Chapter 8.3, Problem 95E

To determine

To solve: the equation.

Answer to Problem 95E

  ${\omega }_0=2^{\frac{1}{6}}\left[\cos \left(\frac{5\pi }{12}\right)+i\sin \left(\frac{5\pi }{12}\right)\right]$

  ${\omega }_1=2^{\frac{1}{6}}\left[\cos \left(\frac{13\pi }{12}\right)+i\sin \left(\frac{13\pi }{12}\right)\right]$

  ${\omega }_2=2^{\frac{1}{6}}\left[\cos \left(\frac{21\pi }{12}\right)+i\sin \left(\frac{21\pi }{12}\right)\right]$

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

Explanation of Solution

Given:

  $z^3+1=-i$

Calculation:

The equation can be rearranged and written as $z^3=-1-i$ . Therefore, the solutions to the equation are the cube roots of the complex number $-1-i$ .

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 $-1-1i$ . The radius/modulus is

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

Now, let's find $\theta$ .

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan \theta =\frac{-1}{-1}\\ \theta ={\tan }^{-1}\left(1\right)\\ \theta =\frac{5\pi }{4}\end{array}$

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

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right),\\ z=\sqrt{2}\left(\cos \left(\frac{5\pi }{4}\right)+i\sin \left(\frac{5\pi }{4}\right)\right)\end{array}$

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

  $z=\sqrt{2}\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)$

We get

  $\begin{array}{l}{\omega }_k={\sqrt{2}}^{\frac{1}{3}}\left[\cos \left(\frac{\frac{5\pi }{4}+2\pi k}{3}\right)+i\sin \left(\frac{\frac{5\pi }{4}+2\pi k}{3}\right)\right]\\ {\omega }_k=2^{\frac{1}{6}}\left[\cos \left(\frac{5\pi }{12}+\frac{2\pi k}{3}\right)+i\sin \left(\frac{5\pi }{12}+\frac{2\pi k}{3}\right)\right]\end{array}$

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 $k=0$ .

The cube roots of $-1-i$ are

  $\begin{array}{l}\boxed{{\omega }_0=2^{\frac{1}{6}}\left[\cos \left(\frac{5\pi }{12}\right)+i\sin \left(\frac{5\pi }{12}\right)\right]}\\ \boxed{{\omega }_1=2^{\frac{1}{6}}\left[\cos \left(\frac{13\pi }{12}\right)+i\sin \left(\frac{13\pi }{12}\right)\right]}\\ \boxed{{\omega }_2=2^{\frac{1}{6}}\left[\cos \left(\frac{21\pi }{12}\right)+i\sin \left(\frac{21\pi }{12}\right)\right]}\end{array}$

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

Conclusion:

  ${\omega }_0=2^{\frac{1}{6}}\left[\cos \left(\frac{5\pi }{12}\right)+i\sin \left(\frac{5\pi }{12}\right)\right]$

  ${\omega }_1=2^{\frac{1}{6}}\left[\cos \left(\frac{13\pi }{12}\right)+i\sin \left(\frac{13\pi }{12}\right)\right]$

  ${\omega }_2=2^{\frac{1}{6}}\left[\cos \left(\frac{21\pi }{12}\right)+i\sin \left(\frac{21\pi }{12}\right)\right]$

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