Question Video: Dividing Complex Numbers in Polar Form | Nagwa Question Video: Dividing Complex Numbers in Polar Form | Nagwa

Question Video: Dividing Complex Numbers in Polar Form Mathematics • Third Year of Secondary School

Given that 𝑍₁ = 1 and 𝑍₂ = (cos 3𝜃 + 𝑖 sin 3𝜃)², find the trigonometric form of 𝑍₁/𝑍₂.

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Video Transcript

Given that 𝑍 one equals one and 𝑍 two is equal to cos of three 𝜃 plus 𝑖 sin of three 𝜃 all squared, find the trigonometric form of 𝑍 one divided by 𝑍 two.

There are several formulae we’ll need to consider here. The first is the product formula. And it says for two complex numbers 𝑍 one with the modulus of 𝑟 one and an argument of 𝜃 one and 𝑍 two with the modulus of 𝑟 two and an argument of 𝜃 two, their product 𝑍 one 𝑍 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 𝑍 in polar form, we square the modulus and double the argument. This will allow us to find the value of cos three 𝜃 plus 𝑖 sine of three 𝜃 all squared. We double the arguments and we get cos of six 𝜃 plus 𝑖 sin of six 𝜃.

The quotient formula says that for the two same complex numbers 𝑍 one and 𝑍 two, their quotient 𝑍 one divided by 𝑍 two can be found by dividing the moduli and subtracting the arguments. 𝑍 one is not yet in trigonometric form though; it’s in rectangular form. The rectangular form for a complex number is 𝑍 equals 𝑎 plus 𝑏𝑖.

If we compare the general rectangular form to our complex number 𝑍 one, we can see that the value for 𝑎 is one and the value for 𝑏 is zero. So we need to find a way now to represent the real and complex components of our number in terms of 𝑟 and 𝜃, essentially writing in trigonometric form.

The modulus 𝑟 is given by the square root of 𝑎 squared plus 𝑏 squared. This comes from the Pythagorean theorem. To find 𝜃, we use the formula tan 𝜃 is equal to 𝑏 over 𝑎. So let’s substitute what we know about our complex number 𝑍 one into these formulae.

The modulus is the square root of one squared plus zero squared, which is just one. tan 𝜃 is equal to 𝑏 over 𝑎, which is zero over one. Zero divided by one is zero. So to solve this equation for 𝜃, we can find the arc tangent of zero, which is simply zero. And now that we have the values for 𝑟 and 𝜃 let’s substitute them into the formula for the general polar or trigonometric form of the complex number 𝑍 one.

Doing this, we can see that 𝑍 one is equal to one multiplied by cos of zero plus 𝑖 sine of zero. Let’s retain this and clear a little bit of space. We’re looking to find the value of 𝑍 one divided by 𝑍 two. Let’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 𝜃. 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 don’t need to write that.

And we’re done: the quotient 𝑍 one divided by 𝑍 two is equal to cos of two 𝜋 minus six 𝜃 plus 𝑖 sin of two 𝜋 minus six 𝜃.

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