So in recent videos, we've been talking about how you can take complex numbers in polar form and how you can do different operations on them. So let's say I had this complex number right here in polar form, and I wanted to multiply it by itself. Well, we learned how to multiply complex numbers by taking the product of the \( r \) values and adding the angles. But it turns out there's another way that we could go about doing this, which is a much simpler approach. What you could do instead is take this complex number and raise it to a power of 2. And raising complex numbers to powers is something we're going to be talking about in this video. This is a skill you're going to need to have, and there's a very intuitive shortcut to do this. So let's just jump right into an example.

So if you wanted to take a complex number and raise it to some kind of power, what you want to do is, whatever that power is, we'll call it \( n \), you're going to take the \( r \) value out in front and you're going to raise it to the \( n \)th power, and then you're going to multiply this \( n \) by the angle theta. So by doing this, you'll be able to complete this operation, which we call de Moivre's theorem. De Moivre's theorem is just the equation that we named for this situation. So let's go ahead and do this. First, I'm going to take this \( r \) and I'm going to raise to the \( n \). Now I can see that the \( r \) we have is 3, so \( 3^2 \) would be \( 3^2 \), and then this is all going to be multiplied by the cosine of \( n \) times our angle. So we're multiplying the angles here. So we're going to have \( n \) which is 2 multiplied by 15 degrees, And then what we're going to do is have \( i \) times the sine of this whole thing as well. So doing this we're going to have +isin(n × 15°). So this right here would be the operation and that's how we can apply De Moivre's theorem to this example. So let's go ahead and simplify things. Now, \( 3^2 \), that's the same thing as 9. And then what I can do is recall that we talked about in previous videos, we can take this whole cosine and sine thing and we can reduce it to just cis. This is an abbreviated version. So you can see this is the same thing as \( 9 \) cis, and then we have \( 2 \times 15 \), which is 30 degrees. So this right here would be the solution to our problem, and that's how you can raise complex numbers to a power. Now as you can see, it gives us the exact same result as if we were to just multiply this number by itself. So this is the shortcut to doing these types of problems.

Now this does beg actually an interesting question, which is why would we necessarily need this shortcut? I mean, yeah, it might be slightly faster, but wouldn't it be easier just to remember one equation like this? Well, there are actually a lot of situations where this shortcut is actually going to be needed to solve the problem or is going to be way simpler than trying to just multiply things out. And to understand when this happens, well, let's take a look at this example down here. So in this example, we're asked to evaluate the following expression. And here we have this complex number where we're given the angle in radians, and notice that we're now raised to the 3rd power. So if I were to try and take this complex number and multiply it by itself 3 times in a row, that would be really tedious. We'd have so much going on. But using this de Moivre's theorem shortcut, we can actually do this very quick. So let's try this. So what I'm first going to do is take 4, the \( r \) value, I'm going to raise it to a power of \( n \). So it's going to be \( 4^3 \), and then this cosine and sine will become cis, and then we're going to have \( n \times \) our angle. So that means that we're going to have this, this \( n \) value 3 multiplied by our angle \( \frac{\pi}{6} \). Now \( 4^3 \) that's the same thing as 64 and then that's going to be cis and \( 3 \times \frac{\pi}{6} \), well, this 3 is actually going to reduce with that 6 because 6 is the same thing as 2 times 3, so we can just cancel a 3. So that means we're going to have cis \( \frac{\pi}{2} \). And this right here is the solution to this example. So that is how you can solve these types of problems. As you can see, de Moivre's theorem allowed us to do this very quickly when we just applied this shortcut, and we're able to get our answers. So hope you found this video helpful. Thanks for watching, and let's get some more practice.