Chapter 8.3, Problem 85E

To determine

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

Answer to Problem 85E

The eighth roots are ${\omega }_0=1$ , ${\omega }_1=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ , ${\omega }_2=i$ , ${\omega }_3=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ , ${\omega }_4=-1$ , ${\omega }_5=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$ , ${\omega }_6=-i$ and ${\omega }_7=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$

Explanation of Solution

Given information: The eighth roots of $1$

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

  $z=1$

Polar form of complex number is

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

Calculating the value of $r$ :

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

Calculating the value of $\theta :$

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

Hence, complex number in polar form is

  $z=1\left(\cos 0+isin0\right)$

In order to find eighth 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, $n=8$ and $z=1\left(\cos 0+isin0\right)$

Thus,

  $\begin{array}{l}{\omega }_k=1^{\frac{1}{8}}\left[\cos \left(\frac{0+2\pi k}{8}\right)+i\sin \left(\frac{0+2\pi k}{8}\right)\right]\\ {\omega }_k={\left(1^8\right)}^{\frac{1}{8}}\left[\cos \left(0+\frac{2\pi k}{8}\right)+i\sin \left(0+\frac{2\pi k}{8}\right)\right]\\ {\omega }_k=1\left[\cos \frac{\pi k}{4}+i\sin \frac{\pi k}{4}\right]\\ {\omega }_k=\cos \frac{\pi k}{4}+i\sin \frac{\pi k}{4}\end{array}$

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

Thus,

  $\begin{array}{l}0\le k\le 8-1\\ 0\le k\le 7\end{array}$

Hence, $k=0,1,2,3,4,5,6,7$

When $k=0$

  $\begin{array}{l}{\omega }_0=\cos \frac{\pi \left(0\right)}{4}+i\sin \frac{\pi \left(0\right)}{4}\\ {\omega }_0=\cos 0+i\sin 0\\ {\omega }_0=1+0i\\ {\omega }_0=1\end{array}$

When $k=1$

  $\begin{array}{l}{\omega }_1=\cos \frac{\pi \left(1\right)}{4}+i\sin \frac{\pi \left(1\right)}{4}\\ {\omega }_1=\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\\ {\omega }_1=\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\\ {\omega }_1=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\end{array}$

When $k=2$

  $\begin{array}{l}{\omega }_2=\cos \frac{\pi \left(2\right)}{4}+i\sin \frac{\pi \left(2\right)}{4}\\ {\omega }_2=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\\ {\omega }_2=0+i1\\ {\omega }_2=i\end{array}$

When $k=3$

  $\begin{array}{l}{\omega }_3=\cos \frac{\pi \left(3\right)}{4}+i\sin \frac{\pi \left(3\right)}{4}\\ {\omega }_3=\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\\ {\omega }_3=-\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\\ {\omega }_3=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\end{array}$

When $k=4$

  $\begin{array}{l}{\omega }_4=\cos \frac{\pi \left(4\right)}{4}+i\sin \frac{\pi \left(4\right)}{4}\\ {\omega }_4=\cos \pi +i\sin \pi \\ {\omega }_4=-1+i0\\ {\omega }_4=-1\end{array}$

When $k=5$

  $\begin{array}{l}{\omega }_5=\cos \frac{\pi \left(4\right)}{4}+i\sin \frac{\pi \left(5\right)}{4}\\ {\omega }_5=\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\\ {\omega }_5=-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\\ {\omega }_5=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\end{array}$

When $k=6$

  $\begin{array}{l}{\omega }_6=\cos \frac{\pi \left(6\right)}{4}+i\sin \frac{\pi \left(6\right)}{4}\\ {\omega }_6=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\\ {\omega }_6=0-i1\\ {\omega }_6=-i\end{array}$

When $k=7$

  $\begin{array}{l}{\omega }_7=\cos \frac{\pi \left(7\right)}{4}+i\sin \frac{\pi \left(7\right)}{4}\\ {\omega }_7=\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\\ {\omega }_7=\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\\ {\omega }_7=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\end{array}$

When $k=4$

  $\begin{array}{l}{\omega }_4=\cos \frac{\pi \left(4\right)}{4}+i\sin \frac{\pi \left(4\right)}{4}\\ {\omega }_4=\cos \pi +i\sin \pi \\ {\omega }_4=-1+i0\\ {\omega }_4=-1\end{array}$

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

  

Conclusion:

Hence, eighth roots are ${\omega }_0=1$ , ${\omega }_1=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ , ${\omega }_2=i$ , ${\omega }_3=-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}$ , ${\omega }_4=-1$ , ${\omega }_5=-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$ , ${\omega }_6=-i$ and ${\omega }_7=\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}$