Chapter 8.3, Problem 96E

To determine

To solve: the equation.

Answer to Problem 96E

  ${\omega }_0=1$

  ${\omega }_1=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

  ${\omega }_2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

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

Explanation of Solution

Given:

  $z^3-1=0$

Formula Used:

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

Calculation:

The equation can be written as $z^3=1$ . 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,

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

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 we know our complex number point straight along the positive real axis. Using these values of $rand\theta$ , write our complex number in polar form.

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

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 $n=3$ (because we have a cube root) and the complex number

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

Thus,

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

Now, let $k$ take on the integer values from $0ton-1$ . Since $n=3$ , go from $0to\text{ 2 }\left(3-1=2\right)$ . Let's start with $k=0$ .

  $\begin{array}{l}{\omega }_0=\cos \left(\frac{2\pi \left(0\right)}{3}\right)+i\sin \left(\frac{2\pi \left(0\right)}{3}\right)\\ {\omega }_0=\cos \left(0\right)+i\sin \left(0\right)\\ {\omega }_0=1+0\\ \boxed{{\omega }_0=1}\end{array}$

Now, let k equal 1.

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

Finally, let k equal 2.

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

The cube roots of 1 are

  $\boxed{{\omega }_0=1}$

  $\boxed{{\omega }_1=-\frac{1}{2}+\frac{\sqrt{3}}{2}i}$

  $\boxed{{\omega }_2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i}$

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

Conclusion:

  ${\omega }_0=1$

  ${\omega }_1=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$

  ${\omega }_2=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$

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