Chapter 8.3, Problem 91E

To determine

To solve : The given equation

Answer to Problem 91E

Here, $\pm \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , these four complex numbers are the solutions to the original equations.

Explanation of Solution

Given: $z^4+1=0$

Calculation:

This equation can be written as $z^4=-1$ . Therefore, the solutions to the equations are the fourth roots of $-1$ .

Use the Rule of $nth$Roots of Complex Numbers. This rule states:

If $z=r\left(\cos \theta +i\sin \theta \right)$ and $n$ is a positive integer, then $z$ has the $n$ distinct $nth$roots

  $\begin{array}{l}{\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]\\ fork=0,1,2,\dots \left(n-1\right).\end{array}$

Let’s begin by using two formulas,

  $r=\left|z\right|=\sqrt{a^2+b^2}$and $\tan \theta =\frac{b}{a},$

in order to write our complex number in polar from. Our complex number is

  $-1=-1+0i$ .

The radius/modulus is

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

Now, let’s find $\theta$ .

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan \theta =\frac{0}{-1}\\ \tan \theta =0\\ \theta =\pi \end{array}$

choose this value of theta because our complex number points straight along the negative real axis. Using these values of $r$ and $\theta$ , write our complex number in polar form.

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right),\\ z=1\left(\cos \left(\pi \right)+i\sin \left(\pi \right)\right)\end{array}$

Now, apply Rule of $nth$ Roots of complex Numbers to this $z$ .

  ${\omega }_k=r^{\frac{1}{n}}\left[\cos \left(\frac{\theta +2\pi k}{n}\right)+i\sin \left(\frac{\theta +2\pi k}{n}\right)\right]$

Applying this formula with $n=4$ (because there is a fourth root) and the complex number

  $z=1\left(\cos \left(\pi \right)+i\sin \left(\pi \right)\right)$

Thus,

  $\begin{array}{l}{\omega }_k=1^{\frac{1}{4}}\left[\cos \left(\frac{\pi +2\pi k}{4}\right)+i\sin \left(\frac{\pi +2\pi k}{4}\right)\right]\\ {\omega }_k=1\left[\cos \left(\frac{\pi }{4}+\frac{2\pi k}{4}\right)+i\sin \left(\frac{\pi }{4}+\frac{2\pi k}{4}\right)\right]\\ {\omega }_k=\cos \left(\frac{\pi }{4}+\frac{\pi k}{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{\pi k}{2}\right)\end{array}$

Now, let $k$ take on the integer values form $0$ to $n-1$ . Since $n=4$ , go from $0to3\left(4-1=3\right)$ . Let’s start with $k=0$ .

  $\begin{array}{l}{\omega }_0=\cos \left(\frac{\pi }{4}+\frac{\pi \left(0\right)}{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{\pi \left(0\right)}{2}\right)\\ {\omega }_0=\cos \left(\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{4}\right)\\ {\omega }_0=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

Continue by finding all of the remaining roots.

  $\begin{array}{l}{\omega }_1=\cos \left(\frac{\pi }{4}+\frac{\pi \left(1\right)}{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{\pi \left(1\right)}{2}\right)\\ {\omega }_1=\cos \left(\frac{\pi }{4}+\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{\pi }{2}\right)\\ {\omega }_1=\cos \left(\frac{3\pi }{4}\right)+i\sin \left(\frac{3\pi }{4}\right)\\ {\omega }_1=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

  $\begin{array}{l}{\omega }_2=\cos \left(\frac{\pi }{4}+\frac{\pi \left(2\right)}{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{\pi \left(2\right)}{2}\right)\\ {\omega }_2=\cos \left(\frac{\pi }{4}+\pi \right)+i\sin \left(\frac{\pi }{4}+\pi \right)\\ {\omega }_2=\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\\ {\omega }_2=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

  $\begin{array}{l}{\omega }_3=\cos \left(\frac{\pi }{4}+\frac{\pi \left(3\right)}{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{\pi \left(3\right)}{2}\right)\\ {\omega }_3=\cos \left(\frac{\pi }{4}+\frac{3\pi }{2}\right)+i\sin \left(\frac{\pi }{4}+\frac{3\pi }{2}\right)\\ {\omega }_3=\cos \left(\frac{7\pi }{4}\right)+i\sin \left(\frac{7\pi }{4}\right)\\ {\omega }_3=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i\end{array}$

The fourth roots of $-1$ are, in a compact form representing four combinations of the real and imaginary terms,

  $\boxed{\pm \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i}$

These four complex numbers are the solutions to the original equations.

Conclusion:

Therefore, $\pm \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ , these four complex numbers are the solutions to the original equations.