 Welcome back to our lecture series math 10-6 teacher genometry for students at Southern Utah University. As usual, I'll be a professor today, Dr. Angel Misseldine. In our previous lecture, we talked about doing products, exponents, and quotients of complex numbers. In lecture 34, we are going to talk about the final operation that is roots of complex numbers, radicals of complex numbers, where we use the polar form to simplify these calculations. Now, radicals take a little bit more explanation, which is why we're dedicating this entire lecture to do conversation about roots of complex numbers. So before we jump straight into the deep end of the pool, let's talk about what it means to be a root in the first place. It has to do with solving polynomial equations. When you have an equation like x squared equals 25, if we solve it, we normally would say something like, oh, the solutions equal plus or minus 5. There's two numbers that solve this equation, because notice that 5 squared is 25, but negative 5 squared is also equal to, of course, 25. So you have two solutions, and there are two real solutions to this equation. In contrast, if you look at the equation x squared equals negative 1, the way that we usually solve this equation is you would take the square root of both sides, negative 1. There's a positive and a negative choice. Now, the square root of negative 1 is just i. So we see there are two solutions plus or minus i. So there are two square roots of negative 1. There's i, and there's negative i, just like there's two square roots of 25, which is 5, and negative 5. Now, in the first case, the square roots are both real numbers, and in the second case, the both square roots are imaginary numbers, but there's 2. And the significance of the 2 there is that we have this quadratic expression. x squared equals whatever. There should be two square roots. Well, what happens if we move on to something like a cube root of some kind? Let's take, for example, x cubed equals 8. Now, we're very accustomed to the real solution. We usually say something like, oh, x equals the cube root of 8, which is equal to 2, and this is the real solution. That's right. But this is not all of the cube roots there. Just like the square roots, there should be three cube roots. There's gonna be three cube roots of 8. Why have you never heard of them before? Well, unlike the square root where both square roots were real numbers for 25, when it comes to 8, there are three cube roots, but only one of them is a real number. The other two are, in fact, complex, non-real complex roots. And so what are they gonna be? It turns out that the other one, so you have x equals 2 is the real root, but you also get negative 1 plus i root 3, and you also get negative 1 minus i root 3. These are the two non-real cube roots of 8, right? So to prove that these are, in fact, the cube roots, what we could do is we can cube them and show that the cube of these numbers is equal to 8. Now the cube isn't so horrible. I'm just gonna do this in the old-fashioned Cartesian way. So you get negative 1 plus i root 3, negative 1 plus i root 3, and then negative 1 plus i root 3. And so let's foil out the first one right here. You're gonna end up with a positive 1, negative i root 3. You're gonna get a negative i root 3, and then you're gonna end up with a negative 3, like so, times negative 1 plus i root 3. And then let's combine like terms before we multiply the second one out here. You get 1 minus 3, which is a negative 2, and you're gonna get negative 2 i root 3, like so, times that by negative 1 plus i root 3. I'm actually gonna factor out a 2, a negative 2, in fact, from the first one I noticed there. So I get 1 plus i root 3 times negative 1 plus i root 3, like so. So then when you do the next product, you have that negative 2 right there, you're gonna get a negative 1, you're gonna get a i root 3, you're gonna get a negative i root 3, just already recognizing those will cancel out, and then you're gonna get a negative 3, like so. That gives us negative 2 times negative 4, which sure enough is in fact equal to 8. So this is in fact an authentic cube root of 8. This is a complex number, which cubes to give us 8. I'm not gonna do the calculation for the other one, but you could practice on your own that negative 1 minus i root 3, raised to the third power is likewise equal to 8. All right? Now I did this in the Cartesian method, which if you've been watching this lecture series, you know that's, as I claimed, the inefficient way to do that, right? By Du Bois's theorem, we actually could simplify this if we did this in polar form. What is the polar form of this number right here? In fact, well, if we think, let me put it a little bit higher on the screen there. If you were to take the modulus of this number, the modulus of negative 1 plus i root 3, this is gonna equal the square root of 1 squared plus the, well, the square root of 3 squared. That gives us the square root of 1 plus 3, which is the square root of 4, which is 2, right? And then the argument in this situation, the arg of negative 1 plus i root 3, this would be arctangent, arctangent of, we're gonna get the square root of 3 over negative 1, right? So that gives you arctangent of negative root 3. So negative root 3, let's just take, let's consider when tangent equals the square root of 3 real quick. The square root of 3 would happen when sine is root 3 over 2 and cosine is 1 half. So we're thinking like theta equals pi thirds right here, but the quadrant, we have a negative right here and a positive. So we're actually living in the second quadrant right here. So instead of pi thirds, we really are thinking of two pi thirds. So that's our argument right there, two pi thirds, like so. And so let's come up to, let's come up with our number here. This is the same thing as two e to the i of two pi thirds. A similar calculation would show that this one is equal to two e to the four, the four pi i over three. And of course, two is just equal to two times e to the i times zero, like so. So one thing you should notice here is when you look at these numbers in their polar form, all of their moduli are the same and what's that moduli too, right? Two was the real cube root of eight, okay? But then they're angles, right? You have zero pi thirds and excuse me, zero two pi thirds. Like look at these angles here. You had zero two pi thirds and four pi thirds. Notice if you added two pi thirds to the last one, you get back zero, then again, again, again, these are just all have a difference of two pi thirds. This is just the arithmetic sequence of adding two pi thirds over and over and over and over again, where two pi thirds notice, you're taking the standard period of a sine and cosine and you divide it by three, right? Because we're taking a cube root. If I'm looking for the cube root of a number, right? You would write a one third power and this actually is suggestive of the true strategy that we have to employ here. That actually gives us the following formula right here. Let z equal r e to the i theta. Let this be a complex number in polar form. Then it'll have n distinct nth roots of z. This is a consequence of the fundamental theorem of algebra, which I'll say why that isn't just a second. So there's gonna be n nth roots of z. And so let's call them w for a moment, where w sub k, k here is just some index variable to dummy variable keeping track of it. You have like the zero, the one, the two, the three, the four, the five going on through that. Wk is the modulus for each of these is gonna be the same. You're gonna take the nth root of r. R is a positive real number, so it has an nth root. The modulus will just be the nth root in the usual real sense. But then the exponent here, the argument, you're gonna take theta plus two pi k and divide it by n. Why is that? Well, when you take an nth root, that's the one over n exponent. So you're just gonna divide the x one by n. But you're changing the period when you divide it by n. And as such, you have to also consider the period. You don't just take theta over n. You have to take two pi k divided by n as well to get the general solution. This is what we did when we had modified periods previously. All right? And so why does this thing work? Well, basically we're trying to solve the equation z to the n, excuse me, x to the n equals z. We're trying to solve this equation like so. And so to do that, x would equal z to the one over n. But z here, and we put it into polar form, will look like r e to the i theta one over n. But I'm gonna rewrite theta as theta plus two pi k. Cause again, if you replace theta with anything coterminal to theta, that would still give you the same complex number. And then when you divide by n here, well, you're gonna get r to the one over n power, which that, of course, is the same thing as just the nth root of r. Then you get e to the i, you're gonna take this theta plus two pi k, and then you divide that by n, like so. And so this is how you compute complex roots. You do it in the polar form. You just have to make sure your variable k goes all the way from zero up to n. Let me show you an example to show you how this really works here. Let's take the complex number z equals 16 e to the pi i over three. It's already in polar form, so we can actually compute the fourth roots of this pretty easily, and we need four of these things, okay? So z to the one fourth here, there's gonna be four of these things. So we have to first take the fourth root of 16, and then we're gonna take e to the pi, well, I'll just put the i over here, pi pi thirds plus, of course, two pi k, and we divide that by four, all right? The fourth root of 16 is two, and then here we end up with pi twelfths, pi twelfths plus pi halves k. Where did the pi halves come from? Well, two pi divided by four is pi halves. Where did the pi twelfths come from? Well, that's pi thirds divided by four. So we have to consider that. So there's gonna be four options. So the first root, we start with the index zero right there, so that's when you take k to be zero. That's gonna be two e to the i pi over 12. We'll come back to that one in a second. The second, the next one, so w one, that's when k equals one, you get two e to the i, we're gonna take pi twelfths, and we're gonna add to it pi halves, like so, which I'm actually gonna write that as six pi halves, or six pi twelfths, excuse me. Then your next root would be w two, which is two e i times pi twelfths. Now we have to add pi, because we took two times pi halves, so we're gonna add pi, which again, to make life a little bit easier, we're gonna write 12 pi over 12, and then the last one here, w three is gonna be two e to the i times pi twelfths plus, now we have to add three pi halves, which again, to make life a little bit easier, I want my denominator to be a 12, so we're gonna get in the numerator here, we need times that by six, we got 18 pi twelfths, right? So, and so what we see on the screen right now, they're not completely simplified yet, but these are the four roots of z in the situation, okay? And so we can of course simplify these things and put them in the Cartesian form, we really like that. So you're gonna get two times cosine of pi twelfths. Now, pi twelfths is the same thing as 15 degrees, in case you're wondering there, we have to do sine of pi twelfths, for which remember here that cosine of 15 degrees, that's actually the same thing as the square root of six plus the square root of two over four, and then the other one, let me scooch it over just a teeny bit, scooch. For the next one here, you're gonna end up with square root of six minus square root of two over four, like so. And then of course you can put the two in there, distribute the two throughs, so I'm actually gonna end up with the following, I'm gonna write this as square root of six plus square root of two over two, and then you get i times the square root of six minus square root of two over two, like so. And so this would then be the first fourth root, this is a number which have raised to the fourth power is equal to the number we started off with, 16e to the pi i thirds. What if we do this with the second one, w one there? So for w one, the complex number is gonna be two times cosine of, in this case, seven pi twelfths plus i sine of seven pi twelfths. This is an angle in the second quadrant, this would turn out to be similar calculation, you're gonna end up with a negative root six plus root two over two, and then plus root six plus root two over two and throw in an i over there. For the next one, w two, this one computes to be two times cosine of, add the angles together, pi twelfths plus 12 pi twelfths is 13 pi twelfths. This is an angle in the third quadrant, we get i sine of 13 pi twelfths. This time it's gonna equal negative root six minus root two over two. So the x-coordinate should be negative, but in the third quadrant, the y-coordinate should also be negative. You get negative root six plus root two over two i, like so. And then for the last one, w three, we're gonna end up with two times cosine of what we get pi twelfths plus 18 pi twelfths, that's 19 pi twelfths, like so, plus i sine of 19 pi twelfths. And so this will be an angle in the fourth quadrant, we end up with root six minus root two over two plus negative root six minus root two over two times i. And so now we've discovered the four fourth roots of the number 16e to the pi i over three. And so while it looks, there's a lot going on here, I'll give you that, but it's a very straightforward calculation, we just have to find sort of the principle root, which is this one right here, we found the modulus and we just found this reference angle, and then we have to compute all the other ones by adding to the period. In this case, we added pi halves until we got four distinct roots.