Question Video: Finding the Quotient of a Complex Number Raised to a Power in Polar Form Mathematics

Given that 𝑍₁ = cos (2πœ‹/3) + 𝑖 sin (2πœ‹/3) and 𝑍₂ = cos (πœ‹/6) + 𝑖 sin (πœ‹/6), find (𝑍₁⁡/𝑍₂²).

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Video Transcript

Given that 𝑍 sub one is equal to the cos of two πœ‹ by three plus 𝑖 sin of two πœ‹ by three and 𝑍 sub two is equal to the cos of πœ‹ by six plus 𝑖 sin of πœ‹ by six, find 𝑍 sub one to the fifth power divided by 𝑍 sub two squared.

In this question, we’re given two complex numbers, 𝑍 sub one and 𝑍 sub two, in polar form. We need to use this to determine 𝑍 sub one to the fifth power divided by 𝑍 sub two squared. And there’s many different ways we can approach answering this question. Since we need to find 𝑍 sub one to the fifth power and 𝑍 sub two squared and both of these are given in polar form, let’s start by recalling de Moivre’s theorem.

One version of de Moivre’s theorem tells us for any integer value of 𝑛 and real number πœƒ, cos πœƒ plus 𝑖 sin πœƒ all raised to the 𝑛th power is equal to cos of π‘›πœƒ plus 𝑖 sin of π‘›πœƒ. And we can use this to evaluate 𝑍 sub one to the fifth power and 𝑍 sub two squared.

Let’s start with 𝑍 sub one raised to the fifth power. This is the cos of two πœ‹ by three plus 𝑖 sin of two πœ‹ by three all raised to the fifth power. And our exponent is an integer value of five. De Moivre’s theorem then tells us we can evaluate this by multiplying each of the arguments by five. And since five times two πœ‹ by three is 10πœ‹ by three, this gives us the cos of 10πœ‹ by three plus 𝑖 sin of 10πœ‹ by three. We can also use this to find 𝑍 sub two all squared. That’s the cos of πœ‹ by six plus 𝑖 sin of πœ‹ by six all squared. This time, our exponent is two, which is once again an integer. And de Moivre’s theorem tells us we can evaluate this by multiplying the argument by two. And since two times πœ‹ by six is πœ‹ by three, this gives us the cos of πœ‹ by three plus 𝑖 sin of πœ‹ by three. This then gives us that 𝑍 sub one raised to the fifth power divided by 𝑍 sub two squared is equal to the cos of 10πœ‹ by three plus 𝑖 sin of 10πœ‹ by three all divided by the cos of πœ‹ by three plus 𝑖 sin of πœ‹ by three.

This is then the quotient of two complex numbers given in polar form. And we can evaluate this by recalling how we find the quotient of two complex numbers given in polar form. If the complex number in our denominator is nonzero, then π‘Ÿ times the cos of πœƒ plus 𝑖 sin of πœƒ all divided by 𝑠 times the cos of πœ‘ plus 𝑖 sin of πœ‘ is equal to π‘Ÿ over 𝑠 multiplied by the cos of πœƒ minus πœ‘ plus 𝑖 sin of πœƒ minus πœ‘. In other words, we find the quotient of their moduli and we find the difference in their arguments.

And in the quotient of the two complex numbers we’re given, we can see there’s no coefficient. In other words, their moduli are both one. So we only need to find the difference in the arguments. We get cos of 10πœ‹ by three minus πœ‹ by three plus 𝑖 sin of 10πœ‹ by three minus πœ‹ by three. And 10πœ‹ by three minus πœ‹ by three is equal to nine πœ‹ by three, which simplifies to give us three πœ‹. And remember, we can add and subtract integer multiples of two πœ‹ from the argument. This gives us an argument of πœ‹, which simplifies to give us cos of πœ‹ plus 𝑖 sin of πœ‹.

And it’s worth noting we can simplify this further. The cos of πœ‹ is negative one and the sin of πœ‹ is zero. So this simplifies to give us negative one. However, we’ll leave our answer in polar form because this is the form 𝑍 sub one and 𝑍 sub two is given in.

Therefore, we were able to show if 𝑍 sub one is cos of two πœ‹ by three plus 𝑖 sin of two πœ‹ by three and 𝑍 sub two is cos of πœ‹ by six plus 𝑖 sin of πœ‹ by six, then 𝑍 sub one to the fifth power divided by 𝑍 sub two squared in polar form is cos of πœ‹ plus 𝑖 sin of πœ‹.

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