Video: Solving a Two-Variable Linear Equation with Complex Coefficients

Given that π‘₯ + 𝑦𝑖 = ((4 + 2𝑖)/(1 βˆ’ 2𝑖))^(1/2), find all possible real values of π‘₯ and 𝑦.

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Video Transcript

Given that π‘₯ plus 𝑦𝑖 is equal to four plus two 𝑖 over one minus two 𝑖 all to the power of a half, find all possible real values of π‘₯ and 𝑦.

We’re given a complex number 𝑧 is equal to π‘₯ plus 𝑦𝑖. And we’re told that this is the square root of another complex number four plus two 𝑖 over one minus two 𝑖. The first thing we’re going to do is to simplify the complex number inside the parentheses. And how we do this is we multiply both the numerator and denominator of our complex number by the complex conjugate of the denominator. We can do this because this is actually equal to one, and anything multiplied by one is just itself. And we note that the denominator is one minus two 𝑖. Therefore, the complex conjugate is one plus two 𝑖.

Now, multiplying out our numerators, it’s four times one, which is four, plus four times two 𝑖, which is eight 𝑖, plus two 𝑖 times one, which is two 𝑖, plus two 𝑖 times two 𝑖, which is four 𝑖 squared. And in our denominator, we have one times one plus one times two 𝑖 plus negative two 𝑖 times one plus negative two 𝑖 times positive two 𝑖. That’s negative four 𝑖 squared.

Now, remember that 𝑖 is the square root of negative one, so that 𝑖 squared is actually equal to negative one, so that in our numerator, four 𝑖 squared is negative four. And in our denominator, negative four 𝑖 squared is negative four times negative one, which is four. In our numerator, we have eight 𝑖 plus two 𝑖; that’s 10𝑖. And in our denominator, two 𝑖 minus two 𝑖 is zero. So we have four plus 10𝑖 minus four over one plus four, that is, 10𝑖 over five, which is two 𝑖. And this means that in our parentheses, we have simply two 𝑖, so that 𝑧 is the square root of two 𝑖. This in turn means that 𝑧 squared is equal to two 𝑖.

But remember that 𝑧 is actually π‘₯ plus 𝑦𝑖. So 𝑧 squared is π‘₯ plus 𝑦𝑖 squared, and that’s equal to two 𝑖. And distributing our parentheses on the left-hand side, we have π‘₯ squared plus two 𝑖π‘₯𝑦 plus 𝑦 squared times 𝑖 squared is two 𝑖. And recalling again that 𝑖 squared is negative one, this gives us π‘₯ squared plus two 𝑖π‘₯𝑦 minus 𝑦 squared is equal to two 𝑖.

And making some space, we can compare now real and imaginary parts, so that π‘₯ squared minus 𝑦 squared is equal to zero since there is no real part on the right-hand side. And for the imaginary part, we have two π‘₯𝑦 is equal to two.

Now, let’s just recall that, for a complex number 𝑧 equal to π‘Ž plus 𝑖𝑏, the modulus of 𝑧 is the square root of π‘Ž squared plus 𝑏 squared, so that the modulus of 𝑧 all squared is π‘Ž squared plus 𝑏 squared. And we can note also that that’s equal to the modulus of 𝑧 squared. If we apply this now to our 𝑧 squared, we have the modulus of 𝑧 squared is equal to the modulus of two 𝑖. And that’s equal to the modulus of 𝑧 all squared, which is the modulus of π‘₯ plus 𝑦𝑖 all squared. And since our π‘₯ corresponds with π‘Ž and our 𝑦 corresponds with 𝑏, our modulus of 𝑧 squared is π‘₯ squared plus 𝑦 squared.

Now, the modulus of two 𝑖 is the square root of zero squared, since there’s no real part, plus two squared. And that’s the positive square root of four, which is two. And so we have π‘₯ squared plus 𝑦 squared is equal to two. And we can add this to our collection of equations.

If we call our equation π‘₯ squared minus 𝑦 squared is equal to zero β€œequation one” and our second equation π‘₯ squared plus 𝑦 squared is equal to two β€œequation two”, we have a pair of simultaneous equations in π‘₯ and 𝑦 that we can solve. If we add equations one and two, we have two π‘₯ squared is equal to two, so that π‘₯ squared is equal to one. And taking the square root on both sides, we have π‘₯ equal to positive or negative one.

So now if we go back to our first equation, equation one, which is π‘₯ squared minus 𝑦 squared is equal to zero, this tells us that π‘₯ squared is equal to 𝑦 squared. And if π‘₯ is positive or negative one, then π‘₯ squared is equal to one. And this means that 𝑦 squared is equal to one. And taking square roots on both sides, we have 𝑦 is positive or negative one.

Now, if we look at our third equation, two π‘₯𝑦 is equal to two, dividing through by two, this gives us π‘₯ times 𝑦 is equal to one. And what this tells us is that π‘₯ and 𝑦 must have the same sign in the same root because a positive multiplied by a positive is a positive and a negative multiplied by a negative is also a positive. And so we have that π‘₯ is positive or negative one, 𝑦 is positive or negative one, and they both have to be the same sign in the same root. So that when π‘₯ is positive, one 𝑦 is positive one. And when π‘₯ is negative one, 𝑦 is also negative one.

So if π‘₯ plus 𝑦𝑖 is the complex number four plus two 𝑖 over one minus two 𝑖 all to the power of a half, all the possible real values of π‘₯ and 𝑦 are one, one and negative one, negative one.

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