Chapter 6.6, Problem 117E

a.

To determine

To calculate:

The roots of complex number.

Answer to Problem 117E

The roots of the complex numbers are,

  $\begin{array}{l}\cos 0+i\sin 0\\ \cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\\ \cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\\ \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\\ \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\end{array}$

Explanation of Solution

Given information:

Fifth roots of $1$ .

Formula used:

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

Where,

  $n=$ positive integer,

complex number $z=r\left(\cos \theta +i\sin \theta \right)$ has exactly $n$ distinct $n^{th}$ roots, $k=0,1,\mathrm{2.....},n-1$

Calculation:

Consider the fifth roots of $1$

Its modulus is given by

  $\begin{array}{l}r=\sqrt{1^2+0}\\ r=1\end{array}$

Its argument is

  $\begin{array}{l}\theta =\arctan \frac{0}{1}\\ \theta =\arctan 0\\ \theta =0\end{array}$

Hence, its trigonometric form comes out to be

  $z=1\left(\cos 0+i\sin 0\right)$

For a positive integer $n$ , the complex number $z=r\left(\cos \pi +i\sin \theta \right)$ has exactly $n$ distinct $n^{th}$ roots given by,

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

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

So, by using the formula

  $\begin{array}{l}z=1\left(\cos 0+i\sin 0\right)\\ =\sqrt[5]{1}\left(\cos \frac{0+2\pi k}{5}+i\sin \frac{0+2\pi k}{5}\right)\\ =1\left(\cos \frac{2\pi k}{5}+i\sin \frac{2\pi k}{5}\right)\end{array}$

So, for $k=0,1,2,3,4$ the square roots ae as follows,

For $0:$

  $\begin{array}{l}\cos \frac{2\pi \times 0}{5}+i\sin \frac{2\pi \times 0}{5}\\ =\cos 0+i\sin 0\end{array}$

For $1:$

  $\begin{array}{l}\cos \frac{2\pi \times 1}{5}+i\sin \frac{2\pi \times 1}{5}\\ =\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\end{array}$

For $2:$

  $\begin{array}{l}\cos \frac{2\pi \times 2}{5}+i\sin \frac{2\pi \times 2}{5}\\ =\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\end{array}$

For $3:$

  $\begin{array}{l}\cos \frac{2\pi \times 3}{5}+i\sin \frac{2\pi \times 3}{5}\\ =\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\end{array}$

For $4:$

  $\begin{array}{l}\cos \frac{2\pi \times 4}{5}+i\sin \frac{2\pi \times 4}{5}\\ =\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\end{array}$

b.

To determine

To calculate:

Given roots in standard form.

Answer to Problem 117E

The complex roots are in the standard form as,

  $\begin{array}{l}1\\ 0.3090+i09511\\ -0.8090+0.5878i\\ -0.8090-0.5878i\\ 0.3090-0.9511i\end{array}$

Explanation of Solution

Given information:

Fifth roots of $1$ .

The roots are,

  $\begin{array}{l}\cos 0+i\sin 0\\ \cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\\ \cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\\ \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\\ \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\end{array}$

Calculation:

Write the forms is the standard form,

  $\cos 0+i\sin 0=1$

For, $\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}$

  $\begin{array}{l}=\cos 72+i\sin 72\\ =0.3090+i09511\end{array}$

For, $\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}$

  $\begin{array}{l}=\cos \left(\pi -\frac{\pi }{5}\right)+i\sin \left(\pi -\frac{\pi }{5}\right)\\ =-\cos \frac{\pi }{5}+i\sin \frac{\pi }{5}\\ =-0.8090+0.5878i\end{array}$

For, $\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}$

  $\begin{array}{l}=\cos \left(\pi +\frac{\pi }{5}\right)+i\sin \left(\pi +\frac{\pi }{5}\right)\\ =-\cos \frac{\pi }{5}-i\sin \frac{\pi }{5}\\ =-0.8090-0.5878i\end{array}$

For, $\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}$

  $\begin{array}{l}=\cos \left(2\pi -\frac{2\pi }{5}\right)+i\sin \left(2\pi -\frac{2\pi }{5}\right)\\ =\cos \frac{2\pi }{5}-i\sin \frac{2\pi }{5}\\ =0.3090-0.9511i\end{array}$

c.

To determine

To graph:

The given roots graphically.

Explanation of Solution

Given information:

The roots are,

  $\begin{array}{l}\cos 0+i\sin 0\\ \cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\\ \cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\\ \cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\\ \cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\end{array}$

Graph:

  

Interpretation:

In above graph, complex roots are described.