Chapter 6.6, Problem 82E

Solution

To find: All solutions of the given equation.

The solutions of the given equation are $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , $-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ and $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ .

Given information:

The equation is $x^4+1=0$ .

Formula used:

The polar form of a complex number $z=a+bi$ is given by,

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

Here, $r=\sqrt{a^2+b^2}$ and $\tan \theta =\frac{b}{a}$ .

The $n\text{th}$ roots of a complex number $z=r\left(\cos \theta +i\sin \theta \right)$ which are $n$ distinct complex numbers are given by,

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

Here, $k=0,1,2,\mathrm{...},n-1$

Calculation:

Solve for $x$ by subtracting $1$ to both sides.

  $\begin{array}{c}{x}^4+1-1=0-1\\ x^4=-1\\ x=\sqrt[4]{-1}\end{array}$

In polar form $-1$ can be written as follows.

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

From the above complex number,

  $\begin{array}{c}r=1\\ \theta =0\\ n=4\end{array}$

Use $k=0,1,2\text{and}3$ to determine the cube roots.

Substitute $1$ for $r$ , $\pi$ for $\theta$ , $4$ for $n$ , and $0$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find first root.

  $\begin{array}{c}{z}_1=\sqrt[4]{1}\left(\cos \frac{\pi +2\pi \cdot 0}{4}+i\sin \frac{\pi +2\pi \cdot 0}{4}\right)\\ =1\left(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}\right)\\ =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

Substitute $1$ for $r$ , $\pi$ for $\theta$ , $4$ for $n$ , and $1$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find second root.

  $\begin{array}{c}{z}_2=\sqrt[4]{1}\left(\cos \frac{\pi +2\pi \cdot 1}{4}+i\sin \frac{\pi +2\pi \cdot 1}{4}\right)\\ =1\left(\cos \frac{3\pi }{4}+i\sin \frac{3\pi }{4}\right)\\ =-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

Substitute $1$ for $r$ , $\pi$ for $\theta$ , $4$ for $n$ , and $2$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find third root.

  $\begin{array}{c}{z}_3=\sqrt[4]{1}\left(\cos \frac{\pi +2\pi \cdot 2}{4}+i\sin \frac{\pi +2\pi \cdot 2}{4}\right)\\ =1\left(\cos \frac{5\pi }{4}+i\sin \frac{5\pi }{4}\right)\\ =-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\end{array}$

Substitute $1$ for $r$ , $\pi$ for $\theta$ , $4$ for $n$ , and $3$ for $k$ in expression $\sqrt[n]{r}\left(\cos \frac{\theta +2\pi k}{n}+i\sin \frac{\theta +2\pi k}{n}\right)$ to find fourth root.

  $\begin{array}{c}{z}_4=\sqrt[4]{1}\left(\cos \frac{\pi +2\pi \cdot 3}{4}+i\sin \frac{\pi +2\pi \cdot 3}{4}\right)\\ =1\left(\cos \frac{7\pi }{4}+i\sin \frac{7\pi }{4}\right)\\ =\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i\end{array}$

Therefore, the solutions of the given equation are $\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , $-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ and $\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$ .