Chapter 8.3, Problem 89E

To determine

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

Answer to Problem 89E

The forth roots are ${\omega }_0=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , ${\omega }_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , ${\omega }_2=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ and ${\omega }_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$

Explanation of Solution

Given information: The forth 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 $r$ :

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

Calculating the value of $\theta :$

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

Hence, complex number in polar form is

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

In order to find fourth 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, $n=4$ and $z=\cos \pi +isin\pi$

Thus,

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

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

Thus,

  $\begin{array}{l}0\le k\le 4-1\\ 0\le k\le 3\end{array}$

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

When $k=0$

  $\begin{array}{l}{\omega }_0=\cos \left(\frac{\pi }{4}+\frac{2\pi \left(0\right)}{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{2\pi \left(0\right)}{4}\right)\\ {\omega }_0=\cos \left(\frac{\pi }{4}+0\right)+i\sin \left(\frac{\pi }{4}+0\right)\\ {\omega }_0=\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{4}\right)\\ {\omega }_0=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\\ {\omega }_0=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

When $k=1$

  $\begin{array}{l}{\omega }_1=\cos \left(\frac{\pi }{4}+\frac{2\pi \left(1\right)}{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{2\pi \left(1\right)}{4}\right)\\ {\omega }_1=\cos \left(\frac{\pi }{4}+\frac{2\pi }{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{2\pi }{4}\right)\\ {\omega }_1=\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right)\\ {\omega }_1=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\\ {\omega }_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

When $k=2$

  $\begin{array}{l}{\omega }_2=\cos \left(\frac{\pi }{4}+\frac{2\pi \left(2\right)}{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{2\pi \left(2\right)}{4}\right)\\ {\omega }_2=\cos \left(\frac{\pi }{4}+\frac{4\pi }{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{4\pi }{4}\right)\\ {\omega }_2=\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\\ {\omega }_2=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\\ {\omega }_2=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\end{array}$

When $k=3$

  $\begin{array}{l}{\omega }_3=\cos \left(\frac{\pi }{4}+\frac{2\pi \left(3\right)}{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{2\pi \left(3\right)}{4}\right)\\ {\omega }_3=\cos \left(\frac{\pi }{4}+\frac{6\pi }{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{6\pi }{4}\right)\\ {\omega }_3=\cos \left(\frac{7\pi }{4}\right)+i\sin \left(\frac{7\pi }{4}\right)\\ {\omega }_3=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\\ {\omega }_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\end{array}$

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

  

Conclusion:

Hence, forth roots are ${\omega }_0=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , ${\omega }_1=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , ${\omega }_2=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ and ${\omega }_3=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$