Chapter 6.5, Problem 152E

a

To determine

The fifth roots of the complex number.

Answer to Problem 152E

The fifth roots of $4\left(1-i\right)$ are $2\left(\cos \frac{7\pi }{20}+i\sin \frac{7\pi }{20}\right)$ , $2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$ , $2\left(\cos \frac{23\pi }{20}+i\sin \frac{23\pi }{20}\right)$ , $2\left(\cos \frac{31\pi }{20}+i\sin \frac{31\pi }{20}\right)$ and $2\left(\cos \frac{39\pi }{20}+i\sin \frac{39\pi }{20}\right)$ .

Explanation of Solution

Given information:

The complex number is $4\left(1-i\right)$

Calculation:

Calculate the value of $r$ .

  $\begin{array}{c}r=\left|4-4i\right|\\ =\sqrt{{\left(4\right)}^2+{\left(-4\right)}^2}\\ =\sqrt{32}\end{array}$

Calculate the value of $\theta$ .

  $\begin{array}{c}\tan \theta =\frac{b}{a}\\ \tan \theta =\frac{-4}{4}\\ \theta ={\tan }^{-1}\left(-1\right)\\ \theta =\frac{7\pi }{4}\end{array}$

The trigonometric form of $z$ .

  $z=\sqrt{32}\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

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.

  $4\left(1-i\right)$

The fourth roots of $4\left(1-i\right)$ is to be evaluated $n=5$ and $r=\sqrt{32}$ by the $n^{th}$ roots formula the roots are

  $z_k=\sqrt[5]{32}\left(\cos \frac{\frac{7\pi }{4}+2\pi k}{5}+i\sin \frac{\frac{7\pi }{4}+2\pi k}{5}\right)$ ……. (2)

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

  $\begin{array}{c}{z}_0=\sqrt[5]{32}\left(\cos \frac{\frac{7\pi }{4}+2\pi \left(0\right)}{5}+i\sin \frac{\frac{7\pi }{4}+2\pi \left(0\right)}{5}\right)\\ z_0=2\left(\cos \frac{7\pi }{20}+i\sin \frac{7\pi }{20}\right)\end{array}$

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

  $\begin{array}{c}{z}_1=\sqrt[5]{32}\left(\cos \frac{\frac{7\pi }{4}+2\pi \left(1\right)}{5}+i\sin \frac{\frac{7\pi }{4}+2\pi \left(1\right)}{5}\right)\\ z_1=2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\end{array}$

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

  $\begin{array}{c}{z}_2=\sqrt[5]{32}\left(\cos \frac{\frac{7\pi }{4}+2\pi \left(3\right)}{5}+i\sin \frac{\frac{7\pi }{4}+2\pi \left(3\right)}{5}\right)\\ z_2=2\left(\cos \frac{23\pi }{20}+i\sin \frac{23\pi }{20}\right)\end{array}$

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

  $\begin{array}{c}{z}_3=\sqrt[5]{32}\left(\cos \frac{\frac{7\pi }{4}+2\pi \left(3\right)}{5}+i\sin \frac{\frac{7\pi }{4}+2\pi \left(3\right)}{5}\right)\\ z_3=2\left(\cos \frac{31\pi }{20}+i\sin \frac{31\pi }{20}\right)\end{array}$

Substitute $4$ for $k$ in equation (2).

  $\begin{array}{c}{z}_4=\sqrt[5]{32}\left(\cos \frac{\frac{7\pi }{4}+2\pi \left(4\right)}{5}+i\sin \frac{\frac{7\pi }{4}+2\pi \left(4\right)}{5}\right)\\ z_4=2\left(\cos \frac{39\pi }{20}+i\sin \frac{39\pi }{20}\right)\end{array}$

Therefore, the fifth roots of $4\left(1-i\right)$ are $2\left(\cos \frac{7\pi }{20}+i\sin \frac{7\pi }{20}\right)$ , $2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$ , $2\left(\cos \frac{23\pi }{20}+i\sin \frac{23\pi }{20}\right)$ , $2\left(\cos \frac{31\pi }{20}+i\sin \frac{31\pi }{20}\right)$ and $2\left(\cos \frac{39\pi }{20}+i\sin \frac{39\pi }{20}\right)$ .

b

To determine

The graph for the complex roots.

Answer to Problem 152E

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

Explanation of Solution

Given information:

The complex number is $4\left(1-i\right)$

Calculation:

The fifth roots of $4\left(1-i\right)$ are $2\left(\cos \frac{7\pi }{20}+i\sin \frac{7\pi }{20}\right)$ , $2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$ , $2\left(\cos \frac{23\pi }{20}+i\sin \frac{23\pi }{20}\right)$ , $2\left(\cos \frac{31\pi }{20}+i\sin \frac{31\pi }{20}\right)$ and $2\left(\cos \frac{39\pi }{20}+i\sin \frac{39\pi }{20}\right)$ .

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 152E

The standard forms of the roots are $0.9078+1.782i$ , $-1.4142+1.4142i$ , $-1.782-0.9078i$ , $0.3128-1.9754i$ and $1.9754-0.3128i$ .

Explanation of Solution

Given information:

The fifth roots of $4\left(1-i\right)$ are $2\left(\cos \frac{7\pi }{20}+i\sin \frac{7\pi }{20}\right)$ , $2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$ , $2\left(\cos \frac{23\pi }{20}+i\sin \frac{23\pi }{20}\right)$ , $2\left(\cos \frac{31\pi }{20}+i\sin \frac{31\pi }{20}\right)$ and $2\left(\cos \frac{39\pi }{20}+i\sin \frac{39\pi }{20}\right)$ .

Calculation:

Calculate the standard form of the root $z_0$ .

  $\begin{array}{c}{z}_0=2\left(\cos \frac{7\pi }{20}+i\sin \frac{7\pi }{20}\right)\\ =0.9078+1.782i\end{array}$

Calculate the standard form of the root $z_1$ .

  $\begin{array}{c}{z}_1=2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ =-1.4142+1.4142i\end{array}$

Calculate the standard form of the root $z_2$ .

  $\begin{array}{c}{z}_2=2\left(\cos \frac{23\pi }{20}+i\sin \frac{23\pi }{20}\right)\\ =-1.782-0.9078i\end{array}$

Calculate the standard form of the root $z_3$ .

  $\begin{array}{c}{z}_3=2\left(\cos \frac{31\pi }{20}+i\sin \frac{31\pi }{20}\right)\\ =0.3128-1.9754i\end{array}$

Calculate the standard form of the root $z_4$ .

  $\begin{array}{c}{z}_4=2\left(\cos \frac{39\pi }{20}+i\sin \frac{39\pi }{20}\right)\\ =1.9754-0.3128i\end{array}$

Therefore, the standard forms of the roots are $0.9078+1.782i$ , $-1.4142+1.4142i$ , $-1.782-0.9078i$ , $0.3128-1.9754i$ and $1.9754-0.3128i$ .