Chapter 6, Problem 132RE

To determine

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

Answer to Problem 132RE

  $\begin{array}{l}2\sqrt{2}\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)=2.613+1.082i\\ 2\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)=-1.082+2.613i\end{array}$

  $\begin{array}{l}2\sqrt{2}\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=-2.613-1.082i\\ 2\sqrt{2}\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=1.082-2.613i\end{array}$

Explanation of Solution

Given information:

  $x^4-64i=0$

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:

  $\begin{array}{l}{x}^4-64i=0\\ x^4=64i\end{array}$

Let $x^4=z$

  $\begin{array}{l}z=-64i=a+bi\\ a=0\\ b=-64\end{array}$

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{(0)}^2+{\left(-64\right)}^2}\\ r=64\end{array}$

  $\begin{array}{l}\tan \theta =\frac{64}{0}\\ \theta ={\tan }^{-1}\infty \\ \theta =\frac{\pi }{2}\end{array}$

As the complex number belongs to 1^st quadrant.

Therefore, $\theta =\frac{\pi }{2}$

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

Now putting value of $n=4,r=64,\theta =\frac{\pi }{2}$ in the equation

  $\begin{array}{l}k=0:2\sqrt{2}\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)=\boxed{2.613+1.082i}\\ k=1:2\sqrt{2}\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)=\boxed{-1.082+2.613i}\end{array}$

  $\begin{array}{l}k=2:2\sqrt{2}\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=\boxed{-2.613-1.082i}\\ k=3:2\sqrt{2}\left(\cos \frac{9\pi }{8}+i\sin \frac{9\pi }{8}\right)=\boxed{1.082-2.613i}\end{array}$