Chapter 6.5, Problem 144E

a

To determine

The cube roots of the complex number.

Answer to Problem 144E

The cube roots of $-4\sqrt{2}\left(1+i\right)$ are $2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ and $2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ .

Explanation of Solution

Given information:

The complex number is $-4\sqrt{2}\left(1+i\right)$

Calculation:

Calculate the value of $r$ .

  $\begin{array}{c}r=\left|-4\sqrt{2}+4\sqrt{2i}\right|\\ =\sqrt{{\left(-4\sqrt{2}\right)}^2+{\left(4\sqrt{2}\right)}^2}\\ =8\end{array}$

Calculate the value of $\theta$ .

  $\begin{array}{c}\theta =\arctan \left(-1\right)\\ =\frac{3\pi }{4}\end{array}$

The trigonometric form of $z$ .

  $z=8\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

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.

  $-4\sqrt{2}\left(1+i\right)$

The fourth roots of $-4\sqrt{2}\left(1+i\right)$ is to be evaluated $n=3$ and $r=8$ by the $n^{th}$ roots formula the roots are

  $z_k=\sqrt[3]{8}\left(\cos \frac{\frac{3\pi }{4}+2\pi k}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi k}{3}\right)$ ……. (2)

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

  $\begin{array}{c}{z}_0=\sqrt[3]{8}\left(\cos \frac{\frac{3\pi }{4}+2\pi \left(0\right)}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi \left(0\right)}{3}\right)\\ z_0=2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\end{array}$

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

  $\begin{array}{c}{z}_1=\sqrt[3]{8}\left(\cos \frac{\frac{3\pi }{4}+2\pi \left(1\right)}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi \left(1\right)}{3}\right)\\ z_1=2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\end{array}$

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

  $\begin{array}{c}{z}_2=\sqrt[3]{8}\left(\cos \frac{\frac{3\pi }{4}+2\pi \left(2\right)}{3}+i\sin \frac{\frac{3\pi }{4}+2\pi \left(2\right)}{3}\right)\\ z_2=2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\end{array}$

Therefore, the cube roots of $-4\sqrt{2}\left(1+i\right)$ are $2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ and $2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ .

b

To determine

The graph for the complex roots.

Answer to Problem 144E

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

Explanation of Solution

Given information:

The complex number is $-4\sqrt{2}\left(1+i\right)$

Calculation:

The cube roots of $-4\sqrt{2}\left(1+i\right)$ are $2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ and $2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ .

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 144E

The standard forms of the roots are $\sqrt{2}+\sqrt{2i}$ , $-1.9319+0.5176i$ and $0.5176-0.9659i$ .

Explanation of Solution

Given information:

The cube roots of $-4\sqrt{2}\left(1+i\right)$ are $2\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)$ and $2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)$ .

Calculation:

Calculate the standard form of the root $z_0$ .

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

Calculate the standard form of the root $z_1$ .

  $\begin{array}{c}{z}_1=2\left(\cos \frac{11\pi }{12}+i\sin \frac{11\pi }{12}\right)\\ =-1.9319+0.5176i\end{array}$

Calculate the standard form of the root $z_2$ .

  $\begin{array}{c}{z}_2=2\left(\cos \frac{19\pi }{12}+i\sin \frac{19\pi }{12}\right)\\ =0.5176-0.9659i\end{array}$

Therefore, the standard forms of the roots are $\sqrt{2}+\sqrt{2i}$ , $-1.9319+0.5176i$ and $0.5176-0.9659i$ .