Chapter 8.3, Problem 94E

To determine

To solve: the equation.

Answer to Problem 94E

  $\pm 1,\pm \frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

These six complex numbers are the solutions to the original equation.

Explanation of Solution

Given:

  $z^6-1=0$

Calculation:

The equation can be written as $z^6=1$ . Therefore, the solutions to the equation are the sixth roots of 1.

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

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

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

Let's begin by using two formulas,

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

in order to write our complex number in polar form. 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 $\theta$ .

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

Choose this value of theta because our complex number point straight along the positive real axis. Using these values of $rand\theta$ and write our complex number in polar form.

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right),\\ z=1\left(\cos \left(0\right)+i\sin \left(0\right)\right)\end{array}$

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 $n=6$ (because there is a cube root) and the complex number

  $z=1\left(\cos \left(0\right)+i\sin \left(0\right)\right),$

Thus,

  $\begin{array}{l}{\omega }_k=1^{\frac{1}{6}}\left[\cos \left(\frac{0+2\pi k}{6}\right)+i\sin \left(\frac{0+2\pi k}{6}\right)\right]\\ {\omega }_k=1\left[\cos \left(\frac{2\pi k}{6}\right)+i\sin \left(\frac{2\pi k}{6}\right)\right]\\ {\omega }_k=\cos \left(\frac{\pi k}{3}\right)+i\sin \left(\frac{\pi k}{3}\right)\end{array}$

Now, let $k$ take on the integer values from $0ton-1$ . Since $n=6$ , go from $0to\text{ 5 }\left(6-1=5\right)$ . Let's start with $k=0$ .

  $\begin{array}{l}{\omega }_0=\cos \left(\frac{\pi \left(0\right)}{3}\right)+i\sin \left(\frac{\pi \left(0\right)}{3}\right)\\ {\omega }_0=\cos \left(0\right)+i\sin \left(0\right)\\ {\omega }_0=1+0\\ \boxed{{\omega }_0=1}\end{array}$

Continue by finding all the roots.

  $\begin{array}{l}{\omega }_1=\cos \left(\frac{\pi \left(1\right)}{3}\right)+i\sin \left(\frac{\pi \left(1\right)}{3}\right)\\ {\omega }_1=\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\\ \boxed{{\omega }_1=\frac{1}{2}+\frac{\sqrt{3}}{2}i}\\ {\omega }_2=\cos \left(\frac{\pi \left(2\right)}{3}\right)+i\sin \left(\frac{\pi \left(2\right)}{3}\right)\\ {\omega }_2=\cos \left(\frac{2\pi }{3}\right)+i\sin \left(\frac{2\pi }{3}\right)\\ \boxed{{\omega }_2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i}\\ {\omega }_3=\cos \left(\frac{\pi \left(3\right)}{3}\right)+i\sin \left(\frac{\pi \left(3\right)}{3}\right)\\ {\omega }_3=\cos \left(\pi \right)+i\sin \left(\pi \right)\\ {\omega }_3=-1+0i\\ \boxed{{\omega }_3=-1}\\ {\omega }_4=\cos \left(\frac{\pi \left(4\right)}{3}\right)+i\sin \left(\frac{\pi \left(4\right)}{3}\right)\\ {\omega }_4=\cos \left(\frac{4\pi }{3}\right)+i\sin \left(\frac{4\pi }{3}\right)\\ \boxed{{\omega }_4=-\frac{1}{2}-\frac{\sqrt{3}}{2}i}\\ {\omega }_5=\cos \left(\frac{\pi \left(5\right)}{3}\right)+i\sin \left(\frac{\pi \left(5\right)}{3}\right)\\ {\omega }_5=\cos \left(\frac{5\pi }{3}\right)+i\sin \left(\frac{5\pi }{3}\right)\\ \boxed{{\omega }_5=\frac{1}{2}-\frac{\sqrt{3}}{2}i}\end{array}$

To sum up these results in a more compact form, the six roots of 1 are

  $\boxed{\pm 1,\pm \frac{1}{2}\pm \frac{\sqrt{3}}{2}i}$

These six complex numbers are the solutions to the original equation.

Conclusion:

  $\pm 1,\pm \frac{1}{2}\pm \frac{\sqrt{3}}{2}i$

These six complex numbers are the solutions to the original equation.