Chapter 6.6, Problem 109E

(a)

To determine

To calculate:

The cube roots of complex number.

Answer to Problem 109E

The cube roots of complex number are $2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right),2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right)$ and $2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)$

Explanation of Solution

Given information:

Complex number is,

  $8\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$

Formula used:

The formula to calculate the roots of complex number are given by,

  $n\sqrt{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

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

Calculation:

From the formula given above,

  $8\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)$

Substituting in the above formula,

  $\begin{array}{l}=2\sqrt{8}\left(\cos \frac{\frac{2\pi }{3}+2\pi k}{3}+i\sin \frac{\frac{2\pi }{3}+2\pi k}{3}\right)\\ =2\left(\cos \frac{2\pi +6\pi k}{9}+i\sin \frac{2\pi +6\pi k}{9}\right)\end{array}$

Now find the cube roots of complex number for $k=0,1,2$

For $k=0,$

  $\begin{array}{l}=2\left(\cos \frac{2\pi +0}{9}+i\sin \frac{2\pi }{9}\right)\\ =2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)\end{array}$

Now for $k=1,$

  $\begin{array}{l}=2\left(\cos \frac{2\pi +6\pi \times 1}{9}+i\sin \frac{2\pi +6\pi \times 1}{9}\right)\\ =2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right)\end{array}$

For $k=2$ ,

  $\begin{array}{l}=2\left(\cos \frac{2\pi +6\pi \times 2}{9}+i\sin \frac{2\pi +6\pi \times 2}{9}\right)\\ =2\left(\cos \frac{2\pi +12\pi }{9}+i\sin \frac{2\pi +12\pi }{9}\right)\\ =2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)\end{array}$

(b)

To determine

To calculate:

The roots in standard from

Answer to Problem 109E

The roots in standard form are $-1.8793+0.6840$ , $0.3473-1.9696$ and $1.5321+1.2856$

Explanation of Solution

Given information:

The roots are,

  $2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right),2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)$ And $2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)$

Calculation:

Consider the roots,

  $2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right),2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)$ And $2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)$

Now, first write the standard form of $2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right)$

  $\begin{array}{l}=2\times \left(-\cos \frac{\pi }{9}+i\sin \frac{\pi }{9}\right)\\ =2\times \left(-\cos 20+i\sin 20\right)\\ =-1.8793+0.6840\end{array}$

Now, the standard form of $2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)$

  $\begin{array}{l}=2\times \left(\cos \left(2\pi -\frac{4\pi }{9}\right)\right)+i\sin \left(2\pi -\frac{4\pi }{9}\right)\\ =2\times \left(\cos \frac{4\pi }{9}+i\sin \frac{4\pi }{9}\right)\\ =2\left(\cos 80+i\sin 80\right)\\ =0.3473-1.9696\end{array}$

Now, the standard form of $2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)$

  $\begin{array}{l}=2\left(\cos \frac{2\times 180}{9}+i\sin \frac{2\times 180}{9}\right)\\ =2\left(\cos 40+i\sin 40\right)\\ =1.5321+1.2856\end{array}$

(c)

To determine

To graph:

The roots in graphical form

Explanation of Solution

Given information:

Cube Roots of complex number,

  $2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right),2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)$ And $2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)$

Graph:

  

Interpretation:

The graph of the cube roots of complex numbers $2\left(\cos \frac{8\pi }{9}+i\sin \frac{8\pi }{9}\right),2\left(\cos \frac{2\pi }{9}+i\sin \frac{2\pi }{9}\right)$ and $2\left(\cos \frac{14\pi }{9}+i\sin \frac{14\pi }{9}\right)$ are sketched as above.