Chapter 6, Problem 119RE

To determine

To find: All the solutions of the given equation and represent the solutions graphically.

Answer to Problem 119RE

The roots are $4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $4\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$ , $4\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ and $4\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$ which are represented graphically in Figure 1.

Explanation of Solution

Given:

The equation is $x^4+256=0$ .

Calculation:

First express the equation in the form of x .

  $\begin{array}{c}{x}^4+256=0\\ x^4=-256\\ x=\sqrt[4]{-256}\end{array}$

For any complex number $z=r\left(\cos \theta +i\sin \theta \right)$ , the nth root is given by,

  $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right),k=0,1,2,\dots ,n-1$

So, to find the fourth root of $-256$ , let $z=-256$ and convert it into the form $r\left(\cos \theta +i\sin \theta \right)$ .

  $\begin{array}{c}r=\sqrt{a^2+b^2}\\ =\sqrt{{\left(-256\right)}^2+{\left(0\right)}^2}\\ =\sqrt{{\left(256\right)}^2}\\ =256\end{array}$

  $\begin{array}{c}\tan \theta =\frac{b}{a}\\ =-\frac{0}{256}\\ =\pi \end{array}$

So, $-256=256\left(\cos \pi +i\sin \pi \right)$ and the fourth root is calculated as shown below:

  $\sqrt[4]{256}\left(\cos \frac{\pi +2\pi k}{4}+i\sin \frac{\pi +2\pi k}{4}\right)=4\left(\cos \frac{\pi +2\pi k}{4}+i\sin \frac{\pi +2\pi k}{4}\right)$

For $k=0$ ,

  $4\left(\cos \frac{\pi +2\pi \left(0\right)}{4}+i\sin \frac{\pi +2\pi \left(0\right)}{4}\right)=4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$

For $k=1$ ,

  $4\left(\cos \frac{\pi +2\pi \left(1\right)}{4}+i\sin \frac{\pi +2\pi \left(1\right)}{4}\right)=4\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$

For $k=2$ ,

  $4\left(\cos \frac{\pi +2\pi \left(2\right)}{4}+i\sin \frac{\pi +2\pi \left(2\right)}{4}\right)=4\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$

For $k=3$ ,

  $4\left(\cos \frac{\pi +2\pi \left(3\right)}{4}+i\sin \frac{\pi +2\pi \left(3\right)}{4}\right)=4\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$

So, the roots are $4\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)$ , $4\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)$ , $4\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)$ and $4\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)$ .

Following is the graph representing the roots:

  

Figure 1