Chapter 8.3, Problem 99E

To determine

To find: the product of the three cube roots of 1 and product of the $nth$ roots of 1 for any $n$

Answer to Problem 99E

The product of the cube roots is $1$ , the product of the fourth roots is $-1$ , the product of the fifth roots is 1, the product of the sixth roots is $-1$ , and the product of the eighth roots is $-1$ Hence, the product of the $nth$ roots of 1 for any $n$ is ${\left(-1\right)}^{n-1}$

Explanation of Solution

Calculation:

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

  $1,\omega ,\mathrm{...},{\omega }^{n-1}$

Using this formula, find all the cube roots of 1 and find their product. Then do the same for the fourth, fifth, sixth, and eighth roots of 1.

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

Write the expression for $\omega$ :

  $\begin{array}{l}\omega =\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\\ =\frac{1}{2}+\frac{i\sqrt{3}}{2}\\ =\frac{-1+i\sqrt{3}}{2}\end{array}$

This means that for ${\omega }^2$ ,

  $\begin{array}{l}{\omega }^2={\left(\frac{-1+i3}{2}\right)}^2\\ =\frac{1-2i\sqrt{3}+i^2{\sqrt{3}}^2}{4}\\ =\frac{-1-i\sqrt{3}}{2}\end{array}$

Now,

  $\begin{array}{l}1\cdot \omega \cdot {\omega }^2=\frac{-1-i\sqrt{3}}{2}\cdot \frac{-1+i\sqrt{3}}{2}\\ ={\frac{-1-i^2\sqrt{3}}{2}}^2\\ =\frac{-1+3}{2}\\ =1\end{array}$

For the fourth roots, $1,\omega ,{\omega }^2,and{\omega }^3$ . Find the expression for $\omega$ :

  $\omega =\cos \frac{2\pi }{4}+i\sin \frac{2\pi }{4}$

So the product is:

  $\begin{array}{l}1\cdot \omega \cdot {\omega }^2\cdot {\omega }^3={\omega }^6\\ ={\left(\cos \frac{2\pi }{4}+i\sin \frac{2\pi }{4}\right)}^6\\ =\cos \left(6\cdot \frac{2\pi }{4}\right)+i\sin \left(6\cdot \frac{2\pi }{4}\right)\\ =\cos \left(3\pi \right)+i\sin \left(3\pi \right)\end{array}$

After simplification, the product is $-1$ .

For the fifth roots, $1,\omega ,{\omega }^2,{\omega }^{3}and{\omega }^4$ . Find the expression for $\omega$ :

  $\omega =\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}$

So the product is:

  $\begin{array}{l}1\cdot \omega \cdot {\omega }^2\cdot {\omega }^3\cdot {\omega }^4={\omega }^{10}\\ ={\left(\cos \frac{2\pi }{5}+i\sin \frac{2\pi }{5}\right)}^{10}\\ =\cos \left(10\cdot \frac{2\pi }{5}\right)+i\sin \left(10\cdot \frac{2\pi }{5}\right)\\ =\cos \left(4\pi \right)+i\sin \left(4\pi \right)\end{array}$

After simplification, the product is 1.

For the sixth roots, $1,\omega ,{\omega }^2,{\omega }^3\text{, }{\omega }^4\text{, }and{\omega }^5$ . Find the expression for $\omega$

  $\omega =\cos \frac{2\pi }{6}+i\sin \frac{2\pi }{6}$

So the product is:

  $\begin{array}{l}1\cdot \omega \cdot {\omega }^2\cdot {\omega }^3\cdot {\omega }^4\cdot {\omega }^5={\omega }^{15}\\ ={\left(\cos \frac{2\pi }{6}+i\sin \frac{2\pi }{6}\right)}^{15}\\ =\cos \left(15\cdot \frac{2\pi }{6}\right)+i\sin \left(15\cdot \frac{2\pi }{6}\right)\\ =\cos \left(5\pi \right)+i\sin \left(5\pi \right)\end{array}$

After simplification, the product is $-1$ .

For the eighth roots, $1,\omega ,{\omega }^2,{\omega }^3\text{, }{\omega }^4\text{, }{\omega }^5\text{, }{\omega }^6\text{, }{\omega }^7$ . Find the expression for $\omega$ :

  $\omega =\cos \frac{2\pi }{8}+i\sin \frac{2\pi }{8}$

So the product is:

  $\begin{array}{l}1\cdot \omega \cdot {\omega }^2\cdot {\omega }^3\cdot {\omega }^4\cdot {\omega }^5\cdot {\omega }^6\cdot {\omega }^7={\omega }^{28}\\ ={\left(\cos \frac{2\pi }{8}+i\sin \frac{2\pi }{8}\right)}^{28}\\ =\cos \left(28\cdot \frac{2\pi }{8}\right)+i\sin \left(28\cdot \frac{2\pi }{8}\right)\\ =\cos \left(7\pi \right)+i\sin \left(7\pi \right)\end{array}$

After simplification, the product is $-1$ .

In summary, the product of the cube roots is $1$ , the product of the fourth roots is $-1$ , the product of the fifth roots is 1, the product of the sixth roots is $-1$ , and the product of the eighth roots is $-1$ Hence, the product of the $nth$ roots of 1 for any $n$ is ${\left(-1\right)}^{n-1}$ .

Conclusion:

The product of the cube roots is $1$ , the product of the fourth roots is $-1$ , the product of the fifth roots is 1, the product of the sixth roots is $-1$ , and the product of the eighth roots is $-1$ Hence, the product of the $nth$ roots of 1 for any $n$ is ${\left(-1\right)}^{n-1}$