Chapter 6, Problem 51RE

Solution

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

The fifth roots of the given complex number are $1,\left(\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\right),\left(\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\right),\left(\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\right),\text{and}\left(\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\right)$ . Each root is shown on the figure (1).

Given data:

  $\begin{array}{l}z=1\\ n=5\end{array}$

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 $\theta$ .

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

  $\begin{array}{c}\theta ={\cos }^{-1}(1)\\ =0°\end{array}$

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

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

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

  $\begin{array}{l}{z}_1=\cos 0+i\sin 0\\ z_2=\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\end{array}$

  $\begin{array}{l}{z}_3=\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\\ z_4=\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\\ z_5=\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\end{array}$

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

  

Figure (1)

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

Therefore, the fifth roots of the given complex number are $1,\left(\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\right),\left(\cos \frac{4\pi }{5}+i\sin \frac{4\pi }{5}\right),\left(\cos \frac{6\pi }{5}+i\sin \frac{6\pi }{5}\right),\text{and}\left(\cos \frac{8\pi }{5}+i\sin \frac{8\pi }{5}\right)$ .