# Question Video: Solving Two-Variable Linear Equations with Complex Coefficients Mathematics • 12th Grade

If π + ππ = (β3 β 5π)/(β3 + 5π), is it true that πΒ² + πΒ² = 1?

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### Video Transcript

If π plus ππ is equal to negative three minus five π over negative three plus five π. Is it true that π squared plus π squared equals one?

Weβve been given an equation involving complex numbers. Remember, a complex number is of the form π plus ππ. π is known as the real component of the complex number. Whereas π is known as its imaginary component. Itβs quite clear that to be able to work out whether π squared plus π squared is equal to one, we need to work out what π and π are equal to. So what weβre going to do is weβre going to simplify the quotient. Weβre going to divide negative three minus five π by negative three plus five π.

So how do we divide complex numbers? Well, to divide complex numbers, we multiply both the numerator and denominator of the fraction by the complex conjugate of the denominator. Sometimes denoted π§ star or π§ bar, the complex conjugate of the number π plus ππ is π minus ππ. Essentially, you change the sign of the imaginary part. The denominator of our fraction is negative three plus five π. Changing its sign, we find its complex conjugate to be equal to negative three minus five π. Weβre therefore going to multiply both the numerator and denominator of this fraction by negative three minus five π.

Letβs begin with the numerator. Weβll do this in the little space on the right. Notice that this expression looks a lot like the product of two binomials. And in fact, we do treat it a little like one. We distribute the parentheses either using the FOIL method or the grid method. Letβs use the FOIL method. F stands for first. We multiply the first term in the first expression by the first term in the second. Negative three times negative three is nine. O stands for outer. We multiply the outer term in each expression. Negative three multiplied by negative five is positive 15. So get positive 15π. Then I stands for inner. We multiply the inner terms. And when we do, we get one more 15π. Finally, L stands for last. We multiply the last two terms. A negative multiplied by a negative is a positive. So we get 25π squared.

But remember, π is the number which is the solution to the equation π₯ squared equals negative one. We can say that π squared equals negative one. So the numerator becomes nine plus 15π plus 15π plus 25 times negative one. Well, that last term is of course just negative 25. And so we collect like terms. And we see that the numerator is equal to negative 16 plus 30π.

So what about the denominator? Weβll use the same method to distribute our parentheses. And it should become clear quite quickly why we actually multiply the denominator by its complex conjugate. Remember, F stands for first. We get nine. We multiply the outer terms to get plus 15π. Then, the inner terms to get minus 15π. And finally, the last terms to get negative 25π squared. We can now see that 15π minus 15π is zero. And negative 25 times negative one is positive 25. So weβre going to calculate nine plus 25, which is 34. Okay, great! Our fraction becomes negative 16 plus 30π over 34.

By separating the real and imaginary part of our fraction and dividing each bit by 34, we find π plus ππ is equal to negative eight over 17 plus 15 over 17π. π is the real part of our complex number. So it must be equal to negative eight over 17. π is the imaginary part. Itβs the coefficient of π. So itβs 15 over 17. All thatβs left is to substitute these two numbers into π squared plus π squared and see if we actually get one. Thatβs negative eight over 17 squared plus 15 over 17 squared. That gives us 64 over 289 plus 225 over 289, which is indeed equal to one.

So, yes, it is indeed true that π squared plus π squared equals one, in this case.