Chapter 6.6, Problem 128E

To determine

To find: All the solutions of $x^6-64i=0$

Answer to Problem 128E

The sixth roots of $x^6-64i=0$ :

  $\begin{array}{l}\Rightarrow 2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)\\ \Rightarrow 2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)\\ \Rightarrow 2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ \Rightarrow 2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\\ \Rightarrow 2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\\ \Rightarrow 2\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Explanation of Solution

Given information: The complex number $x^6-64i=0$

Formula used: $n^{th}$ roots of the complex number: $\sqrt[n]{r}\left(\cos \frac{\frac{3\pi }{2}+2\pi k}{6}+i\sin \frac{\frac{3\pi }{2}+2\pi k}{6}\right);k=0,1,2,3,\mathrm{...},n-1$

Calculation: Apply the $n^{th}$ roots formula to obtain the solutions of the given number.

  $\begin{array}{l}{x}^6-64i=0\\ x^6=64i\end{array}$

The solutions are the sixth roots of $64i=64\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$ ,

  $\begin{array}{l}\sqrt[6]{64}\left(\cos \frac{\frac{\pi }{2}+2\pi k}{6}+i\sin \frac{\frac{\pi }{2}+2\pi k}{6}\right)=2\left(\cos \frac{\pi +4\pi k}{12}+i\sin \frac{\pi +4\pi k}{12}\right)\\ \therefore 2\left(\cos \frac{\pi +4\pi \left(0\right)}{12}+i\sin \frac{\pi +4\pi \left(0\right)}{12}\right)=2\left(\cos \frac{\pi }{12}+i\sin \frac{\pi }{12}\right)\\ \therefore 2\left(\cos \frac{\pi +4\pi \left(1\right)}{12}+i\sin \frac{\pi +4\pi \left(1\right)}{12}\right)=2\left(\cos \frac{5\pi }{12}+i\sin \frac{5\pi }{12}\right)\\ \therefore 2\left(\cos \frac{\pi +4\pi \left(2\right)}{12}+i\sin \frac{\pi +4\pi \left(2\right)}{12}\right)=2\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ \therefore 2\left(\cos \frac{\pi +4\pi \left(3\right)}{12}+i\sin \frac{\pi +4\pi \left(3\right)}{12}\right)=2\left(\cos \frac{13\pi }{12}+i\sin \frac{13\pi }{12}\right)\\ \therefore 2\left(\cos \frac{\pi +4\pi \left(4\right)}{12}+i\sin \frac{\pi +4\pi \left(4\right)}{12}\right)=2\left(\cos \frac{17\pi }{12}+i\sin \frac{17\pi }{12}\right)\\ \therefore 2\left(\cos \frac{\pi +4\pi \left(5\right)}{12}+i\sin \frac{\pi +4\pi \left(5\right)}{12}\right)=2\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\end{array}$

Graph: The required graph of roots are given below,