# DeMoivre's Theorem - Concept 10,629 views

We know how to multiply complex numbers, but raising complex numbers to a high integer power would involve a lot of computation. Fortunately we have DeMoivre's Theorem, which gives us a more simple solution to raising complex numbers to a power. DeMoivre's Theorem can also be used to calculate the roots of complex numbers. DeMoivre's Theorem uses the trigonometric form of complex numbers.

I want to talk about powers of complex numbers.
Let's start by calculating the square of this complex number, z=r times cosine theta plus i sine theta. So I'm going to get r cosine theta plus i sine theta, times r cosine theta plus i sine theta.
Now the modulus, I can pull out. I can pull r out from both of these guys and I get r squared. r times r. And then I get this cosine theta plus i sine theta squared. So that's cosine squared theta and then the other real term is going to be plus i squared sine squared theta. And since I'm squaring, it will be twice the mixed product, i sine theta cosine theta. So plus i times 2 sine theta cosine theta. This is going to give me r squared, cosine squared and then i squared's a negative one, so this will be minus sine squared plus i times 2 sine theta cosine theta.
Now you may recognize that this cosine squared minus sine squared is exactly cosine 2 theta, cosine 2 theta. Plus and 2 sine theta cosine theta is sine 2 theta. These are the double angle identities for sine and cosine. So what I have here is the square of of Z, right? To square Z all i have to do is square the modulus, and double the angle.
Let's generalize this result. The genralization of this result is actually called DeMoivre's Theorem. And it says if z is r cosine theta plus i sine theta then Z to the n where n is any integer, is r to the n times cosine of n theta plus i sine n theta. So all you have to do is raise the modulus to the nth power and multiply the argument by n.
Let's use DeMoivre's Theorem in a problem. It says simplify, express your answer in rectangular form. Let's start with this guy. Here it's already in trigonometric form, so it's really easy to apply DeMoivre's Theorem. We're going to get the modulus 2 to the 5th power times the cosine of the argument and multiply the argument times 5, so you get 5 pi over 3 plus i sine 5 pi over 3. And I have to put this in rectangular form. So this is going to be 32 times what's the cosine of 5pi over 3? It's one half. One half plus, and the sine of 5 pi over 3 is negative root 3 over 2. So i times sorry, negative root 3 over 2. So this is 32 times a half or 16 and then 32 divided by 2. 16 root 3, so minus i times 16 root 3. This is the rectangular form of this complex number to the 5th power.
Let's do another example. 1 plus i to the sixteenth and you really wouldn't want to multiply this out by hand. That's that looks like a pretty big product. So we first need to switch this into trigonometric form. DeMoivre's Theorem requires trigonometric form. So let me just say that the modulus of this is going to be root 2. And 1 plus i has an argument of pi over 4. So this is cosine pi over 4 plus i sine pi over 4. And that's all to the sixteenth power. So this is going to be root 2 to the sixteenth power, and then cosine the argument is going to be multiplied by sixteen, pi over 4 times 16 is 4 pi. So this will be cosine of 4 pi plus i sine of 4 pi.
Now the cosine of 4 pi is the same as the cosine of 2 pi which is the same as the cosine of 0, it's 1. And the sine of 4 pi is 0. So this is just going to be 1+i times 0, times whatever root 2 to the sixteenth is. Now root 2 to the sixteenth is the same as 2 to the eighth. Because root 2 squared is 2. So this is 2 to the eighth. Now what's 2 the eighth? 2 to the fourth is 16. 2 to the eighth is 16 squared, 256. So it's 256 times 1, that's our answer, 256. Now what's interesting about this is that means that one of the sixteenth roots of 256 is 1 plus i.
Anyway, remember when you're using DeMoivre's Theorem, to raise a complex number to a power, you need the complex number to be in trig form. You want to raise the modulus to the power and multiply the argument by the power.