Video: MATH-ALG+GEO-2018-S1-Q4B

If (π‘₯ + 𝑦𝑖)(1 βˆ’ 3𝑖) = 37 [(1/(3 βˆ’ 4πœ”Β²) + (1/(7 + 4πœ”Β²)), find the values of the real numbers π‘₯ and 𝑦.

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Video Transcript

If π‘₯ plus 𝑦𝑖 times one minus three 𝑖 equals 37 times one over three minus four πœ” squared plus one over seven plus four πœ” squared, find the values of the real numbers π‘₯ and 𝑦.

So we have a complex equation to solve. Complex in the sense that it is not any difficult, but it involves complex numbers. We have the complex number 𝑖. In fact, 𝑖 is pure imaginary, which satisfies 𝑖 squared equals negative one. And we also have a complex number πœ” which is a cube root of one.

Let’s start by expanding the left-hand side. We do this as we would for any real equation. We get π‘₯ minus three 𝑖π‘₯ plus 𝑦𝑖 minus three 𝑖 squared 𝑦. Now, how about on the right? We start off by combining these two fractions. To do this, we write the two fractions with a common denominator, which is the product of their denominators. Their sum will have this denominator. And the numerator will be the sum of the numerators. So that’s seven plus four πœ” squared plus three minus four πœ” squared. We notice here that the πœ”-squared terms in the numerator cancel. And the constant terms seven and three will combine. But we’ll leave that to the next line.

Let’s not neglect the left-hand side though. There’s some simplification we can do. We have a factor of 𝑖 squared, and we know that 𝑖 squared is negative one. And so minus three 𝑖 squared 𝑦 becomes minus three times negative one 𝑦, which is minus negative three 𝑦 or plus three 𝑦. Staying with the left-hand side then, we can collect the real terms together and also the imaginary terms, taking out that common factor of 𝑖. And now we continue our simplification of the right-hand side. We’ve already mentioned that the seven and the three in the numerator combine to give 10.

And now we expand the brackets in the denominator. And now we can simplify. Let’s take the denominator and simplify it separately. πœ” cubed is one, and so πœ” to the power of four, which is πœ” cubed times πœ”, is just πœ”. We can also combine the πœ”-squared terms to get 21 minus 16πœ” squared minus 16πœ”. But we’re still not done. There’s another property of πœ” that we can use apart from πœ” cubed equals one. There’s also the fact that πœ” squared plus πœ” plus one equals zero. And hence, πœ” squared is negative πœ” minus one. Making this substitution for πœ” squared, we get 21 minus 16 times negative πœ” minus one minus 16πœ”. Expanding the brackets then, we see that there’s some simplification we can do. The πœ” terms cancel leaving us with just 21 plus 16 which is 37.

So now let’s put this value back in context. The fraction inside the brackets becomes 10 over 37. And the 37 in the denominator cancels with the 37 outside the brackets. And so, we’re left with the equation π‘₯ plus three 𝑦 plus 𝑦 minus three π‘₯ 𝑖 equals 10. We have a complex number on the left-hand side. Let’s make it more obviously complex on the right-hand side by adding zero 𝑖. And now we can compare the real and imaginary parts. Two complex numbers are equal if and only if their real and imaginary parts are equal. As the real parts are equal, π‘₯ plus three 𝑦 must be 10. And as the imaginary parts are equal, one minus three π‘₯ must be zero.

We have two linear equations in π‘₯ and 𝑦 which we can solve simultaneously. Perhaps the easiest way to do this is to add three π‘₯ to both sides of the second equation, finding 𝑦 in terms of π‘₯ and then substituting three π‘₯ for 𝑦 in the first equation to get an equation only in terms of π‘₯. That gives π‘₯ plus three times three π‘₯ equals 10. And three times three π‘₯ is nine π‘₯, and π‘₯ plus nine π‘₯ is 10π‘₯. So we get that 10π‘₯ is 10, and π‘₯ is therefore equal to one. Now we know that 𝑦 is three π‘₯. So as π‘₯ equals one, 𝑦 must be three. And this is our final answer. We’ve found the values of the real numbers π‘₯ and 𝑦 in our equation.

One thing to note is that we could’ve solved this in a slightly different way. Instead of expanding the left-hand side, we could’ve left it how it is. We’d still need to do all the work to show that the right-hand side simplifies to just 10. But having shown this, we could then just have divided both sides by one minus three 𝑖. Multiplying both numerator and denominator by the complex conjugate of one minus three 𝑖 and then comparing the real and imaginary parts, we’d get the same answer. π‘₯ equals one and 𝑦 equals three.

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