### Video Transcript

If π₯ plus π¦π is equal to π plus ππ over π minus ππ, find π₯ squared plus π¦ squared.

To find the value of π₯ and π¦, weβre going to need to begin by evaluating this expression on the right-hand side of our equation. Here, we have the quotient of two complex numbers. To divide these complex numbers, weβd multiply both parts, essentially the numerator and the denominator, by the conjugate of the denominator. The conjugate of π minus ππ is π plus ππ. Remember we simply change the sign between the two terms. Weβre therefore going to multiply π plus ππ and π minus ππ by π plus ππ. Weβll expand or distribute these brackets as normal.

Letβs look at the FOIL method for π plus ππ multiplied by π plus ππ. The F part of FOIL stands for first. We multiply the first term in the first bracket by the first term in the second bracket. π multiplied by π is π squared. O stands for outer. We multiply the outer two terms. Thatβs π multiplied by ππ, which is πππ. We then multiply the inner two terms which is once again πππ. And the L stands for last. Weβre going to multiply the last term in each of the brackets. And doing so, we get π squared π squared. Remember π is the square root of negative one. So π squared is negative one and π squared π squared is equal to negative π squared. Collecting like terms and the expression simplifies to π squared plus two πππ minus π squared.

Weβll now multiply the denominator by π plus ππ. This time, we get π squared plus πππ minus πππ minus π squared π squared. πππ minus πππ is zero and negative π squared π squared becomes negative π squared multiplied by negative one which is simply π squared. And weβve gone some way to dividing π plus ππ by π minus ππ. Remember we want to write this in the form π₯ plus π¦π. So weβre going to need to separate the real and complex components. And when we do, we get π squared minus π squared over π squared plus π squared plus two πππ over π squared plus π squared.

Remember π₯ represents the real component of our complex number. Itβs π squared minus π squared over π squared plus π squared and π¦ represents the imaginary component. Here, itβs the coefficient of π. So itβs two ππ over π squared plus π squared. Now that we have expressions for π₯ and π¦, we can work out the value of π₯ squared plus π¦ squared. To square each of these fractions, we square the numerators and denominators individually.

Remember we do need to be a little bit careful. We canβt just square each of the individual terms. We need to expand these brackets like we did before. And when we do, we get the numerator to be π to the power of four minus two π squared π squared plus π to the power of four. Weβll leave the denominator as it is because it will help us simplify the expression at the end. And π¦ squared is equal to four π squared π squared all over π squared plus π squared all squared.

We can add these two expressions by simply adding their numerators. And since negative two π squared π squared plus four π squared π squared is two π squared π squared, we get that π₯ squared plus π¦ squared is equal to π to the power of four plus two π squared π squared plus π to the power of four all over π squared plus π squared squared. Now in fact, if we refer to the numerator of this expression we see itβs very similar to the expansion of π squared minus π squared all squared.

In fact, we can then factorise this expression. Itβs π squared plus π squared all squared. And that means π₯ squared plus π¦ squared is equal to π squared plus π squared squared over π squared plus π squared squared, which is of course one. So for this example, π₯ squared plus π¦ squared is equal to one.