Question Video: Dividing Complex Numbers in Polar Form and Finding Their Quotient in Algebraic Form Mathematics • 12th Grade

Name: Dividing Complex Numbers in Polar Form and Finding Their Quotient in Algebraic Form
Uploaded: 2019-06-20
Description: Given that 𝑍₁ = 5(cos 5𝜃 + 𝑖 sin 5𝜃), 𝑍₂ = cos 4𝜃 + 𝑖 sin 4𝜃, tan 𝜃 = 4/3, and 𝜃 ∈ (0, 𝜋/2], find 𝑍₁/𝑍₂.

Given that 𝑍₁ = 5(cos 5𝜃 + 𝑖 sin 5𝜃), 𝑍₂ = cos 4𝜃 + 𝑖 sin 4𝜃, tan 𝜃 = 4/3, and 𝜃 ∈ (0, 𝜋/2], find 𝑍₁/𝑍₂.

03:08

Video Transcript

Given that 𝑍 one is equal to five cos five 𝜃 plus 𝑖 sin five 𝜃, 𝑍 two is equal to cos four 𝜃 plus 𝑖 sin four 𝜃, tan 𝜃 is equal to four-thirds, and 𝜃 is greater than zero and less than or equal to 𝜋 by two, find 𝑍 one over 𝑍 two.

We’ve been given two complex numbers written in trigonometric or polar form. And we have a few little extra bits of information. But ultimately, we’re looking to find the quotient of these two numbers. We need to recall what we know about finding the quotient of two complex numbers in polar form. To divide them, we divide their moduli and we subtract their arguments.

The general polar form of a complex number is 𝑟 cos 𝜃 plus 𝑖 sin 𝜃, where 𝑟 is the modulus of this number and 𝜃 is its argument. Comparing our complex numbers with the general form, we can see that the modulus of our first complex number, 𝑍 one, is five. And whilst it’s not explicitly written, we can see that the modulus of our second complex number, 𝑍 two, is one. Then, we can see that the argument of 𝑍 one is five 𝜃. And the argument of 𝑍 two is four 𝜃.

To find the modulus of 𝑍 one over 𝑍 two, we’re going to divide 𝑟 one by 𝑟 two. That’s five divided by one, which is simply five. And to find the argument of 𝑍 one divided by 𝑍 two, we’re going to subtract their arguments. That’s five 𝜃 minus four 𝜃, which is of course 𝜃. And we can now substitute this back into the general polar form of our complex number. And we see that 𝑍 one over 𝑍 two is five cos 𝜃 plus 𝑖 sin 𝜃. But where does this extra piece of information coming?

We’re not going to try to solve the equation tan 𝜃 is equal to four-thirds by finding the arctan of both sides. Instead, we recall that tan 𝜃 is equal to opposite over adjacent. And since 𝜃 is greater than zero and less than or equal to 𝜋 by two, we can form a right-angle triangle with an included angle 𝜃. The side opposite to this angle is four units. And the side adjacent to the angle is three units. We also know that the hypotenuse of this right-angle triangle must be five units.

This is the most well-known Pythagorean triple. Three squared plus four squared equals five squared. And since sin 𝜃 is equal to opposite over hypotenuse, we can say that, for this triangle, sin 𝜃 is equal to four over five. It’s four-fifths. We can also say that since cos 𝜃 is adjacent over hypotenuse, it’s in this triangle three-fifths.

And we can now substitute these values for sin 𝜃 and cos 𝜃 into our complex number. It’s five multiplied by three-fifths plus four-fifths 𝑖. And if we distribute this bracket, we see that the fives cancel. And we’re simply left with three as the real part and four as the imaginary component of this complex number.

𝑍 one over 𝑍 two is therefore equal to three plus four 𝑖.