Chapter 8.3, Problem 98E

To determine

To find: value of the three cube roots of 1 and explain the reason why the sum of the $nth$ roots of 1 for any $n$ is $n$ .

Answer to Problem 98E

The sum of the cube roots is $3$ , the sum of the fourth roots is $4$ , the sum of the fifth roots is $5$ , the sum of the sixth roots is $6$ .

The sum of the $nth$ roots of 1 for any $n$ is $n$ .

Explanation of Solution

Calculation:

The formula $\omega =\cos \frac{2\pi }{n}+i\sin \frac{2\pi }{n}$ gives us the $n$ distinct nth roots of 1, which are

  $1,\omega ,\mathrm{...},{\omega }^{n-1}$ . Using this formula, find all the cube roots of 1 and add them thogether. Then do the same for the fourth, fifth, and sixth roots of 1.

First find the values of the cube roots, which are given by $1,\omega ,and{\omega }^2$ . Write the expression for $\omega and{\omega }^2$ :

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

So the sum is:

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

For the fourth roots, there are $1,\omega ,{\omega }^2,and{\omega }^3$ . Find the expression for ${\omega }^3$ . Make use of the additive property of exponents, which states:

  $x^{a+b}=x^a\cdot x^b$

Thus for ${\omega }^4$

  $\begin{array}{l}{\omega }^4={\omega }^3\cdot \omega \\ =1\cdot 1\\ =1\end{array}$

So the sum is

  $\begin{array}{l}1+\omega +{\omega }^2+{\omega }^3+{\omega }^4=1+1+1+1+1\\ =5\end{array}$

For the fifth roots, there are $1,\omega ,{\omega }^2,{\omega }^3\text{, }{\omega }^4\text{, }and{\omega }^5$ . Find the expression for ${\omega }^5$ . Make use of the additive property of exponents, which states:

  $x^{a+b}=x^a\cdot x^b$

Thus for ${\omega }^5$

  $\begin{array}{l}{\omega }^5={\omega }^4\cdot \omega \\ =1\cdot 1\\ =1\end{array}$

So the sum is

  $\begin{array}{l}1+\omega +{\omega }^2+{\omega }^3+{\omega }^4+{\omega }^5=1+1+1+1+1+1\\ =6\end{array}$

In summary, the sum of the cube roots is $3$ , the sum of the fourth roots is $4$ , the sum of the fifth roots is $5$ , the sum of the sixth roots is $6$ . Hence, the sum of the $nth$ roots of 1 for any $n$ is $n$ .

Conclusion:

The sum of the cube roots is $3$ , the sum of the fourth roots is $4$ , the sum of the fifth roots is $5$ , the sum of the sixth roots is $6$ .

Hence, the sum of the $nth$ roots of 1 for any $n$ is $n$ .