Euler's Formula: A Proof

Hey guys, this is Matt Kids 101, and today I'm going to be making an experimental math video for some math sessions. So leave a comment below if you have any requests or what you think about these math series, or of course, if you have any questions about the video.

So today, I'm going to be proving Euler's identity, which is that e to the I theta is equal to cosine theta plus I times sine theta. I'm gonna do this by solving differential equations by separating variables. There are other ways of doing this; that's how I'm going to be doing it. I think that it's clearer than using a Taylor series.

So I'm gonna let y equal cosine theta plus I sine theta. So Y is equal to the right side of the statement, and I'm gonna prove that it's equal to the left side of the statement. I'm also gonna keep in mind that Y of 0 is equal to 1 because that's gonna come in handy later. Cosine of 0 is 1, plus sine of I sine of 0, which is 0.

Okay, take the derivative of both sides with respect to theta. So it's going to be dy/dtheta is equal to the derivative of cosine, which is minus sine theta, plus I times the derivative of sine, which is cosine theta.

All right, now I'm going to factor out an I, so that is equal to I times... So I squared is going to be negative 1, so if you bring out an I, this is gonna be I sine theta plus cosine. If you look back, this is exactly what we defined y to be, so it's equal to iy.

All right, who's gonna write that there? We're running out of space. So dy/dtheta is equal to Iy. Now I'm gonna bring... I'm gonna divide both sides by Iy and multiply both sides by dtheta, so we have minus I times 1 over y dy is equal to dtheta.

1 over I is equal to minus I. You can say that 1 over I is equal to I over minus 1, and so I squared is minus 1. When you cross multiply, that's an easy way to check that that floats true.

All right, so I'm going to integrate both sides. I'm going to use an indefinite integral, but there might be other means of doing this. All right, now when I do this, I'm gonna get minus I times Ln of Y is equal to theta plus some arbitrary constant.

Now all we have to do is isolate y, solve for the constants, and check what that's equal to. So I'm going to let Alex do that part.

All right, so now I'm gonna be putting the final touches on this proof and showing you that Y is actually equal to e to the I theta. Okay, so first off, we have this situation right here with an ugly I on an ugly constant there, and we got this I on that side and the natural log of Y here.

So the first step is going to be to isolate y. In order to go about this, first, we're gonna get rid of this mind side. To do that, we just are going to multiply both sides by I because if you do that, you have I times I and then minus that, so that's minus -1. So we just get the natural log of Y.

I'm just gonna write what I did there: I multiplied by I is equal to I theta plus I times some arbitrary constant, which is just some other arbitrary constant. In this case, the arbitrary constant is probably going to be complex; we don't know what it is. So I'm just going to keep that as C, but you really know this isn't the same constant; it's the original constant times I.

Now we can just do e to both sides. So if I do this, then that gives us e to the natural log of Y is just Y. So it's y equals e to the I theta plus C in the top part. In order to prove that this is equal to that, all that's left is to prove that this arbitrary constant right here is actually indeed 0.

To go about doing that, we just need to use our initial condition that we said initially up there. So if we say that Y of 0 is 1, then e to the I times 0 plus C, since this is still our Y function, has to equal 1. I times 0 is 0, so e to the C equals 1.

Then we take the natural log of both sides: C equals the natural log of 1, which is equal to 0. So now, using our initial condition of this function, we prove that Y has to be e to the I theta plus 0. Thus, from here, it finally follows that y is equal to e to the I theta.

So that's how you get rid of the arbitrary constant that comes from separating variables, and you get the final equation for y. All right, so now that Alex demonstrated this right before in the video, I'm gonna show us some cool properties that are useful about this.

The first is an equation that you've probably seen before, but what happens if now that we know this is true, what happens if theta is equal to PI? So on the left side, we have e to the I pi equals cosine of PI, which is negative 1, and then I times sine of pi. Well, sine of pi is 0, thus i times 0 equals 0.

So e to the I pi is equal to negative 1 or e to the I pi plus 1 equals 0. This is easy. It's positive 1 equals 0; all the fundamental constants in math multiplying exponentiation, like the whole shebang.

So another cool thing is if you want to take I to a power that might be hard to do, but with this it should be pretty easy to do. So let's take a look. If we have e to the I pi is equal to negative 1, then you might wanna think about negative 1.

That's equal to I squared. All right, so then you can take the square root of both sides. You can take both sides to the 1 over 2. So then e to the I PI over 2 is equal to I.

If I want to take both sides to the I, let's say when I to the I, then I could say I to the I. Normally that would be pretty hard; I would have no idea how to do that. But in this case, it's going to be e to the I squared PI over 2. I just brought both sides to the I, and I squared is negative 1.

Thus, I to the I is e to the minus PI over 2, which is real. So I'm gonna leave you that; we've got to think about all right. Thanks for watching and subscribing.