Video: Simplifying Complex Numbers and Converting Them from Algebraic to Polar Form

Given that 𝑧 = ((6𝑖 βˆ’ 6)(4 + 3𝑖))/(1 + 2𝑖)Β², express the complex number 𝑧 in the form of π‘₯ + 𝑦𝑖, and then determine its trigonometric form.

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Video Transcript

Given 𝑧 is equal to six 𝑖 minus six multiplied by four 𝑖 plus three 𝑖 all over one plus two 𝑖 squared, express the complex number 𝑧 in the form π‘₯ plus 𝑦𝑖, and then determine its trigonometric form.

We have been given a rather complicated-looking complex number. And the first thing we’re asked to do is represent it in the form π‘₯ plus 𝑦𝑖. Now whilst 𝑖 is not a variable, remember it’s the square root of negative one. We can distribute the brackets on the top and the bottom of our fraction as we would normally. Let’s look at using the FOIL method.

F stands for first. We’re going to multiply the first term in the first bracket by the first term in the second bracket. Six 𝑖 multiplied by four is 24𝑖. O stands for outer. We’re going to multiply the outer two terms. Six 𝑖 multiplied by three 𝑖 is 18𝑖 squared. And remember, since 𝑖 is equal to the square root of negative one, 𝑖 squared is just negative one. And we can say that 18𝑖 squared is the same as negative 18.

Then, I stands for inner. We’re going to multiply the inner terms in each bracket. Negative six multiplied by four is negative 24. And finally, L stands for last. We multiply the last term in each bracket. Negative six multiplied by three 𝑖 is negative 18𝑖. Collecting like terms, we can see that the numerator of our fraction becomes six 𝑖 minus 42.

Next, we’re going to repeat this process for the denominator, one plus two 𝑖 squared. And since squaring just means to multiply it by itself, this becomes one plus two 𝑖 multiplied by one plus two 𝑖. Expanding this using the FOIL method, we get one plus two 𝑖 plus two 𝑖 plus four 𝑖 squared, which becomes one plus two 𝑖 plus two 𝑖 minus four.

So, the denominator of our fraction simplifies to negative three plus four 𝑖. And we can see that we can rewrite our complex number as six 𝑖 plus 42 over negative three plus four 𝑖. This still isn’t in the form π‘₯ plus 𝑦𝑖 though. So, how do we divide these two numbers?

We’re going to need to multiply both the numerator and the denominator of this fraction by the conjugate of the denominator. And remember, for a complex number of the form π‘Ž plus 𝑏𝑖, its conjugate, often denoted 𝑧 bar, is π‘Ž minus 𝑏𝑖. Essentially, we change the sign between the two terms. So, the conjugate of negative three plus four 𝑖 is negative three minus four 𝑖. And we’re going to need to expand the brackets on the top and the bottom of this fraction again.

And doing so, we get negative 18𝑖 minus 24 squared plus 126 plus 168𝑖 all over nine plus 16. And, once again, we use the fact that 𝑖 squared is equal to negative one. And negative 24𝑖 squared is positive 24. So, our complex number simplifies to 150 plus 150𝑖 all over 25. And since 150 and 25 share a common factor of 25, we divide through. And we can see that our complex number is now in the form π‘₯ plus 𝑦𝑖. It’s six plus six 𝑖.

Our very last step is to write this number in trigonometric form. For a complex number in algebraic or rectangular form, π‘Ž plus 𝑏𝑖, we write it in trigonometric form as π‘Ÿ cos πœƒ plus 𝑖 sine πœƒ, where π‘Ÿ is the modulus of the complex number and πœƒ is its argument. And we can use these two-conversion formulae to help us find the modulus and the argument. The modulus is the square root of π‘Ž squared plus 𝑏 squared. And the argument is arctan, or inverse tan, of 𝑏 over π‘Ž.

For our complex number, π‘Ž is the constant; It’s six. And 𝑏 is the coefficient of 𝑖; It’s also six. So, we can see that the modulus of our complex number is the square root of six squared plus six squared, which is the square root of 72, or six root two. And the argument is the arctan of six over six, which is πœ‹ by four.

And it’s always worth double-checking this value on the Argand diagram. Remember, the horizontal axis represents the real component and the vertical axis represents the imaginary component of our complex number. Six plus six 𝑖 lies in the first quadrant as shown.

And we always measure of the argument from the horizontal axis in a counterclockwise direction. So, we can see that πœ‹ by four is certainly a sensible value for our argument. All that’s left is for us to substitute the value of the modulus and argument into the general trigonometric form. And we see that it’s six root two multiplied by cos of πœ‹ by four plus 𝑖 sin of πœ‹ by four.

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