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Question Video: Solving Equations Involving Complex Numbers Mathematics • 12th Grade

Let 𝑧₁ = 4π‘₯ + 2𝑦𝑖 and 𝑧₂ = 4𝑦 + π‘₯𝑖, where π‘₯ and 𝑦 ∈ ℝ. Given that 𝑧₁ βˆ’ 𝑧₂ = 5 + 2𝑖, find 𝑧₁ and 𝑧₂.

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Video Transcript

Let 𝑧 one equal four π‘₯ plus two 𝑦𝑖 and 𝑧 two equal four 𝑦 plus π‘₯𝑖, where π‘₯ and 𝑦 are real numbers. Given that 𝑧 one minus 𝑧 two is equal to five plus two 𝑖, find 𝑧 one and 𝑧 two.

Let’s look carefully at what we’ve been given. We’ve been given two complex numbers in terms of π‘₯ and 𝑦. And we know these are complex numbers because we’re told that π‘₯ and 𝑦 are real numbers. That’s an important definition of a complex number. Both the real and imaginary parts of the complex numbers must be made up by real numbers. We’re also told that the difference between these two numbers is five plus two 𝑖.

Let’s recall: to subtract complex numbers, we simply subtract the real parts and then subtract the imaginary parts separately. This means that the real part of 𝑧 one minus 𝑧 two must be equal to the difference between the real parts of 𝑧 one and 𝑧 two. The real part of 𝑧 one minus 𝑧 two is five. The real part of 𝑧 one is four π‘₯, and the real part of our second complex number is four 𝑦. So five is equal to four π‘₯ minus four 𝑦.

Let’s repeat this process for our imaginary numbers. The imaginary part of the difference is two. The imaginary part of 𝑧 one is two 𝑦, and the imaginary part of 𝑧 two is π‘₯. So two equals two 𝑦 minus π‘₯. And now we see we have a pair of simultaneous equations in π‘₯ and 𝑦. We can use any method we’re comfortable with to solve these.

Now I think substitution lends itself quite nicely to these equations. Let’s rearrange this second equation to make π‘₯ the subject. We add π‘₯ to both sides and then subtract two. And we get π‘₯ equals two 𝑦 minus two. We then substitute this into our first equation. And we see that five is equal to four lots of our value of π‘₯, which is two 𝑦 minus two. And then we subtract that four 𝑦.

We distribute these parentheses by multiplying each term by four. And we see that five is equal to eight 𝑦 minus eight minus four 𝑦. Eight 𝑦 minus four 𝑦 is four 𝑦. We’ll solve this equation by adding eight to both sides to get 13 equals four 𝑦. And then we’ll divide through by four. And we see that 𝑦 is equal to 13 over four.

We could substitute this value back into any of our original equations. But it’s sensible to choose the rearranged form of the second equation. π‘₯ is equal to two multiplied by 13 over four minus two. Two multiplied by 13 over four is the same as 13 over two. And two is the same as four over two. 13 over two minus four over two is nine over two. And this is usually where we would stop.

But we’ve been asked to find the complex numbers 𝑧 one and 𝑧 two. So we need to substitute our values for π‘₯ and 𝑦 into each of these. We get 𝑧 one equals four multiplied by nine over two plus two multiplied by 13 over four 𝑖. That’s 18 plus 13 over two 𝑖. 𝑧 two is four multiplied by 13 over four plus nine over two 𝑖. This is 13 plus nine over two 𝑖.

And it’s sensible to check our answer by subtracting 𝑧 two from 𝑧 one and checking we do indeed get five plus two 𝑖. We subtract their real parts. 18 minus 13 is five, as required. And we subtract their imaginary parts. 13 over two minus nine over two is four over two, which simplifies to two. And the imaginary part is two as required.

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