Chapter 6, Problem 133RE

To determine

To Find: All solutions of the equation and represent the solutions graphically

Answer to Problem 133RE

  $\begin{array}{l}\cos 0+i\sin 0=1\\ \cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i\end{array}$

  $\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

  $\begin{array}{l}\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i\\ \cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}=-i\end{array}$

Explanation of Solution

Given information:

  $x^5+x^3-x^2-1=0$

Formula used:

  $\begin{array}{l}{z}_k=\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)\\ k=0,1,2,\mathrm{3......}n-1\end{array}$

Calculation:

  $x^5+x^3-x^2-1=0$

By factorizing we have:

  $\begin{array}{l}{x}^3\left(x^2+1\right)-\left(x^2+1\right)=0\\ \left(x^2+1\right)\left(x^3-1\right)=0\end{array}$

In $\left(x^3-1\right)$

Let $x^3=z$

  $\begin{array}{l}z=1=a+bi\\ a=1\\ b=0\end{array}$

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

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

As the complex number belongs to 1^st quadrant.

Therefore, $\theta =0$

  $z_k=\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

Now putting value of $n=3,r=1,\theta =0$ in the equation

  $\begin{array}{l}k=0:\cos 0+i\sin 0=\boxed{1}\\ \\ k=1:\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}=\boxed{-\frac{1}{2}+\frac{\sqrt{3}}{2}i}\end{array}$

  $\begin{array}{l}\\ k=2:\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3}=\boxed{-\frac{1}{2}-\frac{\sqrt{3}}{2}i}\end{array}$

In $\left(x^2+1\right)$

Let $x^2=z$

  $\begin{array}{l}z=-1=a+bi\\ a=-1\\ b=0\end{array}$

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

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

As the complex number belongs to 2^nd quadrant.

Therefore, $\theta =\pi$

  $z_k=\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$

Now putting value of $n=2,r=1,\theta =\pi$ in the equation

  $\begin{array}{l}\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=\boxed{i}\\ \cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}=\boxed{-i}\end{array}$