Video: EG19M2-ALGANDGEO-Q06

EG19M2-ALGANDGEO-Q06

04:14

Video Transcript

If π‘₯ plus 𝑦𝑖 is equal to π‘Ž plus 𝑏𝑖 over π‘Ž minus 𝑏𝑖, find π‘₯ squared plus 𝑦 squared.

To find the value of π‘₯ and 𝑦, we’re 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, we’d multiply both parts, essentially the numerator and the denominator, by the conjugate of the denominator. The conjugate of π‘Ž minus 𝑏𝑖 is π‘Ž plus 𝑏𝑖. Remember we simply change the sign between the two terms. We’re therefore going to multiply π‘Ž plus 𝑏𝑖 and π‘Ž minus 𝑏𝑖 by π‘Ž plus 𝑏𝑖. We’ll expand or distribute these brackets as normal.

Let’s look at the FOIL method for π‘Ž plus 𝑏𝑖 multiplied by π‘Ž plus 𝑏𝑖. 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‘Ž multiplied by π‘Ž is π‘Ž squared. O stands for outer. We multiply the outer two terms. That’s π‘Ž multiplied by 𝑏𝑖, which is π‘Žπ‘π‘–. We then multiply the inner two terms which is once again π‘Žπ‘π‘–. And the L stands for last. We’re going to multiply the last term in each of the brackets. And doing so, we get 𝑏 squared 𝑖 squared. Remember 𝑖 is the square root of negative one. So 𝑖 squared is negative one and 𝑏 squared 𝑖 squared is equal to negative 𝑏 squared. Collecting like terms and the expression simplifies to π‘Ž squared plus two π‘Žπ‘π‘– minus 𝑏 squared.

We’ll now multiply the denominator by π‘Ž plus 𝑏𝑖. This time, we get π‘Ž squared plus π‘Žπ‘π‘– minus π‘Žπ‘π‘– minus 𝑏 squared 𝑖 squared. π‘Žπ‘π‘– minus π‘Žπ‘π‘– is zero and negative 𝑏 squared 𝑖 squared becomes negative 𝑏 squared multiplied by negative one which is simply 𝑏 squared. And we’ve gone some way to dividing π‘Ž plus 𝑏𝑖 by π‘Ž minus 𝑏𝑖. Remember we want to write this in the form π‘₯ plus 𝑦𝑖. So we’re going to need to separate the real and complex components. And when we do, we get π‘Ž squared minus 𝑏 squared over π‘Ž squared plus 𝑏 squared plus two π‘Žπ‘π‘– over π‘Ž squared plus 𝑏 squared.

Remember π‘₯ represents the real component of our complex number. It’s π‘Ž squared minus 𝑏 squared over π‘Ž squared plus 𝑏 squared and 𝑦 represents the imaginary component. Here, it’s the coefficient of 𝑖. So it’s two π‘Žπ‘ over π‘Ž squared plus 𝑏 squared. Now that we have expressions for π‘₯ and 𝑦, we can work out the value of π‘₯ squared plus 𝑦 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 can’t 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‘Ž to the power of four minus two π‘Ž squared 𝑏 squared plus 𝑏 to the power of four. We’ll leave the denominator as it is because it will help us simplify the expression at the end. And 𝑦 squared is equal to four π‘Ž squared 𝑏 squared all over π‘Ž squared plus 𝑏 squared all squared.

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

In fact, we can then factorise this expression. It’s π‘Ž squared plus 𝑏 squared all squared. And that means π‘₯ squared plus 𝑦 squared is equal to π‘Ž squared plus 𝑏 squared squared over π‘Ž squared plus 𝑏 squared squared, which is of course one. So for this example, π‘₯ squared plus 𝑦 squared is equal to one.

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