Dividing complex numbers in polar form | Precalculus | Khan Academy

So we are given these two complex numbers and we want to know what ( w_1 ) divided by ( w_2 ) is. So pause this video and see if you can figure that out.

All right, now let's work through this together. The form that they've written this in actually makes it pretty straightforward to spot the modulus and the argument of each of these complex numbers. The modulus of ( w_1 ) we can see out here is equal to 8, and the argument of ( w_1 ) we can see is ( \frac{4\pi}{3} ) if we're thinking in terms of radians, so ( \frac{4\pi}{3} ) radians.

Then similarly for ( w_2 ), its modulus is equal to 2 and its argument is equal to ( \frac{7\pi}{6} ).

Now in many videos we have talked about when you multiply one complex number by another, you're essentially transforming it. So you are going to scale the modulus of one by the modulus of the other, and you're going to rotate the argument of one by the argument of the other. I guess you could say you're going to add the angles.

So another way to think about it is if you have the modulus of ( \frac{w_1}{w_2} ), well then you're just going to divide these moduli here. So this is just going to be ( \frac{8}{2} ) which is equal to 4.

And then the argument of ( \frac{w_1}{w_2} ): this is, you could imagine you're starting at ( w_1 ) and then you are going to rotate it clockwise by ( w_2 )'s argument. So this is going to be ( \frac{4\pi}{3} - \frac{7\pi}{6} ).

And let's see what this is going to be. If we have a common denominator, ( \frac{4\pi}{3} ) is the same thing as ( \frac{8\pi}{6} - \frac{7\pi}{6} ) which is going to be equal to ( \frac{\pi}{6} ).

And so we could write this. The quotient ( \frac{w_1}{w_2} ) is going to be equal to, if we wanted to write it in this form, its modulus is equal to 4.

It's going to be ( 4 \times \cos\left(\frac{\pi}{6}\right) + i \times \sin\left(\frac{\pi}{6}\right) ). Now ( \cos\left(\frac{\pi}{6}\right) ) we can figure out. ( \frac{\pi}{6} ) is the same thing as a 30 degree angle, and so the cosine of that is ( \frac{\sqrt{3}}{2} ).

( \frac{\sqrt{3}}{2} ) and the sine of ( \frac{\pi}{6} ) we know from our 30-60-90 triangles is going to be one-half. So this is one-half.

And so if you distribute this 4, this is going to be equal to ( 4 \times \frac{\sqrt{3}}{2} ) is ( 2\sqrt{3} ), and then ( 4 \times \frac{1}{2} ) is 2, so plus ( 2i ), and we are done.