Chapter 6.5, Problem 146E

a

To determine

The fourth roots of the complex number.

Answer to Problem 146E

The fourth roots of $i$ are $4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ , $4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$ and $4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$ .

Explanation of Solution

Given information:

The complex number is $i$

Calculation:

Calculate the value of $r$ .

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

Calculate the value of $\theta$ .

  $\begin{array}{c}\theta =\arctan \left(\infty \right)\\ =\frac{\pi }{2}\end{array}$

The trigonometric form of $z$ .

  $z=1\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$

The $n^{th}$ roots of a complex number $z=r\left(\cos \theta +i\sin \theta \right)$ are given by

  $z_k=\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ ……. (1)

Here, $k=0,1,\mathrm{2......}n-1$ .

Consider the complex number.

  $i$

The fourth roots of $i$ is to be evaluated $n=4$ and $r=1$ by the $n^{th}$ roots formula the roots are

  $z_k=\sqrt[4]{1}\left(\cos \frac{\frac{\pi }{2}+2\pi k}{4}+i\sin \frac{\frac{\pi }{2}+2\pi k}{4}\right)$ ……. (2)

Substitute $0$ for $k$ in equation (2).

  $\begin{array}{c}{z}_0=\sqrt[4]{1}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(0\right)}{4}+i\sin \frac{\frac{\pi }{2}+2\pi \left(0\right)}{4}\right)\\ z_0=\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\end{array}$

Substitute $1$ for $k$ in equation (2).

  $\begin{array}{c}{z}_1=\sqrt[4]{1}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(1\right)}{4}+i\sin \frac{\frac{\pi }{2}+2\pi \left(1\right)}{4}\right)\\ z_1=\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\end{array}$

Substitute $2$ for $k$ in equation (2).

  $\begin{array}{c}{z}_2=\sqrt[4]{1}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(2\right)}{4}+i\sin \frac{\frac{\pi }{2}+2\pi \left(2\right)}{4}\right)\\ z_2=\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\end{array}$

Substitute $3$ for $k$ in equation (2).

  $\begin{array}{c}{z}_3=\sqrt[4]{1}\left(\cos \frac{\frac{\pi }{2}+2\pi \left(3\right)}{4}+i\sin \frac{\frac{\pi }{2}+2\pi \left(3\right)}{4}\right)\\ z_3=\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)\end{array}$

Therefore, the fourth roots of $i$ are $4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ , $4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$ and $4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$ .

b

To determine

The graph for the complex roots.

Answer to Problem 146E

The graph for the complex roots is shown in Figure (1).

Explanation of Solution

Given information:

The complex number is $256i$

Calculation:

The fourth roots of $i$ are $4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ , $4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$ and $4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$ .

Draw the graph for the roots.

  

Figure-(1)

Therefore, the graph for the complex roots is shown in Figure (1).

c

To determine

The standard form of the complex roots.

Answer to Problem 146E

The standard forms of the roots are $0.9239+0.3827i$ , $-0.3827+0.9239i$ , $-0.9239-0.3827i$ and $0.3827-0.9237i$ .

Explanation of Solution

Given information:

The fourth roots of $i$ are $4\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ , $4\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $4\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$ and $4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)$ .

Calculation:

Calculate the standard form of the root $z_0$ .

  $\begin{array}{c}{z}_0=\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\\ =0.9239+0.3827i\end{array}$

Calculate the standard form of the root $z_1$ .

  $\begin{array}{c}{z}_1=\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\\ =-0.3827+0.9239i\end{array}$

Calculate the standard form of the root $z_2$ .

  $\begin{array}{c}{z}_2=\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\\ =-0.9239-0.3827i\end{array}$

Calculate the standard form of the root $z_3$ .

  $\begin{array}{c}{z}_3=4\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)\\ =0.3827-0.9237i\end{array}$

Therefore, the standard forms of the roots are $0.9239+0.3827i$ , $-0.3827+0.9239i$ , $-0.9239-0.3827i$ and $0.3827-0.9237i$ .