# Video: Dividing Complex Numbers in Polar Form

Given that πβ = 1 and πβ = (cos 3π + π sin 3π)Β², find the trigonometric form of πβ/πβ.

03:14

### Video Transcript

Given that π one equals one and π two is equal to cos of three π plus π sin of three π all squared, find the trigonometric form of π one divided by π two.

There are several formulae weβll need to consider here. The first is the product formula. And it says for two complex numbers π one with the modulus of π one and an argument of π one and π two with the modulus of π two and an argument of π two, their product π one π two can be found by multiplying the moduli and adding the arguments.

We can extend this into squaring a complex number and say that to find the square of a complex number π in polar form, we square the modulus and double the argument. This will allow us to find the value of cos three π plus π sine of three π all squared. We double the arguments and we get cos of six π plus π sin of six π.

The quotient formula says that for the two same complex numbers π one and π two, their quotient π one divided by π two can be found by dividing the moduli and subtracting the arguments. π one is not yet in trigonometric form though; itβs in rectangular form. The rectangular form for a complex number is π equals π plus ππ.

If we compare the general rectangular form to our complex number π one, we can see that the value for π is one and the value for π is zero. So we need to find a way now to represent the real and complex components of our number in terms of π and π, essentially writing in trigonometric form.

The modulus π is given by the square root of π squared plus π squared. This comes from the Pythagorean theorem. To find π, we use the formula tan π is equal to π over π. So letβs substitute what we know about our complex number π one into these formulae.

The modulus is the square root of one squared plus zero squared, which is just one. tan π is equal to π over π, which is zero over one. Zero divided by one is zero. So to solve this equation for π, we can find the arc tangent of zero, which is simply zero. And now that we have the values for π and π letβs substitute them into the formula for the general polar or trigonometric form of the complex number π one.

Doing this, we can see that π one is equal to one multiplied by cos of zero plus π sine of zero. Letβs retain this and clear a little bit of space. Weβre looking to find the value of π one divided by π two. Letβs go back to the quotient formula.

We divide the moduli and we get one over one. We subtract the arguments and we get zero minus six π. Remember sine and cosine functions are periodic with the period of two π. So we can add two π to the argument and the complex number itself will remain unchanged. And since the modulus of the complex number is one, we actually donβt need to write that.

And weβre done: the quotient π one divided by π two is equal to cos of two π minus six π plus π sin of two π minus six π.