 Let's talk now about complex division using polar forms, of course. We've talked about this with multiplication and exponents here, where recognizing exponents, of course, are just iterated multiplication. How does division work? Where division is the inverse operation multiplication? Well, it turns out using the polar form, it makes life a whole lot easier to do division here. Take, for example, z1 is equal to r1. It's modulus cosine of theta1 plus i sine theta1 with theta1 is its argument. And then z2 is equal to r2, its modulus, times cosine of theta2 plus i sine of theta2, where in this case theta2 would be the argument of z2. Then it turns out when you divide the two complex numbers, if you take its quotient z1 divided by z2, you're going to divide the moduli r1 divided by r2, and then you're going to subtract the arguments cosine of theta1 minus theta2 plus i sine theta1 minus theta2. And this seems to match up with what we seen previously, right? When it came to multiplication z1 times z2, you would multiply together the moduli r1 r2, and then you added together the arguments there, theta1 plus theta2 plus i sine of theta1 plus theta2. This is what we saw previously. When it came to exponents, right, z to the n right there, you would take the exponent of the modulus, so r to the n, and then you would just take the multiple of the argument in theta plus i sine theta, like so, right? Because notice that exponents is iteration of multiplication, right? But when you iterate addition, you end up with multiplication. So the analog seems to carry forward here. Notice here that you're going to divide the moduli. So whatever operation you're doing, that's exactly what you do the moduli. Multiplication of complex numbers gives you multiplication of moduli. Exponents of complex numbers gives you an exponent of the modulus, and therefore division of complex numbers gives you division of the moduli, all right? But the argument's always one order of operation lower, right? When it came to multiplication of complex numbers, you added the arguments there, one degree lower. When you took the exponents, you multiplied together the arguments. Now we're into inverse operations. When we divide the complex numbers, we're going to subtract the arguments like you see on the screen right there. And the proof of this gets really nasty if you only want to use trigonometric identities, because now you've got to use angle difference identities. It gets very intense, like the multiplication case would have taken. But we're going to take the direction of using Euler's identity, which admittedly in our lecture series, you did not prove Euler's identity, but with a little bit of calculus under our belts, it's a super easy proof. It really is. It's very, very immediately follows from the McLaurin series of sine cosine and e to the x. But that's a conversation for another day, I suppose. If you take z1 divided by z2, if you put them in complex polar form, you'll get r1 e to the i theta 1 on top. That's just this friend right here, r1 e to the i theta 1. And then the denominator would look like r2 e to the i theta 2, which we then see over here, r2 e to the i theta 2. And so then if we just treat this like it's a fraction, you're going to get r1 divided by r2 like so. That's the first part to recognize. And the next part, we have these exponentials with the common base. We can subtract the exponents. So we get e to the i theta 1 minus i theta 2. There's a common divisor of i there. You can factor it out, r1 over r2, and then e to the i over, or well, not over, but times theta 1 minus theta 2. And then if you switch back to the Cartesian form, you end up with r1 over r2 times cosine of theta 1 minus theta 2 plus i sine of theta 1 minus theta 2. And so again, we go through these arguments so you can see what's happening here. It's helping us remember these formulas if we have an understanding where they come from. We subtract the arguments because the arguments are exponents. And when you divide, you subtract exponents. When you think of it in that perspective, it seems very intuitive. Why do we divide the moduli but subtract the arguments? Well, the arguments are exponents, and division means subtract exponents. This is just exponent laws. So let's do a calculation of such. Let's find the quotient of the complex number 20 times cosine of pi over 12 plus i sine of pi over 12, and then divide that by 4 cosine of pi over 12 plus i sine of pi over 12. If you think of this in terms of their polar form, which is in right now, this is just looking like 20 e to the 5 pi i over 12. And we divide that by 4 e to the pi i over 12. So by the formula we just saw, we should divide the moduli. 20 divided by 4 is 5. And then we're going to subtract the exponents there. We get e to the, well, 5 pi over 12 minus pi over 12, that's going to be 4 pi i over 12, which we can simplify that exponent. We end up with e to the pi i over 3. This is the same thing as 5 times cosine of 5 pi thirds plus i sine of pi thirds, like so. And we honestly could just leave it like that, given that's how the original numbers were written as well in that complex form right there, the complex trigonometric form. But pi thirds, that's a special angle. I really just feel like I want to finish it off here. So we're going to end up with cosine of pi thirds is going to be 1 half. Sine of pi thirds is root 3 over 2. And so we end up with 5 halves plus 5 root 3 over 2 times i. And so this is then the complex quotient of those complex numbers in the trigonometric form. Super easy, like so. So you could honestly leave it that way if you want the trigonometric form, but I did also go all the way down to the Cartesian form just for kicks and giggles there. All right, we've saw an example where we multiplied together these numbers before, both in the standard Cartesian form and in this new trigonometric polar form, and we compared the two methods together. I want to try that right now again. So let's consider the quotient first in the traditional Cartesian form, right? So we're going to take z1 to be 1 plus i root 3 over z2, which is negative root 3 plus i. And remember that the usual strategy here is we're going to multiply top and bottom by the conjugate. So we're going to get negative root 3 minus i and then negative root 3 minus i like so. We've foiled out the numerator. So you're going to get negative root 3, 1 times negative root 3, 1 times negative i. It's going to give me negative i. Then you get i root 3 times negative root 3. That's going to give you a negative 3i. And then finally, i root 3 times negative i. That, of course, is going to give you a positive root 3 like so. And then in the denominator, it's a lot easier because it's using the conjugate. I know it's just going to be a sum of squares. You're going to get negative root 3 squared plus 1 squared in that situation. So we end up with in the numerator, the square roots of 3 actually cancel each other out. It would look here. And then you get the complex numbers adding together. You get negative i minus 3i. It's a negative 4i. In the denominator, what do we end up with? We end up with a 3 plus 1, which of course is a 4. And so this is going to turn out to be negative i when we're done and simplified there. So that's the answer we're looking for. Complex division wasn't the worst thing here. It definitely is a little bit harder than a real division there, but we were able to do it using these conjugates and the such. Okay? What about the polar form? Well, we previously have computed the polar forms of these complex numbers. So I'm not going to do the details again. Take the look at the link in the video if you want to see how that happens here. But we computed the moduli of these two numbers. Both of their moduli turn out to be the square root of 2 here. So we get root 2 root 2. Then their arguments are what we have to compute next. The argument for the first number we got was pi thirds. And the argument of the second number turned out to be 5 pi thirds. Excuse me, 5 pi 6 is what we computed previously. So the conversion part is the hardest part you perhaps could argue here because once you get into the polar form, then the quotient is going to be super easy. We get z1 divided by z2. This is going to equal the square root of 2 e to the pi i thirds divided by the square root of 2 e to the 5 pi i sixth like so. You can see very quickly that their moduli are going to cancel each other out. I mean, after all, the modulus of our final number was 1 in that situation. So then we have to subtract the exponents. So we're going to get e to the i times. We have pi thirds minus 5 pi thirds. So I need to write the pi thirds as I had to change the denominator. So I'm going to change it to 2 pi sixth like so. And so we end up with this 2 pi sixth minus 5 pi sixth so 2 pi sixth minus 5 pi sixth like so. That's going to give us when we're done e to the i times well 2 minus 5 is negative 3 pi sixth. And that does of course simplify to be equal to e to the i negative pi halves like so. So I have to compute cosine of negative pi halves, which is zero. And we have to then compute sine of negative pi halves, which is actually negative one, and that gives us the negative i that we were looking for. So in terms of the division process, it's a lot easier. But of course, you have to convert from Cartesian to polar and polar to Cartesian necessary. We could have honestly stopped right here if we were happy with the polar form. That's okay. And so I just want to show you the difference between these two methods. So if you just have to divide two complex numbers, maybe the conversion process is a little bit weighty on us. But when it comes to, of course, repeated division, like we saw with exponents, right, exponents are repeated multiplication. It's so much faster to convert and then do it in polar form. Well, what about division? What if you repeat division over and over and over again? Well, this leads to roots. This leads to radicals. And we'll talk about this more in the next video where it's essentially, well, it's essential to do these in polar form.