Chapter 8.3, Problem 97E

(a)

To determine

To show: $1,w,w^2,w^3,\mathrm{...},w^{n-1}$ are the $n$ distinct nth roots of $1$ .

Answer to Problem 97E

Hence, $1,\omega ,{\omega }^2,\mathrm{...},{\omega }^{n-1}$ are the $ndistinct\text{ n}throotsof1.$

Explanation of Solution

Given:

  $\omega =\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}$

Calculation:

Consider the following formula for positive integer's n:

  $\omega =\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}$

To show $1,\omega ,{\omega }^2,\mathrm{...},{\omega }^{n-1}$ are the $n$ distinct $nth$ distinct roots of 1.

Recall the definition of a root of a complex number $z$ :

If $\omega$ satisfies ${\omega }^n=z$ for some positive integer $n$ , then ${\omega }^n$ is the nth root of $z$ .

Start with ${\omega }^0=1$ .

Clearly $1=1$

For $\omega =cos2\pi +isin2\pi$ :

  $\begin{array}{l}cos2\pi +isin2\pi =1+0\\ =1\end{array}$

For ${\omega }^2={\left(\cos \frac{2\pi }{2}+i\sin \frac{2\pi }{2}\right)}^2$ :

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

For ${\omega }^3={\left(\cos \frac{2\pi }{2}+i\sin \frac{2\pi }{2}\right)}^3$ ,

Use of the additive law of exponents:

  $x^{a+b}=x^a\cdot b^b$

Thus,

  $\begin{array}{l}{\omega }^3={\omega }^2\cdot \omega \\ =1\cdot 1\\ =1\end{array}$

Note that although

  $\begin{array}{l}1=\omega \\ ={\omega }^2\\ ={\omega }^3\end{array}$

Here, each is the distinct 0th root, 1st root, and 2nd (square), and 3rd (cube) root respectively. This means that all the other 0th, 1st, 2nd (square), and 3rd (cube) roots are the same as $1,\omega ,{\omega }^{2}and{\omega }^3$ .

Since it seems that the formula works for lower powers of $\omega$ , check the general case ${\omega }^{n-1}$ by assuming that ${\omega }^{n-2}$ is the distinct $\left(n-2\right)th$ root.

That is, assume that ${\omega }^{n-2}=1$ .

Check ${\omega }^{n-1}$ by rewriting the power in terms of ${\omega }^{n-2}$ .

This will make use of the additive law of exponents:

  $x^{a+b}=x^a\cdot x^b$

Thus,

  ${\omega }^{a+b}={\omega }^a\cdot {\omega }^b$

Substitute ${\omega }^{n-2}=1and\omega =1$ .

Because assumed ${\omega }^{n-2}=1$ after checking $\omega =1$ :

  $\begin{array}{l}{\omega }^{n-1}={\omega }^{n-1}\cdot \omega \\ =1\cdot 1\\ =1\end{array}$

Hence, $1,\omega ,{\omega }^2,\mathrm{...},{\omega }^{n-1}$ are the $ndistinct\text{ n}throotsof1.$

Conclusion:

Hence, $1,\omega ,{\omega }^2,\mathrm{...},{\omega }^{n-1}$ are the $ndistinct\text{ n}throotsof1.$

(b)

To determine

To show: the $n$ distinct nth roots of $z$ are $s,sw,s{w}^2,s{w}^3,\mathrm{...},s{w}^{n-1}$

Answer to Problem 97E

Hence, $s,s\omega ,s{\omega }^2,s{\omega }^3,\mathrm{...},s{\omega }^{n-1}$ are the $n$ distinct $nth$ roots of $z$ .

Explanation of Solution

Given:

  $s^n=z$

Calculation:

Consider a complex number $z\cancel{=}0and{s}^n=z$ ,

To show that the $n$ distinct $nth$ roots of $zares$ , $s\omega ,s{\omega }^2,s{\omega }^3,\mathrm{...},s{\omega }^{n-1}$

First recall the definition of a root of a complex number $z$ :

If $\omega$ satisfies ${\omega }^n=z$ for some positive integer $n$ , then ${\omega }^n$ is the $nth$ root of $z$ .

This means that for any positive integer $n,s^n=z$ is an $nth$ root of $z$ .

To find the distinct $nth$ roots, recall from part (a) which showed that $1,\omega ,{\omega }^2,\mathrm{...},{\omega }^{n-1}$ are the n distinct $nth$ roots of 1.

This means that:

  $\begin{array}{l}1=\omega \\ ={\omega }^2\\ ={\omega }^3\\ =\mathrm{...}\\ ={\omega }^{n-1}\end{array}$

These are $n$ distinct $nth$ roots.

Use this result along with the given formula to find $n$ distinct $nth$ roots $z$ .

For $n=1$ , $s$ is clearly a root from the equation $s^n=z$ :

  $s=z$

For $n=2,use\omega \text{=1}$ :

  $\begin{array}{l}s=z\\ s\cdot 1=z\\ s\cdot \omega =z\end{array}$

In general, for $k=0,1,\mathrm{...}\left(n-1\right)$ ,

  $\begin{array}{l}s=z\\ s\cdot 1=z\\ s\cdot {\omega }^k=z\end{array}$

Which are all distinct.

Hence, $s,s\omega ,s{\omega }^2,s{\omega }^3,\mathrm{...},s{\omega }^{n-1}$ are the $n$ distinct $nth$ roots of $z$ .

Conclusion:

Hence, $s,s\omega ,s{\omega }^2,s{\omega }^3,\mathrm{...},s{\omega }^{n-1}$ are the $n$ distinct $nth$ roots of $z$ .