ 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.

  $\begin{array}{c}{z}_4=\sqrt[4]{1}\left(\cos \frac{0+2\pi \cdot 3}{4}+i\sin \frac{0+2\pi \cdot 3}{4}\right)\\ =1\left(\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\right)\\ =0-i\\ =-i\end{array}$

Therefore, the correct answer is option (E).