 In this video, I want to talk about Dimwavra's theorem, which gives us an effective tool to calculate complex exponents, that is to say, taking an exponent of a complex number. So we've done things like, you know, calculate x squared, x cubed, x to the fifth, what have you. That's nice and easy for real numbers, but for complex numbers it gets a little bit more complicated, right? As we saw in our previous video, when it comes to complex multiplication, it is a lot simpler to compute the product of two complex numbers when you have that complex number in polar form, right? But there's a conversion cost, right? We usually write our complex numbers in the Cartesian form, so it looks like z equals a plus bi. That's how we like it. That's great for addition and subtraction. It gets a little bit more cumbersome when you multiply and divide, but the polar form is really useful in that situation. R, ooh, that's a hideous R. Try that again. R equals i theta, like so. In which case, multiplication is a cinch because it's really just typical exponential rules in that situation. But again, there's a cost of conversion. When might the cost actually be worth paying to convert? And that definitely is the case when you start doing exponents of these complex numbers. If you want to do something like z cubed in the terms of the FOIL method, that gets very, very cumbersome because you have to take a plus bi times a plus bi times a plus bi, which admittedly, you can use the binomial theorem to try to simplify that calculation, but even still, it gets very cumbersome very quickly. De Moivre's theorem makes it a lot easier to do these things because we just utilize the polar form of the complex numbers. So if you have a complex number z, which has the form R times cosine theta plus i sine theta, which remember here, R would be the modulus of the complex number. Theta is the argument, and this right here is just R e to the i theta, like so. Then it turns out that z to the n, this exponent, here we're going to take z to be an integer. We see the exponent here is z to the n, it's going to equal R to the n, so you just take the exponent, the same exponent of the modulus, you take that power. And then you're going to take cosine of n theta plus i sine n theta. So you're going to take the exponent of the modulus, but then you're just going to multiply the argument by that value. And why is that? Well, when you think of this in terms of the exponential form, the polar form of the complex number, it's very, very simple. So take z to the n. We're going to write z to the n as R e to the i theta to the n, for which then by exponent rules, this is R to the n times e to the i theta to the n. And then using more exponent rules, when you have an exponent to an exponent, you then multiply them together, you get R to the n times e to the i theta times n, which I'm going to write that as e to the i n theta. And then utilizing Euler's identity, what is e to the i n theta? Well, you're going to end up with an R to the n that sits in front, and then you get cosine of n theta plus i sine of n theta. And so then we see exactly why we are going to multiply the arguments by n. It's just a consequence of exponent rules, right? If you don't utilize Euler's identity, this proof gets a lot more complicated, a lot more messy, and you have to do some pretty intense trigonometric identities. For the most part, we've proved a formula in our previous video that we can then derive DeMauro's theorem from it pretty quickly, but that argument we did in the previous video would have been intense with trigonometric identities. But with Euler's identity, it's pretty straightforward. It's just their exponent rules, and so these are natural consequence of those exponent rules. Let's see an example. Let's compute 1 plus i to the 10th power. You do not want to do 1 plus i times 1 plus i times 1 plus i times 1 plus i times 1 plus i times 1 plus i do that 10 times. That'd be very complicated. Even, like I said, with the binomial theorem, it would still get very complicated because you end up with these 11 terms where you can combine them together. DeMauro's theorem is definitely the way we want to go here. And so we have to first convert our number into complex into the trigonometric form of the complex number. So if we take the number z equals 1 plus i, we have to first find the modulus. This is the square root of 1 squared plus 1 squared, which is the square root of 2. We then have to find the argument, go pirate for a moment, arg. We need to find arc tangent of 1 over 1, so just arc tangent of 1 there. That's going to happen at a 45 degree angle, which quadrant we're in. We're in positive, positive quadrant. So that's the first quadrant. So this is going to give us pi force. We should always use radians when we're talking about complex numbers and their arguments. Therefore, our number z is equal to the square root of 2 times e to the pi i over 4 power. So now we want to calculate, if we're doing 1 plus i to the 10th, we're really just trying to calculate z to the 10th power. And by the previous theorem, we're going to get the square root of 2 to the 10th power. And then we're going to get e times pi, well, 10 pi i over 4 times the angle by 10 in that situation. Now, when you take the square root of 2, notice if you square the square root of 2, that gives you back 2. So this is going to give you 2 to the 5th. And then, of course, 10 and 4 have a common divisor of 2. So you're going to get e to the 5 pi i over 2. So 2 to the 5th is going to be 32. So you take 2 times 2 times 2, you get up to 32. And so then we have to do cosine of 5 pi over 2 plus i sine of 5 pi over 2. And you'll look at that thing there. And so it's like 5 pi over 2, that's actually bigger than 2 pi, right? In fact, 5 pi over 2, 5 pi over 2, that's the same thing as 4 pi over 2 plus just pi over 2, like so, which 5 pi over 2 is the same thing as 2 pi, so I can just ignore that. And so with that consideration in mind, we end up with 32, we have cosine of just pi halves. Cosine of pi halves, that's just going to equal 0. And then sine of pi halves, that's just going to equal 1, like so. So we see the final product turns out to be 32i. And so in this situation, when you have exponents, and your exponent doesn't even have to be relatively big. If you start taking exponents of complex numbers, you're going to do yourself a huge favor by using the polar form. Essentially, you do want to use DeMauro's theorem in a situation like this. Use the polar form of a complex number, and this will help you when you start doing these complex exponents. It's so much more effective this way than perhaps the old-fashioned ways with Cartesian coordinates.