Question Video: Simplifying Complex Number Expressions Using Conjugates and Converting Them to Polar Form Mathematics

Simplify (6 βˆ’ 6𝑖)/βˆ’2𝑖, giving your answer in both algebraic and trigonometric form.


Video Transcript

Simplify six minus six 𝑖 over negative two 𝑖, giving your answer in both algebraic and trigonometric form.

To divide two complex numbers in rectangular form, that’s the general form 𝑧 equals π‘Ž plus 𝑏𝑖, we first need to find the complex conjugate of the denominator. For a complex number π‘Ž plus 𝑏𝑖, its complex conjugate is π‘Ž minus 𝑏𝑖. The denominator of our fraction is negative two 𝑖. We can think of that as zero minus two 𝑖, and this means that the complex conjugate of the denominator is zero plus two 𝑖 is actually just two 𝑖. To evaluate this fraction then, we multiply both the numerator and the denominator by two 𝑖. Six multiplied by two 𝑖 is 12𝑖, and negative six 𝑖 multiplied by two 𝑖 is negative 12 𝑖 squared. Negative two 𝑖 multiplied by two 𝑖 is negative four 𝑖 squared. Remember though, 𝑖 squared is just negative one, so our fraction simplifies to 12 𝑖 plus 12 all over four.

We can divide through by this four and we get three 𝑖 plus three. Notice I’ve actually written it as three plus three 𝑖 to match the general algebraic form for a complex number. In algebraic or rectangular form then, the complex number is 𝑧 is equal to three plus three 𝑖. And if we compare this to the general form, we can see that π‘Ž, the real component, has a value of three and 𝑏, the imaginary component, also has a value of three. We need to find a way to write this complex number in trigonometric or polar form; that’s 𝑧 is equal to π‘Ÿ cos πœƒ plus 𝑖 sin πœƒ. π‘Ÿ is the modulus of a complex number 𝑧, and πœƒ is the argument in polar form or trigonometric form. πœƒ can be in degrees or radians. In exponential form though, it does need to be in radians, so we need to find a way to represent the real and component parts of our number 𝑧 in terms of π‘Ÿ and πœƒ.

In fact, we can use this formulae to help us: the modulus π‘Ÿ is equal to the square root of π‘Ž squared plus 𝑏 squared; and to find πœƒ, we use tan πœƒ is equal to 𝑏 over π‘Ž. Let’s substitute what we know about our complex number into these formulae. π‘Ž is three and 𝑏 is three. So the modulus π‘Ÿ is the square root three squared plus three squared, which is the square root of 18. This simplifies to three root two. Tan πœƒ is three divided by three or one. We can solve this by finding the arctangent of one, which is πœ‹ over four. And then we can substitute all values for π‘Ÿ and πœƒ into the general trigonometric form of a complex number. And doing so, we get that 𝑧 is equal to three root two cos of πœ‹ over four plus 𝑖 sin of πœ‹ over four. And remember, in our algebraic or rectangular form, we said that 𝑧 was equal to three plus three 𝑖.

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