Chapter 8.3, Problem 88E

To determine

To calculate: To find the indicated roots and graph the roots in the complex plane

Answer to Problem 88E

The fifth roots are ${\omega }_0=\cos \frac{\pi }{10}+i\sin \frac{\pi }{10}$ , ${\omega }_1=i$ , ${\omega }_2=\cos \frac{9\pi }{10}+i\sin \frac{9\pi }{10}$ , ${\omega }_3=\cos \frac{13\pi }{10}+i\sin \frac{13\pi }{10}$ and ${\omega }_4=\cos \frac{17\pi }{10}+i\sin \frac{17\pi }{10}$

Explanation of Solution

Given information: The fifth roots of $i$

Formula Used:

Complex number is a number that can be expressed in the form of $a+bi$ , where $a$ and $b$ are real numbers

  $z=a+bi$

If $z=r\left(\cos \theta +isin\theta \right)$ and $n$ is a positive integer, then $z$ has the $n$ distinct $n^{th}$ roots

  ${\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]$ for $k=0,1,2,\mathrm{3....},n-1$

Calculation:

Complex number is given as

  $z=i$

Polar form of complex number is

  $z=r\left(\cos \theta +isin\theta \right)--(1)$

Calculating the value of $r$ :

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

Calculating the value of $\theta :$

  $\begin{array}{l}\tan \theta =\frac{1}{0}\\ \theta ={\tan }^{-1}\frac{1}{0}\\ \theta =\frac{\pi }{2}\end{array}$

Hence, complex number in polar form is

  $\begin{array}{l}z=1\left(\cos \frac{\pi }{2}+isin\frac{\pi }{2}\right)\\ z=\cos \frac{\pi }{2}+isin\frac{\pi }{2}\end{array}$

In order to find fifth roots of above complex number, apply nth Rule of Complex numbers

  ${\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]$

Here, $n=5$ and $z=\cos \frac{\pi }{2}+isin\frac{\pi }{2}$

Thus,

  $\begin{array}{l}{\omega }_k=1^{\frac{1}{5}}\left[\cos \left(\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+2\pi k}{5}\right)+i\sin \left(\frac{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+2\pi k}{5}\right)\right]\\ {\omega }_k=1\left[\cos \left(\frac{\pi }{10}+\frac{2\pi k}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi k}{5}\right)\right]\\ {\omega }_k=\cos \left(\frac{\pi }{10}+\frac{2\pi k}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi k}{5}\right)\end{array}$

Now, the value of $k$ is from $0$ to $n-1$

Thus,

  $\begin{array}{l}0\le k\le 5-1\\ 0\le k\le 4\end{array}$

Hence, $k=0,1,2,3,4$

When $k=0$

  $\begin{array}{l}{\omega }_0=\cos \left(\frac{\pi }{10}+\frac{2\pi \left(0\right)}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi \left(0\right)}{5}\right)\\ {\omega }_0=\cos \left(\frac{\pi }{10}+0\right)+i\sin \left(\frac{\pi }{10}+0\right)\\ {\omega }_0=\cos \frac{\pi }{10}+i\sin \frac{\pi }{10}\end{array}$

When $k=1$

  $\begin{array}{l}{\omega }_1=\cos \left(\frac{\pi }{10}+\frac{2\pi \left(1\right)}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi \left(1\right)}{5}\right)\\ {\omega }_1=\cos \left(\frac{\pi }{10}+\frac{2\pi }{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi }{5}\right)\\ {\omega }_1=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\\ {\omega }_1=0+i1\\ {\omega }_1=i\end{array}$

When $k=2$

  $\begin{array}{l}{\omega }_2=\cos \left(\frac{\pi }{10}+\frac{2\pi \left(2\right)}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi \left(2\right)}{5}\right)\\ {\omega }_2=\cos \left(\frac{\pi }{10}+\frac{4\pi }{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{4\pi }{5}\right)\\ {\omega }_2=\cos \frac{9\pi }{10}+i\sin \frac{9\pi }{10}\end{array}$

When $k=3$

  $\begin{array}{l}{\omega }_3=\cos \left(\frac{\pi }{10}+\frac{2\pi \left(3\right)}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi \left(3\right)}{5}\right)\\ {\omega }_3=\cos \left(\frac{\pi }{10}+\frac{6\pi }{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{6\pi }{5}\right)\\ {\omega }_3=\cos \frac{13\pi }{10}+i\sin \frac{13\pi }{10}\end{array}$

When $k=4$

  $\begin{array}{l}{\omega }_4=\cos \left(\frac{\pi }{10}+\frac{2\pi \left(4\right)}{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{2\pi \left(4\right)}{5}\right)\\ {\omega }_4=\cos \left(\frac{\pi }{10}+\frac{8\pi }{5}\right)+i\sin \left(\frac{\pi }{10}+\frac{8\pi }{5}\right)\\ {\omega }_4=\cos \frac{17\pi }{10}+i\sin \frac{17\pi }{10}\end{array}$

Now, plotting the above points on the complex plane. All points lie on a circle of radius $r=1$

  

Conclusion:

Hence, fifth roots are ${\omega }_0=\cos \frac{\pi }{10}+i\sin \frac{\pi }{10}$ , ${\omega }_1=i$ , ${\omega }_2=\cos \frac{9\pi }{10}+i\sin \frac{9\pi }{10}$ , ${\omega }_3=\cos \frac{13\pi }{10}+i\sin \frac{13\pi }{10}$ and ${\omega }_4=\cos \frac{17\pi }{10}+i\sin \frac{17\pi }{10}$