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Question Video: Solving Equations Including Complex Conjugates Mathematics • 12th Grade

Solve 𝑧𝑧^(βˆ—) + 𝑧^(βˆ—) βˆ’ 𝑧 = 4 + 2𝑖.


Video Transcript

Solve 𝑧𝑧 star plus 𝑧 star minus 𝑧 equals four plus two 𝑖.

Now, let’s just see what we’ve got here. 𝑧 and 𝑧 star denote complex numbers. 𝑧 itself will be in the form π‘Ž plus 𝑏𝑖. π‘Ž and 𝑏 are real constants. π‘Ž represents the real part of the complex number whereas 𝑏 represents its imaginary part. 𝑧 star is the complex conjugate. And that’s sometimes represented using π‘Ž bar as well. Now, to find the complex conjugate of a complex number, we simply change the sign of the imaginary parts. And so if 𝑧, our complex number, is π‘Ž plus 𝑏𝑖, 𝑧 star is π‘Ž minus 𝑏𝑖.

And there are some interesting properties of a complex number and its conjugate. Firstly, when we find their product, we end up with a purely real number. So let’s replace 𝑧 with π‘Ž plus 𝑏𝑖 and 𝑧 star with π‘Ž minus 𝑏𝑖 in our complex equation. When we do, we find 𝑧𝑧 star to be π‘Ž plus 𝑏𝑖 times π‘Ž minus 𝑏𝑖. And then, we add π‘Ž minus 𝑏𝑖 and subtract 𝑧, which is π‘Ž plus 𝑏𝑖. And of course, this is all equal to four plus two 𝑖.

Let’s distribute some of our parentheses. We’ll begin here by multiplying the first term in each expression. That’s π‘Ž times π‘Ž, which is equal to π‘Ž squared. We’ll multiply the outer term in each expression. That gives us negative π‘Žπ‘π‘–. We’ll then multiply the inner terms to get π‘Žπ‘π‘–. And finally, we’ll multiply the last term. That gives us negative 𝑏𝑖 all squared or negative 𝑏 squared 𝑖 squared plus π‘Ž minus 𝑏𝑖 just remains the same. And then, we’re subtracting that third term, that third expression π‘Ž plus 𝑏𝑖. So we get minus π‘Ž minus 𝑏𝑖.

Let’s see if we can neaten this up a little bit. We can see that π‘Ž minus π‘Ž is zero. So those cancel. Similarly, negative π‘Žπ‘π‘– plus π‘Žπ‘π‘– is zero. We then recall that 𝑖 squared is equal to negative one. So we use this. And we write negative 𝑏 squared 𝑖 squared as negative negative one 𝑏 squared. Similarly, we can collect like terms. And we have negative two 𝑏𝑖 here. Simplifying a little further and we find π‘Ž squared plus 𝑏 squared minus two 𝑏𝑖 is all equal to four plus two 𝑖.

Now, this bit is really important. We essentially have two complex numbers. On the left-hand side, the real part of the complex number is simply π‘Ž squared plus 𝑏 squared whereas its imaginary part, remember, that’s the coefficient of 𝑖, is negative two 𝑏. On the right-hand side, the real part of our complex number is four. And its imaginary part is two. And so, for the complex numbers on each side of our equation to be equal, their real parts must be equal. And their imaginary parts must separately be equal. That is, π‘Ž squared plus 𝑏 squared must be equal to four. And negative two 𝑏 must be equal to two.

Well, we can now solve that second equation. We’ll divide through by negative two. And when we do, we find that 𝑏 is equal to negative one. Let’s substitute that into our first equation. When we do, we find π‘Ž squared plus negative one squared equals four. Well, negative one squared is simply one.

We’ll subtract one from both sides of this equation to find that π‘Ž squared is equal to three. And then, we square root both sides, remembering that we’re going to need to take both the positive and negative square root of three. And we find π‘Ž is either positive or negative root three. Now, if we go back to our original equation, we said that 𝑧 was equal to π‘Ž plus 𝑏𝑖. Well, π‘Ž is either positive or negative root three. And 𝑏 is negative one.

So this means the two solutions to our equation are 𝑧 equals root three minus 𝑖 or 𝑧 equals negative root three minus 𝑖.

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