Chapter 6.5, Problem 142E

(a)

To determine

To find: The indicated root of the given expression

Answer to Problem 142E

The roots of the complex numbers are $5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ , $5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$ , $5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$ and $2\left(\cos \frac{13\pi }{30}+i\sin \frac{13\pi }{30}\right)$ .

Explanation of Solution

Given information:

The given expression is:

  $\sqrt[4]{625i}$

Formula used:

Formula of $n^{th}$ root:

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

Consider the general form of complex number:

  $\begin{array}{l}z=a+ib\\ z=r\left(\cos \theta +i\sin \theta \right)\end{array}$

Calculate the value of r :

  $r=\sqrt{a^2+b^2}$

Substitute $0$ for $a$ and $625$ for $b$ in the above equation:

  $\begin{array}{c}r=\sqrt{0^2+{625}^2}\\ =625\end{array}$

Find the value of $\tan \theta$ using the equation $\tan \theta =\frac{b}{a}$ .

Substitute $0$ for $a$ and $625$ for $b$ in the above equation:

  $\begin{array}{c}\tan \theta =\frac{625}{0}\\ =625\end{array}$

Consider the angle to lie in the first quadrant:

  $\theta =\frac{\pi }{2}$

Consider the formula of $n^{th}$ root:

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

Substitute $\frac{\pi }{2}$ for $\theta$ in the above expression:

  $\begin{array}{l}\sqrt[4]{625}\left(\cos \frac{\frac{\pi }{2}+2\pi k}{4}+i\sin \frac{\frac{\pi }{2}+2\pi k}{4}\right)\\ =\sqrt[4]{625}\left(\cos \frac{\pi +2\pi k}{8}+i\sin \frac{\pi +2\pi k}{8}\right)\end{array}$

Substitute $0$ for $k$ in the above expression to find the first root:

  $\begin{array}{l}\sqrt[4]{625}\left(\cos \frac{\pi +2\pi 0}{8}+i\sin \frac{\pi +2\pi 0}{8}\right)\\ =5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\end{array}$

Substitute $1$ for $k$ in the above expression to find the second root:

  $\begin{array}{c}\sqrt[4]{625}\left(\cos \frac{\pi +2\pi 1}{8}+i\sin \frac{\pi +2\pi 1}{8}\right)\\ =5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)\end{array}$

Substitute $2$ for $k$ in the above expression to find the third root:

  $\begin{array}{c}\sqrt[4]{625}\left(\cos \frac{\pi +2\pi 2}{8}+i\sin \frac{\pi +2\pi 2}{8}\right)\\ =5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)\end{array}$

Substitute $3$ for $k$ in the above expression to find the forth root:

  $\begin{array}{c}\sqrt[4]{625}\left(\cos \frac{\pi +2\pi 3}{8}+i\sin \frac{\pi +2\pi 3}{8}\right)\\ =5\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)\end{array}$

Hence, the roots of the complex numbers are $5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ , $5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$ , $5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$ and $2\left(\cos \frac{13\pi }{30}+i\sin \frac{13\pi }{30}\right)$ .

(b)

To determine

To graph: The roots of the complex number from the above results.

Explanation of Solution

Given information:

The given expression is:

  $5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$

  $5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$

  $2\left(\cos \frac{13\pi }{30}+i\sin \frac{13\pi }{30}\right)$ .

Graph:

Consider the given expression as:

  $5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$

  $5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$

  $2\left(\cos \frac{13\pi }{30}+i\sin \frac{13\pi }{30}\right)$ .

Graph the above equations:

  

(c)

To determine

To find: The roots of the above expression in standard form

Answer to Problem 142E

The standard form of the complex numbers are $4.6194+1.9134i$ , $-1.9134+4.619i$ , $-4.6194-1.9134i$ and $1.9134-4.6194i$ .

Explanation of Solution

Given information:

The given expression is:

  $5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$

  $5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$

  $2\left(\cos \frac{13\pi }{30}+i\sin \frac{13\pi }{30}\right)$ .

Consider the given expression as:

  $5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$

  $5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)$

  $5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)$

  $2\left(\cos \frac{13\pi }{30}+i\sin \frac{13\pi }{30}\right)$ .

Substitute $0.9239$ for $\cos \frac{\pi }{8}$ and $0.3827$ for $\cos \frac{\pi }{8}$ in the above expression:

  $\begin{array}{c}5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)=5\left(0.9239+0.3827i\right)\\ =4.6194+1.9134i\end{array}$

Substitute $-0.3827$ for $\cos \frac{5\pi }{8}$ and $0.9239$ for $\sin \frac{5\pi }{8}$ in the above expression:

  $\begin{array}{c}5\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)=5\left(-0.3827+0.9239i\right)\\ =-1.9134+4.619i\end{array}$

Substitute $-0.9239$ for $\cos \frac{9\pi }{8}$ and $-0.3827$ for $\sin \frac{9\pi }{8}$ in the above expression:

  $\begin{array}{c}5\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=5\left(-0.9239-0.3827i\right)\\ =-4.6194-1.9134i\end{array}$

Substitute $0.3827$ for $\cos \frac{13\pi }{8}$ and $-0.9239$ for $\sin \frac{13\pi }{8}$ in the above expression:

  $\begin{array}{c}2\left(\cos \frac{13\pi }{8}+i\sin \frac{13\pi }{8}\right)=5\left(0.3827-0.9239i\right)\\ =1.9134-4.6194i\end{array}$

Hence, the standard form of the complex numbers are $4.6194+1.9134i$ , $-1.9134+4.619i$ , $-4.6194-1.9134i$ and $1.9134-4.6194i$ .