Multiplying complex numbers in polar form | Precalculus | Khan Academy

We're given two different complex numbers here and we want to figure out what is the product. Pause this video and see if you can figure that out.

All right, now let's work on this together. So we know from the form that it's written here that the modulus of ( w_1 ) is equal to 3, and we know that the argument of ( w_1 ) is equal to 330 degrees. By the same line of reasoning, we know that the modulus of ( w_2 ) is equal to 2, and that the argument of ( w_2 ) is going to be equal to, we can see that right over here, 120 degrees.

Now, when you multiply complex numbers, you could view it as one transforming the other. We've seen this in multiple examples. So let's imagine that we are transforming ( w_2 ) by multiplying it by ( w_1 ). So what is going to happen?

Well, let me write it here. So what's the resulting modulus of ( w_1 \times w_2 )? Well, we're just going to scale up ( w_2 )'s modulus by ( w_1 )'s modulus, or essentially we're just going to multiply the two. So this is going to be equal to 6, ( 3 \times 2 ).

And then the argument of ( w_1 \times w_2 ). If we start at ( w_2 )'s argument, which is 120 degrees, and then we rotate it by ( w_1 )'s argument, well then you're going to add these two angles. That gets you to 450 degrees. So this is equal to 450 degrees, which is more than a complete rotation.

If we wanted to give it an angle between 0 and 360 degrees, if we just subtract 360 from that, that is going to be equal to 90 degrees. So we can rewrite this here, or we can rewrite the product as ( w_1 \times w_2 ) is equal to its modulus 6 times cosine of its argument 90 degrees plus ( i ) times sine of its argument.

Now we know what the cosine and sine of 90 degrees is. Cosine of 90 degrees is equal to 0, and sine of 90 degrees is equal to 1. So all of this simplifies quite nicely. All you're left with is a 6 times ( i ). So this is equal to ( 6i ), and we are done.