>> If we have a number that is in polar form, than we can find the Nth root of Z, which is the Nth root of this expression, by using this pattern. It is the Nth root of R times cosign -- now, notice we have theta up here -- but it's the cosign of this fraction. Theta plus 2 pi K -- I'll tell you about K in a minute -- over N. Now, the N is the same as this Nth root business, you see? So if it was the 4th root, that would be a 4, 6th root, that would be a 6, and so on. Okay, and then the same fraction occurs over here with sign, as we might expect. Now, K is simply a little counting mechanism. And K takes on various values up to N minus 1. Now, K starts at the value 0, and it goes up to N minus 1. And it turns out that all of these roots will occur by letting K take on a separate value each time. Now, notice there are N different values that K takes on here, and there will be N values for this Nth root notion. Alright, let's take a problem. Suppose we're talking about the problem we had before. Now, with this problem we had before, this idea of the 6th roots of 1, well, 1 is not in complex form. Let's write it in complex form. So the Z we're talking about is 1 plus 0 I. Now, let's put it into polar form. Alright, polar form. We need an R and a theta. So R is the square root of 1 squrared plus 0 squared. Oh, that's simply 1. Then to find theta, we use tangent theta as B over A. Well, it's B over A, that's 0 over 1, or 0. Let' see, inverse tangent of 0 is 0 degrees. So theta is 0 degrees. So we have an R of 1 and a theta of 0 degrees. Sounds like a really degenerate kind of expression in polar form. So it's 1 times cosign 0 degrees plus I sign 0 degrees. Now, we apply this idea, this pattern, like this. We think, "Okay. The notion is that we want the 6th root of Z--" I'm gonna go ahead and replace this with a 6. We've already talked about the general case here, but this specific case involving the 6th root of Z is the 6th root of the R, the 6th root of 1 is just 1 -- it'll kind of go out here -- times the cosign of the angle theta plus 2 pi K over 6. Now, the 6 is corresponding with the 6th root idea. Alright. And then the same fraction appears over here. And then we let K take on values 0 through 5. So away we go. We say, "Alright, let's let K be 0." If K is 0, then we look at this, and we have, "Alright, if K is 0, then this term goes out. We have 0 -- 6. Oh, that's the cosign of 0." And then the same way with the sign. The really nice thing about this is that once you evaluate for one of these fractions, you have automatically evaluated for the other. So you can kind of bring down in a lot of situations. So we have -- let's see -- the cosign of 0 plus I sign 0, and the cosign of 0 -- you see, and then the sign of 0 turns out to be 0 and that turns out to be 1. So we have 1 as one of the six roots of 1, as we found before. We found 1 to be one of those guys. Now, if K is one, here's what happens. We have a 2 pi times 1. So we have 2 pi in the numerator, 6 in the denominator, 2 pi over 6 -- oh, that's pi over 3. So the fraction here becomes pi over 3. And over here also, pi over 3. So it's cosign pi over 3, plus I sign pi over 3. Cosign over pi over 3 is one half. The sign of pi over 3 is the square root of 3 over 2. And now writing this with the common denominator, it's 1 plus root 3 I over 2. And that's one of the items that we had before. And then we come back over here and let K take on the next value: 2. And if K is 2, here's how we evaluate. We come up here to this form, and we let K be 2. 2 times 2 pi, 4 pi. 4 pi over 6, reduces to 2 pi over 3. Both fractions then are 2 pi over 3. Cosign of 2 pi over 3, think about it a little bit. It's negative one-half. The sign of 2 pi over 3 is root 3 over 2. Using our common denominator here, we have negative 1 plus root 3 I all over 2. Again, another one of the zeros that we found earlier. Well, then we let K take on the other values: the values 3, 4 and 5, and that will give us all 6 of the 6 roots.