Video: Evaluating Algebraic Expressions with Complex Variables

Given that π‘₯ = (2 βˆ’ 𝑖) and 𝑦 = (2 + 𝑖), determine the value of π‘₯Β² βˆ’ π‘₯𝑦 + 𝑦².


Video Transcript

Given that π‘₯ equals two minus 𝑖 and 𝑦 equals two plus 𝑖, determine the value of π‘₯ squared minus π‘₯𝑦 plus 𝑦 squared.

In this question, we’ve been given two complex numbers. π‘₯ is equal to two minus 𝑖. The real part of π‘₯ is two. And its imaginary part is negative one. Remember, that’s the coefficient of 𝑖. 𝑦 is two plus 𝑖. This time, its real part is two. And its imaginary part is plus one. This means π‘₯ and 𝑦 are complex conjugates of one another.

We know that, to find the conjugate of a complex number, often denoted by a bar or a star, we simply change the sign of the imaginary part. And there are some interesting properties of a complex number and its conjugate. In particular, the product of a complex number and its conjugate is a real number. And that means we should be expecting that π‘₯ times 𝑦 gives us a real result. Let’s substitute π‘₯ and 𝑦 into the expression π‘₯ squared minus π‘₯𝑦 plus 𝑦 squared. That’s two minus 𝑖 squared minus two minus 𝑖 times two plus 𝑖 plus two plus 𝑖 squared.

We’re going to individually distribute each pair of parentheses. Before we do though, it’s sensible to write two minus 𝑖 all squared as two minus 𝑖 times two minus 𝑖 and similarly two plus 𝑖 all squared as two plus 𝑖 times two plus 𝑖. And then, we can use whatever method we have for distributing parentheses. I’m going to use the FOIL method.

We multiply the first term in each binomial. Two times two gives us four. We then multiply the outer terms. Two multiplied by negative 𝑖 is negative two 𝑖. We multiply the inner terms. Negative 𝑖 times two is negative two 𝑖. And then, we multiply the last terms. Negative 𝑖 times negative 𝑖 is 𝑖 squared.

We’re now going to repeat this with the next set of parentheses. Remember though, since we’re going to be subtracting the resulting expression, we should keep this in brackets for now. Two times two gives us four. Two times 𝑖 gives us two 𝑖. Negative 𝑖 times two gives us negative two 𝑖. And negative 𝑖 times 𝑖 gives us negative 𝑖 squared.

We repeat this for the final pair of parentheses. This time, we’re going to be adding the resulting expression. So we don’t need to add a second set of brackets. Let’s neaten things up a little. Negative two 𝑖 minus two 𝑖 plus four 𝑖 is zero. We also know that 𝑖 squared is equal to negative one. So we have negative one here, negative one here, and negative one here.

We should also notice that two minus two 𝑖 is zero. So we end up with four minus negative one here. Remember, we were expecting π‘₯𝑦 to be a real number since it’s the product of a complex number and its conjugate. So our result is four minus one minus four minus negative one plus four minus one, which simplifies to three minus five plus three, which is simply one.

And so π‘₯ squared minus π‘₯𝑦 plus 𝑦 squared is equal to one.

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