 In this video I want to solve the polynomial equation x cubed plus one equals zero. I want to solve this for all of its complex roots. Notice if I subtract one from both sides I end up with x cubed is equal to negative one and therefore solving this polynomial equation is equivalent to finding all of the complex cube roots of negative one. Therefore I'm going to solve this equation using the technique of complex roots. So let's take our number negative one here. This is going to be our complex number z. If we put this into its polar form its modulus is going to equal one. Its argument is then going to equal pi. That is to say e to the pi i is equal to negative one. That's the observation we're using here. Therefore x if we take the cube root of both sides will give us e to the well we have to replace we have to replace this angle with anything that is coterminal to it. So we actually take you have to take pi plus 2 pi k you have to consider all those possibilities there. So you're going to cut that by a third in which case that's going to give us pi thirds plus 2 pi thirds k. And so we are going to end up with three possible complex roots. So the first one the principal root we'll call it w zero here. This is going to be e to the i pi thirds plus 2 pi thirds times zero like so. The next one would be w one. This is when that value k is equal to one in that situation. So you get i pi thirds plus 2 pi thirds times one. And then the last one we'll do that one in red. We'll call it w two. That's equal to e to the i pi thirds plus 2 pi thirds times two like so. And so then we just need to simplify these numbers here. Well of course with the first one since you have 2 pi thirds times zero just disappears. So we just have to consider the complex number e to the pi i over three. This is the same thing as cosine of pi thirds plus i sine of pi thirds which is the same thing as one half plus i root three over two. So this is our first cube root of negative one. What about w one here? What's the next one? In this case you times by the one you end up with e to the i. We have to add these together. That's going to be a three pi over three which I can simplify that thing. That's going to be e to just the pi i. Oh that's equal to negative one. We saw that earlier. And this is actually the real this is the real cube root that we're used to like take the cube root of negative one. But yeah negative one cubed is negative one so it's its own cube root as well. We detected that one. Clearly this is a non real cube root of negative one. And what about the last one here w two? We're going to take two times two times two pi thirds. That of course is going to give us four pi thirds. Like so add those exponents together. We get e to the i times five pi thirds like so. And so the last one is going to give us e excuse me it's going to give us cosine of five pi thirds plus i sine of five pi thirds. So now remember the angle five pi thirds. This is an angle in the fourth quadrant that references pi thirds. And so this is the same thing as cosine of pi thirds. In fact minus i sine of pi thirds. Which just kind of feels like what we did before right. This is going to equal one half minus i root three over two. Like so. Which let's come up for a second. All right. So look at our complex roots here. So we had one half plus i root three over two. We had negative one which is the real one which could is easier to predict. And then we have one half negative i root three over two. I want you to notice that these numbers right here are actually complex conjugates to each other. Notice the change of signs like so. We actually get as a consequence of the fundamental theorem calculus that whenever you have a real polynomial like in this case x cubed plus one equals zero. If you're trying to solve a polynomial equation and you get complex roots they always come in they always come in conjugate pairs. And so as you're cycling through these roots like you this k value as it ranges from zero to two. When you get through the second half the second half is just going to be conjugates of the first half. And so that actually can speed up the process dramatically here as we're trying to look for these roots. Solving these polynomials right. I actually could have skipped this entire step here. Could have ignored all of this knowing that hey if this was a solution then its conjugate would have been another solution as well. So once I did this one the first one and then I figured out negative one I could have been like oh conjugate got me the last one. So I could actually sped up this process. But for the sake of example I went through all of them here.