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

Solve 2𝑧 βˆ’ 𝑧 bar = 5 in β„‚.

01:41

Video Transcript

Solve two 𝑧 minus 𝑧 bar equals five in β„‚.

Here we have a complex number. And we can say that that could be of the form π‘Ž plus 𝑏𝑖, where π‘Ž and 𝑏 are real numbers. 𝑧 bar is its conjugate. That’s π‘Ž minus 𝑏𝑖. And β„‚ is used to denote the set of complex numbers. Let’s substitute 𝑧 and 𝑧 bar into our equation.

When we do, we see that two multiplied by π‘Ž plus 𝑏𝑖 minus π‘Ž minus 𝑏𝑖 equals five. Then, we distribute the brackets by multiplying the real and imaginary part by the number on the outside. For the first bracket, that’s two multiplied by π‘Ž and two multiplied by 𝑏𝑖. And for the second bracket, that’s negative one multiplied by π‘Ž plus negative one multiplied by negative 𝑏𝑖. So we get two π‘Ž plus two 𝑏𝑖 minus π‘Ž plus 𝑏𝑖 equals five. And of course, we can collect like terms. And we see that π‘Ž plus three 𝑏𝑖 is equal to five.

Now, this number is purely real. Or we can say it’s a complex number whose imaginary part is zero. And once we’ve identified that, we can equate the real and imaginary parts. We see that π‘Ž must be equal to five. And three 𝑏 must be equal to zero. In fact, if three 𝑏 is equal to zero, 𝑏 must also be equal to zero. We’re solving for 𝑧. And we’ve established that π‘Ž β€” its real part is equal to five. And 𝑏 β€” its imaginary part is equal to zero. So we could say that 𝑧 is equal to five plus zero 𝑖 though we don’t need to write the imaginary part. So we say that 𝑧 is simply equal to five.

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