Chapter 6, Problem 52RE

Solution

To find: The $n\text{th}$ roots of the complex number for the specified value of $n$ . Also graph the roots.

The sixth roots of the given complex number are $\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right),i,\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right),\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right),-i$ and $\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ . The roots are shown in figure (1).

Given data:

  $\begin{array}{l}z=-1\\ n=6\end{array}$

Formula used:

If $z=r(\cos \theta +i\sin \theta )$ then the $n$ distinct complex numbers $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ where $k=0,1,2,3,\mathrm{....},n-1$ are the $n^{th}$ roots of the complex number $z$ .

Calculation:

Compute $r$ and $\theta$ .

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

  $\begin{array}{c}\theta ={\cos }^{-1}(-1)\\ =\pi \end{array}$

Rewrite the complex number $z=-1$ in polar form.

  $\begin{array}{c}z=-1+0i\\ =\cos \pi +i\sin \pi \end{array}$

Compute the sixth roots of $z=\cos \pi +i\sin \pi$ are the complex numbers $\left(\cos \frac{\pi +2\pi k}{6}+i\sin \frac{\pi +2\pi k}{6}\right)$ for $k=0,1,2,3,4,5$ .

  $\begin{array}{l}{z}_1=\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\\ z_2=\cos \frac{\pi +2\pi }{6}+i\sin \frac{\pi +2\pi }{6}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\\ z_3=\cos \frac{\pi +4\pi }{6}+i\sin \frac{\pi +4\pi }{6}=\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\end{array}$

  $\begin{array}{l}{z}_4=\cos \frac{\pi +6\pi }{6}+i\sin \frac{\pi +6\pi }{6}=\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\\ z_5=\cos \frac{\pi +8\pi }{6}+i\sin \frac{\pi +8\pi }{6}=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\\ z_6=\cos \frac{\pi +10\pi }{6}+i\sin \frac{\pi +10\pi }{6}=\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\end{array}$

Plot the graph of the complex numbers $z_1,z_2,z_3,z_4\text{,}{\text{z}}_5,\text{and}{z}_6$ .

  

Figure (1)

From the graph, it can be observed that the graph of all sixth roots of given complex number is a straight line with different slopes.

Therefore, the sixth roots of the given complex number are $\left(\cos \frac{\pi }{6}+i\sin \frac{\pi }{6}\right),i,\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right),\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right),-i$ and $\left(\cos \frac{11\pi }{6}+i\sin \frac{11\pi }{6}\right)$ .