Chapter 6.6, Problem 87E

To determine

All solutions of the equationand represent the solutions graphically.

Answer to Problem 87E

The solutions are $x=0.38+0.92i,x=-0.92+0.38i,x=-0.38-0.92i\text{ and }x=0.92-0.38i$

Explanation of Solution

Given info:

  $x^4+i=0$

Formula used:

The trigonometric form of the complex number $z=a+ib$ is

  $z=r\left(\cos \theta +i\sin \theta \right)$

Where $r=\sqrt{a^2+b^2}$ and $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

For a positive integer n the complex number $z=r\left(\cos \theta +i\sin \theta \right)$ has exactly n distinct nth roots given by

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

where k = 0, 1, 2, . . . , ,n -1.

  $z^n={\left(r\left(\cos \theta +i\sin \theta \right)\right)}^n=r^n\left(\cos n\theta +i\sin n\theta \right)$

Calculation:

We have

  $\begin{array}{l}{x}^4+i=0\\ x^4=-i\\ x=\sqrt[4]{-i}\\ a=0,b=-1\end{array}$

  $\begin{array}{l}r=\sqrt{a^2+b^2}=\sqrt{0^2+{\left(-1\right)}^2}=1\\ \theta ={\tan }^{-1}\left(\frac{-1}{0}\right)=\frac{3\pi }{2}\end{array}$

  $\begin{array}{l}-i=\cos \frac{3\pi }{2}+i\sin \frac{3\pi }{2}\\ z_k=\sqrt[n]{r}\left(\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right)\end{array}$

  $n=4,r=1,\theta =\frac{3\pi }{2}$

  $z_0=\sqrt[4]{1}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 0}{4}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 0}{4}\right)\right)$

  $z_0=\cos \frac{3\pi }{8}+i\sin \frac{3\pi }{8}=0.38+0.92i$

  $z_1=\sqrt[4]{1}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 1}{4}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 1}{4}\right)\right)$

  $z_1=\cos \frac{7\pi }{8}+i\sin \frac{7\pi }{8}=-0.92+0.38i$

  $z_2=\sqrt[4]{1}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 2}{4}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 2}{4}\right)\right)$

  $z_2=\cos \frac{11\pi }{8}+i\sin \frac{11\pi }{8}=-0.38-0.92i$

  $z_3=\sqrt[4]{1}\left(\cos \left(\frac{\frac{3\pi }{2}+2\pi \times 3}{4}\right)+i\sin \left(\frac{\frac{3\pi }{2}+2\pi \times 3}{4}\right)\right)$

  $z_3=\cos \frac{15\pi }{8}+i\sin \frac{15\pi }{8}=0.92-0.38i$

The solutions are $x=0.38+0.92i,x=-0.92+0.38i,x=-0.38-0.92i\text{ and }x=0.92-0.38i$

Graphical representation

  

Conclusion:

Thus,the solutions are $x=0.38+0.92i,x=-0.92+0.38i,x=-0.38-0.92i\text{ and }x=0.92-0.38i$ .