Chapter 6.5, Problem 139E

a

To determine

The roots of the complex number.

Answer to Problem 139E

The fourth roots of the complex number $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ are $3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ , $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ , $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ and $3\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ .

Explanation of Solution

Given information:

The complex number is $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$

Calculation:

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.

  $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$

The fourth roots of $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ is to be evaluated $n=4$ and $r=81$ by the $n^{th}$ roots formula the roots are

  $z_k=\sqrt[4]{81}\left(\cos \frac{\frac{2\pi }{3}+2\pi k}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi k}{4}\right)$ ……. (2)

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

  $\begin{array}{c}{z}_0=\sqrt[4]{81}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(0\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(0\right)}{4}\right)\\ z_0=3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\end{array}$

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

  $\begin{array}{c}{z}_1=\sqrt[4]{81}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(1\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(1\right)}{4}\right)\\ z_1=3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\end{array}$

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

  $\begin{array}{c}{z}_2=\sqrt[4]{81}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(2\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(2\right)}{4}\right)\\ z_2=3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\end{array}$

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

  $\begin{array}{c}{z}_3=\sqrt[4]{81}\left(\cos \frac{\frac{2\pi }{3}+2\pi \left(3\right)}{4}+i\sin \frac{\frac{2\pi }{3}+2\pi \left(3\right)}{4}\right)\\ z_3=3\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)\end{array}$

Therefore, the fourth roots of $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ are $3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ , $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ , $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ and $3\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ .

b

To determine

The graph for the complex roots.

Answer to Problem 139E

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

Explanation of Solution

Given information:

The complex number is $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$

Calculation:

The fourth roots of $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ are $3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ , $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ , $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ and $3\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ .

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 139E

The standard forms of the complex roots are $2.598+1.5i$ , $-1.5+2.598i$ , $-2.598-1.5i$ and $1.5-2.598i$ .

Explanation of Solution

Given information:

The fourth roots of $81\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ are $3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)$ , $3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$ , $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ and $3\left(\cos \frac{5\pi }{3}+i\sin \frac{5\pi }{3}\right)$ .

Calculation:

Calculate the standard form of the root $z_0$ .

  $\begin{array}{c}{z}_0=3\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right)\\ =3\left(0.866+i\left(0.5\right)\right)\\ =2.598+1.5i\end{array}$

Calculate the standard form of the root $z_1$ .

  $\begin{array}{c}{z}_1=3\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\\ =3\left(-0.5+i\left(0.866\right)\right)\\ =-1.5+2.598i\end{array}$

Calculate the standard form of the root $z_2$ .

  $\begin{array}{c}{z}_2=3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\\ =3\left(-0.866+i\left(-0.5\right)\right)\\ =-2.598-1.5i\end{array}$

Calculate the standard form of the root $z_3$ .

  $\begin{array}{c}{z}_3=3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)\\ =3\left(0.5+i\left(-0.866\right)\right)\\ =1.5-2.598i\end{array}$

Therefore, the standard forms of the roots are $2.598+1.5i$ , $-1.5+2.598i$ , $-2.598-1.5i$ and $1.5-2.598i$ .