# Video: 8th Roots of Unity in Cartesian Form

Which of the following is one of the 8th roots of unity in Cartesian form? [A] ((β2)/2) β ((β2)/2 π) [B] β2 β β2π [C] ((β3)/2) β (1/2 π) [D] 2β2 + 2β2π [E] (β1/2) β ((β3)/2 π)

03:20

### Video Transcript

Which of the following is one of the eighth roots of unity in Cartesian form? Is it a) root two over two minus root two over two π? b) Root two minus root two π. c) Root three over two minus a half π. d) Two root two plus two root two π. Or e) Negative one-half minus root three over two π.

We begin by recalling what we mean by the eighth roots of unity. Well, the πth roots of unity are the solutions to the equation π₯ to the πth power equals one. This equation has π-roots. And, in trigonometric form, these roots or these solutions are given by π₯ equals cos of two ππ over π plus π sin of two ππ over π for values of π between zero to π minus one.

Now, of course, in this question, weβre finding the eighth roots of unity. So weβre going to let π be equal to eight. We can then say that π₯ must be equal to cos of two ππ over eight plus π sin of two ππ over eight for values of π from zero through to π minus one. Well, thatβs of course seven. Letβs simplify two-eighths to be equal to one-quarter. And we see that our solutions given by π₯ equals cos of ππ over four plus π sin of ππ over four.

Weβre now going to list each root by letting π be equal to zero, one, two, three all the way through to seven. The first root is a result of letting π be equal to zero. So we get π₯ equals cos of zero π over four plus π sin of zero π over four. Zero π over four is of course zero. Cos of zero is equal to one. And sin of zero is equal to zero. So we find π₯ to be equal to one plus zero π. So the first solution to the equation π₯ to the eighth power equals one is one, which makes a lot of sense.

Our next root, the second root, is found by letting π be equal to one. This time, π₯ is equal to cos of one π over four plus π sin of one π over four. Now, of course, we donβt need these two ones here. And cos of π by four is root two over two. And sin of π by four is also root two over two. So we find our second solution is π₯ is equal to the square root of root two over two plus the square root of two over two π.

Letβs repeat this process for the third root, this time letting π be equal to two. And our solution is π₯ equals cos of two π by four plus π sin of two π by four. Of course, two π by four is equivalent to π by two. And cos of π by two is zero whilst sin of π by two is one. So π₯ is equal to zero plus one π, which is simply equal to π. The fourth root is found by letting π be equal to three. So we get π₯ equals cos of three π by four plus π sin of three π by four, which is equal to negative root two over two plus root two over two π.

Our final four solutions are found by letting π be equal to four, five, six, and seven. And when π is equal to four, we get π₯ equals negative one. When π is equal to five, we get π₯ equals negative root two over two minus root two over two π. When π is equal to six, we get π₯ is equal to negative π. And our eighth root and the eighth solution to the equation π₯ to the eighth power equals one is found by letting π be equal to seven. And we get π₯ equals the square root of two over two minus the square root of two over two π.

We go back to our list of solutions. And we see that the only one that appears in the roots that we created is a. Itβs the square root of two over two minus the square root of two over two π.