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 𝑖? b) Root two minus root two 𝑖. c) Root three over two minus a half 𝑖. d) Two root two plus two root two 𝑖. Or e) Negative one-half minus root three over two 𝑖.
We begin by recalling what we mean by the eighth roots of unity. Well, the 𝑛th roots of unity are the solutions to the equation 𝑥 to the 𝑛th power
equals one. This equation has 𝑛-roots. And, in trigonometric form, these roots or these solutions are given by 𝑥 equals cos
of two 𝑘𝜋 over 𝑛 plus 𝑖 sin of two 𝑘𝜋 over 𝑛 for values of 𝑘 between zero to
𝑛 minus one.
Now, of course, in this question, we’re finding the eighth roots of unity. So we’re going to let 𝑛 be equal to eight. We can then say that 𝑥 must be equal to cos of two 𝑘𝜋 over eight plus 𝑖 sin of
two 𝑘𝜋 over eight for values of 𝑘 from zero through to 𝑛 minus one. Well, that’s of course seven. Let’s simplify two-eighths to be equal to one-quarter. And we see that our solutions given by 𝑥 equals cos of 𝑘𝜋 over four plus 𝑖 sin of
𝑘𝜋 over four.
We’re now going to list each root by letting 𝑘 be equal to zero, one, two, three all
the way through to seven. The first root is a result of letting 𝑘 be equal to zero. So we get 𝑥 equals cos of zero 𝜋 over four plus 𝑖 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 𝑥 to be equal to one plus zero 𝑖. So the first solution to the equation 𝑥 to the eighth power equals one is one, which
makes a lot of sense.
Our next root, the second root, is found by letting 𝑘 be equal to one. This time, 𝑥 is equal to cos of one 𝜋 over four plus 𝑖 sin of one 𝜋 over
four. Now, of course, we don’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 𝑥 is equal to the square root of root two over two
plus the square root of two over two 𝑖.
Let’s repeat this process for the third root, this time letting 𝑘 be equal to
two. And our solution is 𝑥 equals cos of two 𝜋 by four plus 𝑖 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 𝑥 is equal to zero plus one 𝑖, which is simply equal to 𝑖. The fourth root is found by letting 𝑘 be equal to three. So we get 𝑥 equals cos of three 𝜋 by four plus 𝑖 sin of three 𝜋 by four, which is
equal to negative root two over two plus root two over two 𝑖.
Our final four solutions are found by letting 𝑘 be equal to four, five, six, and
seven. And when 𝑘 is equal to four, we get 𝑥 equals negative one. When 𝑘 is equal to five, we get 𝑥 equals negative root two over two minus root two
over two 𝑖. When 𝑘 is equal to six, we get 𝑥 is equal to negative 𝑖. And our eighth root and the eighth solution to the equation 𝑥 to the eighth power
equals one is found by letting 𝑘 be equal to seven. And we get 𝑥 equals the square root of two over two minus the square root of two
over two 𝑖.
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. It’s the square root of two over two minus the square root of two over two 𝑖.
