Chapter 6.5, Problem 137E

(a)

To determine

To find: The indicated root of the given expression

Answer to Problem 137E

The roots of the complex numbers are $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$ and $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$ .

Explanation of Solution

Given information:

The given expression is:

  $\sqrt[]{5}\left(\cos 120°+i\sin 120°\right)$

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 given expression as:

  $\sqrt[]{5}\left(\cos 120°+i\sin 120°\right)$

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 $120°$ for $\theta$ in the above expression:

  $\sqrt[2]{r}\left(\cos \frac{120+2\pi k}{n}+i\sin \frac{120+2\pi k}{n}\right)$

Substitute $0$ for $k$ in the above expression:

  $\begin{array}{c}\sqrt[2]{r}\left(\cos \frac{120+2\pi \left(0\right)}{n}+i\sin \frac{120+2\pi \left(0\right)}{n}\right)\\ =\sqrt{5}\left(\cos 60°+i\sin 60°\right)\end{array}$

Substitute $1$ for $k$ in the above expression:

  $\begin{array}{c}\sqrt[2]{r}\left(\cos \frac{120+2\pi \left(1\right)}{n}+i\sin \frac{120+2\pi \left(1\right)}{n}\right)\\ =\sqrt{5}\left(\cos 240°+i\sin 240°\right)\end{array}$

Hence, the roots of the complex numbers are $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$ and $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$

(b)

To determine

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

Explanation of Solution

Given information:

The given expression is:

  $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$

  $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$

Graph:

Consider the given expression as:

  $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$

  $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$

Graph the above equations:

  

(c)

To determine

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

Answer to Problem 137E

The roots of the complex numbers are $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$ and $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$ .

Explanation of Solution

Given information:

The given expression is:

  $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$

  $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$

Consider the given expression as:

  $\sqrt{5}\left(\cos 60°+i\sin 60°\right)$

  $\sqrt{5}\left(\cos 240°+i\sin 240°\right)$

Substitute $\frac{1}{2}$ for $\cos 60°$ and $\frac{\sqrt{3}}{2}$ for $\sin 60°$ in the above expression:

  $\begin{array}{c}\sqrt{5}\left(\cos 60°+i\sin 60°\right)=\sqrt{5}\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)\\ =\frac{\sqrt{5}}{2}+\frac{\sqrt{15}}{2}i\end{array}$

Substitute $-\frac{1}{2}$ for $\cos 240°$ and $-\frac{\sqrt{3}}{2}$ for $\sin 240°$ in the above expression:

  $\begin{array}{c}\sqrt{5}\left(\cos 240°+i\sin 240°\right)=\sqrt{5}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)\\ =-\frac{\sqrt{5}}{2}-\frac{\sqrt{15}}{2}i\end{array}$

Hence, the standard form of the complex numbers are $\frac{\sqrt{5}}{2}+\frac{\sqrt{15}}{2}i$ and $-\frac{\sqrt{5}}{2}-\frac{\sqrt{15}}{2}i$ .