Chapter 8.3, Problem 93E

To determine

To solve: the equation.

Answer to Problem 93E

  $\begin{array}{l}{\omega }_0=2\left[\cos \left(\frac{\pi }{18}\right)+i\sin \left(\frac{\pi }{18}\right)\right]\\ {\omega }_1=2\left[\cos \left(\frac{13\pi }{18}\right)+i\sin \left(\frac{13\pi }{18}\right)\right]\\ {\omega }_2=2\left[\cos \left(\frac{25\pi }{18}\right)+i\sin \left(\frac{25\pi }{18}\right)\right]\end{array}$

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

Explanation of Solution

Given:

  $z^3-4\sqrt{3}-4i=0$

Calculation:

The equation can be rearranged and written as $z^3=4\sqrt{3}+4i$ . Therefore, the solutions to the equation are the cube roots of the complex number $4\sqrt{3}+4i$ .

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 $4\sqrt{3}+4i$ . The radius/modulus is

  $\begin{array}{l}r=\sqrt{a^2+b^2}\\ r=\sqrt{{\left(4\sqrt{3}\right)}^2+{\left(4\right)}^2}\\ r=\sqrt{48+16}\\ r=\sqrt{64}\\ r=8\end{array}$

Now, let's find $\theta$ .

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan \theta =\frac{4}{4\sqrt{3}}\\ \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)\\ \theta =\frac{\pi }{6}\end{array}$

Choose this value of theta because our complex number is in Quadrant I. Using these values of $rand\theta$ , write our complex number in polar form.

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right),\\ =8\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)\end{array}$

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 there is a cube root) and the complex number

  $z=8\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right),$

Thus,

  $\begin{array}{l}{\omega }_k=8^{\frac{1}{3}}\left[\cos \left(\frac{\frac{\pi }{6}+2\pi k}{3}\right)+i\sin \left(\frac{\frac{\pi }{6}+2\pi k}{3}\right)\right]\\ {\omega }_k=2\left[\cos \left(\frac{\pi }{18}+\frac{2\pi k}{3}\right)+i\sin \left(\frac{\pi }{18}+\frac{2\pi k}{3}\right)\right]\end{array}$

Now, let $k$ take on the integer values from $0ton-1$ . Since $n=3$ , go from $0to2\left(3-1=2\right)$ . Let's start with $k=0$ .

  $\begin{array}{l}{\omega }_0=2\left[\cos \left(\frac{\pi }{18}+\frac{2\pi \left(0\right)}{3}\right)+i\sin \left(\frac{\pi }{18}+\frac{2\pi \left(0\right)}{3}\right)\right]\\ {\omega }_0=2\left[\cos \left(\frac{\pi }{18}\right)+i\sin \left(\frac{\pi }{18}\right)\right]\\ Now,dok=1.\\ {\omega }_1=2\left[\cos \left(\frac{\pi }{18}+\frac{2\pi \left(1\right)}{3}\right)+i\sin \left(\frac{\pi }{18}+\frac{2\pi \left(1\right)}{3}\right)\right]\\ {\omega }_1=2\left[\cos \left(\frac{\pi }{18}+\frac{2\pi }{3}\right)+i\sin \left(\frac{\pi }{18}+\frac{2\pi }{3}\right)\right]\\ {\omega }_1=2\left[\cos \left(\frac{13\pi }{18}\right)+i\sin \left(\frac{13\pi }{18}\right)\right]\\ Finally,dok=2\\ {\omega }_2=2\left[\cos \left(\frac{\pi }{18}+\frac{2\pi \left(2\right)}{3}\right)+i\sin \left(\frac{\pi }{18}+\frac{2\pi \left(2\right)}{3}\right)\right]\\ {\omega }_2=2\left[\cos \left(\frac{\pi }{18}+\frac{4\pi }{3}\right)+i\sin \left(\frac{\pi }{18}+\frac{4\pi }{3}\right)\right]\\ {\omega }_2=2\left[\cos \left(\frac{25\pi }{18}\right)+i\sin \left(\frac{25\pi }{18}\right)\right]\end{array}$

The cube roots of $4\sqrt{3}+4i$ are

  $\begin{array}{l}\boxed{{\omega }_0=2\left[\cos \left(\frac{\pi }{18}\right)+i\sin \left(\frac{\pi }{18}\right)\right]}\\ \boxed{{\omega }_1=2\left[\cos \left(\frac{13\pi }{18}\right)+i\sin \left(\frac{13\pi }{18}\right)\right]}\\ \boxed{{\omega }_2=2\left[\cos \left(\frac{25\pi }{18}\right)+i\sin \left(\frac{25\pi }{18}\right)\right]}\end{array}$

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

Conclusion:

  $\begin{array}{l}{\omega }_0=2\left[\cos \left(\frac{\pi }{18}\right)+i\sin \left(\frac{\pi }{18}\right)\right]\\ {\omega }_1=2\left[\cos \left(\frac{13\pi }{18}\right)+i\sin \left(\frac{13\pi }{18}\right)\right]\\ {\omega }_2=2\left[\cos \left(\frac{25\pi }{18}\right)+i\sin \left(\frac{25\pi }{18}\right)\right]\end{array}$

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