Chapter 8.3, Problem 92E

To determine

To solve: The given equation

Answer to Problem 92E

Here, the solutions to the equations are the eighth roots of $i$ . They are

  ${\omega }_0=\cos \left(\frac{\pi }{16}\right)+i\sin \left(\frac{\pi }{16}\right)$ , ${\omega }_1=\cos \left(\frac{5\pi }{16}\right)+i\sin \left(\frac{5\pi }{16}\right)$ , ${\omega }_2=\cos \left(\frac{9\pi }{16}\right)+i\sin \left(\frac{9\pi }{16}\right)$ , ${\omega }_3=\cos \left(\frac{13\pi }{16}\right)+i\sin \left(\frac{13\pi }{16}\right)$ , ${\omega }_4=\cos \left(\frac{17\pi }{16}\right)+i\sin \left(\frac{17\pi }{16}\right)$ , ${\omega }_5=\cos \left(\frac{21\pi }{16}\right)+i\sin \left(\frac{21\pi }{16}\right)$ , ${\omega }_6=\cos \left(\frac{25\pi }{16}\right)+i\sin \left(\frac{25\pi }{16}\right)$ , ${\omega }_7=\cos \left(\frac{29\pi }{16}\right)+i\sin \left(\frac{29\pi }{16}\right)$

Explanation of Solution

Given: $z^8-i=0$

Calculation:

This equation can be written as $z^8-i=0$ . Therefore, the solutions to the equations are the eighth roots of $i$ .

As discussed in this section, 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

  $i=0+1i$ .The radius/modulus is

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

Now, let’s find $\theta$ .

  $\begin{array}{l}\tan \theta =\frac{b}{a}\\ \tan \theta =\frac{1}{0}\\ \tan \theta isundefined\\ \theta =\frac{\pi }{2}\end{array}$

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.

  $\begin{array}{l}z=r\left(\cos \theta +i\sin \theta \right),\\ z=1\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right)\end{array}$

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 $n=8$ (because there is a eighth root) and the complex number

  $z=1\left(\cos \left(\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{2}\right)\right),$

Thus,

  $\begin{array}{l}{\omega }_k=1^{\frac{1}{8}}\left[\cos \left(\frac{\pi /2+2\pi k}{8}\right)+i\sin \left(\frac{\pi /2+2\pi k}{8}\right)\right]\\ {\omega }_k=1\left[\cos \left(\frac{\pi }{16}+\frac{\pi k}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi k}{4}\right)\right]\\ {\omega }_k=\cos \left(\frac{\pi }{16}+\frac{\pi k}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi k}{4}\right)\end{array}$

Now, let $k$ take on the integer values form $0$ to $n-1$ . Since $n=8$ , go from $0to7\left(8-1=7\right)$ . Let’s start with $k=0$ .

  $\begin{array}{l}{\omega }_0=\cos \left(\frac{\pi }{16}+\frac{\pi \left(0\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(0\right)}{4}\right)\\ {\omega }_0=\cos \left(\frac{\pi }{16}\right)+i\sin \left(\frac{\pi }{16}\right)\end{array}$

Continue by finding the rest of the roots.

  $\begin{array}{l}{\omega }_1=\cos \left(\frac{\pi }{16}+\frac{\pi \left(1\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(1\right)}{4}\right)\\ {\omega }_1=\cos \left(\frac{\pi }{16}+\frac{\pi }{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi }{4}\right)\\ {\omega }_1=\cos \left(\frac{5\pi }{16}\right)+i\sin \left(\frac{5\pi }{16}\right)\end{array}$

  $\begin{array}{l}{\omega }_2=\cos \left(\frac{\pi }{16}+\frac{\pi \left(2\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(2\right)}{4}\right)\\ {\omega }_2=\cos \left(\frac{\pi }{16}+\frac{\pi }{2}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi }{2}\right)\\ {\omega }_2=\cos \left(\frac{9\pi }{16}\right)+i\sin \left(\frac{9\pi }{16}\right)\end{array}$

  $\begin{array}{l}{\omega }_3=\cos \left(\frac{\pi }{16}+\frac{\pi \left(3\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(3\right)}{4}\right)\\ {\omega }_3=\cos \left(\frac{\pi }{16}+\frac{3\pi }{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{3\pi }{4}\right)\\ {\omega }_3=\cos \left(\frac{13\pi }{16}\right)+i\sin \left(\frac{13\pi }{16}\right)\end{array}$

  $\begin{array}{l}{\omega }_4=\cos \left(\frac{\pi }{16}+\frac{\pi \left(4\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(4\right)}{4}\right)\\ {\omega }_4=\cos \left(\frac{\pi }{16}+\pi \right)+i\sin \left(\frac{\pi }{16}+\pi \right)\\ {\omega }_4=\cos \left(\frac{17\pi }{16}\right)+i\sin \left(\frac{17\pi }{16}\right)\end{array}$

  $\begin{array}{l}{\omega }_5=\cos \left(\frac{\pi }{16}+\frac{\pi \left(5\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(5\right)}{4}\right)\\ {\omega }_5=\cos \left(\frac{\pi }{16}+\frac{5\pi }{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{5\pi }{4}\right)\\ {\omega }_5=\cos \left(\frac{21\pi }{16}\right)+i\sin \left(\frac{21\pi }{16}\right)\end{array}$

  $\begin{array}{l}{\omega }_6=\cos \left(\frac{\pi }{16}+\frac{\pi \left(6\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(6\right)}{4}\right)\\ {\omega }_6=\cos \left(\frac{\pi }{16}+\frac{3\pi }{2}\right)+i\sin \left(\frac{\pi }{16}+\frac{3\pi }{2}\right)\\ {\omega }_6=\cos \left(\frac{25\pi }{16}\right)+i\sin \left(\frac{25\pi }{16}\right)\end{array}$

  $\begin{array}{l}{\omega }_7=\cos \left(\frac{\pi }{16}+\frac{\pi \left(7\right)}{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{\pi \left(7\right)}{4}\right)\\ {\omega }_7=\cos \left(\frac{\pi }{16}+\frac{7\pi }{4}\right)+i\sin \left(\frac{\pi }{16}+\frac{7\pi }{4}\right)\\ {\omega }_7=\cos \left(\frac{29\pi }{16}\right)+i\sin \left(\frac{29\pi }{16}\right)\end{array}$

The eight roots of $i$ are,

  $\boxed{{\omega }_0=\cos \left(\frac{\pi }{16}\right)+i\sin \left(\frac{\pi }{16}\right)}$

  $\boxed{{\omega }_1=\cos \left(\frac{5\pi }{16}\right)+i\sin \left(\frac{5\pi }{16}\right)}$

  $\boxed{{\omega }_2=\cos \left(\frac{9\pi }{16}\right)+i\sin \left(\frac{9\pi }{16}\right)}$

  $\boxed{{\omega }_3=\cos \left(\frac{13\pi }{16}\right)+i\sin \left(\frac{13\pi }{16}\right)}$

  $\boxed{{\omega }_4=\cos \left(\frac{17\pi }{16}\right)+i\sin \left(\frac{17\pi }{16}\right)}$

  $\boxed{{\omega }_5=\cos \left(\frac{21\pi }{16}\right)+i\sin \left(\frac{21\pi }{16}\right)}$

  $\boxed{{\omega }_6=\cos \left(\frac{25\pi }{16}\right)+i\sin \left(\frac{25\pi }{16}\right)}$

  $\boxed{{\omega }_7=\cos \left(\frac{29\pi }{16}\right)+i\sin \left(\frac{29\pi }{16}\right)}$

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

Conclusion:

Therefore, the solutions to the equations are the eighth roots of $i$ . They are ${\omega }_0=\cos \left(\frac{\pi }{16}\right)+i\sin \left(\frac{\pi }{16}\right)$ , ${\omega }_1=\cos \left(\frac{5\pi }{16}\right)+i\sin \left(\frac{5\pi }{16}\right)$ , ${\omega }_2=\cos \left(\frac{9\pi }{16}\right)+i\sin \left(\frac{9\pi }{16}\right)$ , ${\omega }_3=\cos \left(\frac{13\pi }{16}\right)+i\sin \left(\frac{13\pi }{16}\right)$ , ${\omega }_4=\cos \left(\frac{17\pi }{16}\right)+i\sin \left(\frac{17\pi }{16}\right)$ , ${\omega }_5=\cos \left(\frac{21\pi }{16}\right)+i\sin \left(\frac{21\pi }{16}\right)$ , ${\omega }_6=\cos \left(\frac{25\pi }{16}\right)+i\sin \left(\frac{25\pi }{16}\right)$ , ${\omega }_7=\cos \left(\frac{29\pi }{16}\right)+i\sin \left(\frac{29\pi }{16}\right)$