Chapter 6, Problem 40CLT

To determine

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

Answer to Problem 40CLT

  $\begin{array}{l}3\left(\cos \frac{\pi }{5}+i\sin \frac{\pi }{5}\right)=2.43+1.76i\\ 3\left(\cos \frac{3\pi }{5}+i\sin \frac{3\pi }{5}\right)=-0.93+2.43i\end{array}$

  $\begin{array}{l}3\left(\cos \pi +i\sin \pi \right)=-3\\ 3\left(\cos \frac{7\pi }{5}+i\sin \frac{7\pi }{5}\right)=-0.93-2.86i\\ 3\left(\cos \frac{9\pi }{5}+i\sin \frac{9\pi }{5}\right)=2.43-1.76i\end{array}$

Explanation of Solution

Given information:

  $x^5+243=0$

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:

  $\begin{array}{l}{x}^5+243=0\\ x^5=-243\end{array}$

Let $x^5=z$

  $\begin{array}{l}z=-243=a+bi\\ a=-243\\ b=0\end{array}$

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{(-243)}^2+0^2}\\ r=243\end{array}$

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan \theta =\frac{0}{-243}\\ \tan \theta =0\end{array}$

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 $n=5,r=243,\theta =\pi$ in the equation

  $\begin{array}{l}k=0:\sqrt[5]{243}\left(\cos \frac{\pi +2\pi \left(0\right)}{5}+i\sin \frac{\pi +2\pi \left(0\right)}{5}\right)\\ 3\left(\cos \frac{\pi }{5}+i\sin \frac{\pi }{5}\right)=\boxed{2.43+1.76i}\\ \\ k=1:\sqrt[5]{243}\left(\cos \frac{\pi +2\pi \left(1\right)}{56}+i\sin \frac{\pi +2\pi \left(1\right)}{5}\right)\\ 3\left(\cos \frac{3\pi }{5}+i\sin \frac{3\pi }{5}\right)=\boxed{-0.93+2.86i}\end{array}$

  $\begin{array}{l}\\ k=2:\sqrt[5]{243}\left(\cos \frac{\pi +2\pi \left(2\right)}{5}+i\sin \frac{\pi +2\pi \left(2\right)}{5}\right)\\ 3\left(\cos \pi +i\sin \pi \right)=\boxed{-3}\\ \\ k=3:\sqrt[5]{243}\left(\cos \frac{\pi +2\pi \left(3\right)}{5}+i\sin \frac{\pi +2\pi \left(3\right)}{5}\right)\\ 3\left(\cos \frac{7\pi }{5}+i\sin \frac{7\pi }{5}\right)=\boxed{-0.93-2.86i}\end{array}$

  $\begin{array}{l}k=4:\sqrt[5]{243}\left(\cos \frac{\pi +2\pi \left(4\right)}{5}+i\sin \frac{\pi +2\pi \left(4\right)}{5}\right)\\ 3\left(\cos \frac{9\pi +5}{5}+i\sin \frac{9\pi +5}{5}\right)=\boxed{2.43-1.76i}\end{array}$