 Welcome back to our lecture series Math 10-6, teacher-genometry for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misildine. Let's talk some more about complex numbers. You'll remember from a previous video, we've introduced Euler's identity, which tells us how to deal with imaginary exponents. If you see something of the form e to the i theta, this is actually the same thing as cosine theta plus i times sine of theta, where theta is some angle and radians right there. This gives us the most general complex form when we throw on some modulus r, where r could be any positive real number or zero, I suppose. And this gives us the polar form or the trigonometric form of a complex number. So what's the point of the polar form? Well, when it comes to adding and subtracting complex numbers, there's really not a huge advantage. Actually, the polar form makes it a little bit more difficult to add and subtract these things. But when it comes to multiplication and division, the polar form is actually the superior method for doing this complex algebra. And I'd like to illustrate that to you here in lecture 33. So let's first talk about multiplication. Suppose we have two complex numbers. We'll call it z one and z two. Remember, we like to use the complex variable z instead of x or y when we talk about these complex numbers here. So imagine that z one is given so that it's modulus is r one. Again, we like to use r here because we're thinking of it as the radius of a circle in the complex plane. So z one equals r one times cosine theta one plus i sine theta one, right? So this thing right here is just equal to r one times e to the i theta one. Just the theta one right there, just the polar form of that complex number. Likewise, z two is given as r two times cosine theta two plus i sine theta two. Again, using Euler's identity, this is equal to r two times e to the i theta, theta two, like so. So when it comes to multiplying these together, if you multiply together z one with z two, you're gonna multiply together the moduli. Okay, that makes sense, r one times r two. But then you're gonna add together the arguments. You're gonna take cosine of theta one plus theta two and then plus i sine of theta one plus theta two. So you just add together the arguments. You add together the angles and you multiply together the moduli. Why is that the case? Well, if you try to do it without Euler's identity, this becomes a very, very intense trigonometric identity type argument there. There's a lot that goes on there, but you don't need Euler's identity. It can be done with just trigonometric identities alone. You have to use some angle identities. It's quite involved there. Look in your trig book. You probably can find the proof there. We will of course use the superior argument of Euler's identity, which I confess in our lecture series, we didn't prove it because we don't have calculus. But for most of you, you'll eventually take calculus and therefore you'll be quite qualified to see that proof. So if you just do a little bit of suspension of belief, then I think we're gonna be just fine here. So z one is equal to r one e to the i theta one and z two is equal to r two e to the i theta two. Theta two, like so. And so if we recognize that these are in fact, imaginary exponents, then just by usual rules of multiplication, we can commute. We're gonna put the r one and r two together in moduli. We're gonna put the exponentials together as well. e to the i theta one, e to the i theta two. And so in terms of multiplying the moduli, we're already done, right? Isn't that what we were looking for? An r one, r two, we already have that. So that makes sense. When you multiply together two complex numbers, you multiply together the moduli. But why are we adding the arguments? Well, in terms of these exponential expressions, when you multiply together an exponential, you add the exponent. So yeah, I remember learning that. You get e to the i theta one plus i theta two. But as there's that common factor of i, you can factor it out. So you get r one, r two, and you're gonna get e to the i times theta one plus theta two, like so. And then if we were to translate this from its trigonometric form back to the expanded form, the Cartesian form there, you'll end up with r one, r two, times cosine of, in this case, the angle is gonna be theta one plus theta two. And then you add to that i sine of theta one plus theta two, like so. So we can go back and put it into this cosine i sine form like we see right here. And so we see that, okay, we add the angles as a consequence of just usual exponent rules. If you have an imaginary exponent, we do that. And I like to think of this argument instead, not have some, again, crazy trigonometric identity proof that no one will remember. This is something that you can remember. It's like, oh, you multiply together exponentials by adding the exponents. That's why we add the angles because theta, the angle, the argument is the exponent of this complex exponential expression. So let's show you how you can do these calculations. Well, if your numbers are already given in polar form, then it's very easy to multiply them together. We're gonna multiply together their moduli, so you get three times five, and then you're gonna add together their arguments. All right, so you're gonna have e to the i times two pi over nine plus pi over 18. That's how we go from there. So we multiply together the moduli, that's easy enough, you'll get three times five, which is 15. And then we have to add together the exponents. Well, since we made a common denominator, 18's what our target should be. So let's replace this with four pi over 18, and then when we add those together, we end up with e to the i over five pi over 18, like so. And then that's the product of these two complex numbers. When they're in polar form, the product is pretty easy. The worst it ever gets is we might have to add the right together fractions, add together fractions, and we found a common denominator. I think I might be a little bit with that. Now, of course, if the complex number is not in its polar form, we have to make a choice. We can multiply them together the old fashioned way, or we can convert them to the polar form and multiply them together. So let's try the Cartesian approach here. If we take z one times z two, in this case, z one is one plus i root three, and z two is negative root three plus i, like so. We then get one plus, well i root three times negative root three plus i, like so. We gotta foil this thing out. So we're gonna get one times negative root three. We'll get one times i. We'll get i root three times negative root three, so it's a negative three i. And then we're gonna get an i root three times i. That's gonna be a negative square root of three. I notice i times i is negative one. Combining like terms, We end up with a negative two root three and then we get a negative two I. So this is the product and that honestly wasn't so bad. I mean, basic foil method, not the worst thing in the world. What would happen if we wanted to, of course, compute the product using the polar form? Well, if that's the case, we have to convert these things first over to the polar form. So let's think about that for a second. If we compute the modules of Z one, that's gonna equal the square root of one square plus the square root of three squared. One squared, of course, is one. The square root of three squared is clearly three. So you get the square root of four here. That gives you a two, all right? What is the argument, the arg of Z one there? You're a pirate for a moment. What's the argument there? You're supposed to take tangent inverse of root three over one, right? When is tangent equal to the square root of three? That happens at pi thirds. Sine is root three over two. Cosine's one half in that situation. So that takes care of Z one. We gotta convert over Z two next. So for Z two, it's modulus like so. Same basic calculation. You're gonna end up with, in this case, the square root, whoops, the square root of negative root three squared plus one squared. That's again gonna be a two. Same basic calculations before. Let's be a pirate again for a moment. Let's calculate arg of Z two. So this would be arc tangent of negative one over root three. Now remember tangent, sine over cosine. So how do you get the fraction one over root three? That happens when sine is one half and cosine is root three over two. That's gonna happen at a pi six reference angle. But let's focus on the quadrant right here, right? The original value, we had X negative root three over two, excuse me, negative square root of three. That's a negative value. And then the Y coordinate is one, so that's positive. So if you have negative X and positive Y, that would suggest you're gonna be in the second quadrant. Of course, our first one was plus and plus, so that put us in the first quadrant, of course. But we're in the second quadrant. So we need the angle which references to pi sixth in the second quadrant. That's gonna be five pi over sixth, like so. And so now we can convert these things over to their polar form. So we get Z one times Z two. This is equal to two E to the I pi thirds times two E to the I five pi thirds, a pi sixth, excuse me. And so then when we put these together, you're gonna get two times two, which is four. Then we have to add together the arguments there. Since I need a common denominator of six, I'm gonna write the first one, rewrite it, of course, as two pi six, like so. So when I add those together, you're gonna end up with seven pi sixth as the angle, like so. And so this gives us the trigonometric form. Or if you wanted to get right in the more expanded form, four times cosine of seven pi sixth plus I sine of seven pi sixth, of course, like that. Oh, I need another parenthesis. It went off the screen. There you go. And so that gives us the product as well. And so which one is better, right? Well, that's the same number. So it really doesn't matter, I suppose. It just depends on which form do you need. Like again, with multiplication, it really is easier to do it in the polar form. Although in this situation, since we start off in the Cartesian form, there is a cost of transitioning. But are these even the same numbers, right? Let's think about that for a second. Cosine of seven pi over six, seven pi over six. This is, of course, an angle in the third quadrant. It will reference to pi sixth. So this would be the same thing as negative four cosine of pi sixth. And same thing over here in the third quadrant, sine is negative, so you get negative I, sorry, negative four I, I need to distribute that four there. Negative four I sine of pi sixth, like so. So cosine of pi sixth is root three over two. And then we get sine of pi sixth, which is gonna be one half. And that then simplifies to give us the negative two root three plus, excuse me, minus two I, like we saw before. So the two products in fact do agree with each other. And it might look like the polar approach is more difficult. And that's only because we go through all these conversions to show that we converted from the polar to the Cartesian so we could see it's the same result. Also, we had to convert from Cartesian to polar so that we could actually do it in the polar method. So don't get deceived by the translation process. It really is much simpler to multiply in the polar form. But of course, since there is a cost to translation when it comes to doing complex products, we have to make a decision. Where is it more efficient just to do it the old fashioned way, the Cartesian way? Or maybe it's the translation worth it. It might not seem like it on this example, but we'll get to examples very soon where it'll be highly advantageous to do exactly that translation.