Question Video: Solving a Linear Equation Involving a Complex Number and its Complex Conjugate Mathematics • 12th Grade

Video Transcript

Solve two 𝑧 minus 𝑧 bar equals five in the set of complex numbers.

In this question, we have an equation involving a complex number 𝑧. We would say in general that 𝑧 is of the form 𝑎 plus 𝑏𝑖, where 𝑎 and 𝑏 are real constants. 𝑧 bar, which is sometimes written as 𝑧 star, is the complex conjugate of 𝑧. And we find this by changing the sign of the imaginary parts. So, the conjugate of our general complex number 𝑎 plus 𝑏𝑖 is 𝑎 minus 𝑏𝑖.

Let’s substitute these complex numbers into our original equation. When we do, we get two times 𝑎 plus 𝑏𝑖 minus 𝑎 minus 𝑏𝑖 equals five. We’re now going to distribute our parentheses. Two times 𝑎 is two 𝑎, and two times 𝑏𝑖 is two 𝑏𝑖. Negative one times 𝑎 is negative 𝑎, and negative one times negative 𝑏𝑖 is positive 𝑏𝑖. So, our equation is two 𝑎 plus two 𝑏𝑖 minus 𝑎 plus 𝑏𝑖 equals five.

Let’s collect like terms. We get two 𝑎 minus 𝑎, which is 𝑎, and two 𝑏𝑖 plus 𝑏𝑖, which is three 𝑏𝑖. So, we get 𝑎 plus three 𝑏𝑖 equals five. What we’re going to do next is equate the real and imaginary parts of both sides of our equation. On the left-hand side, the real part is 𝑎. And on the right, the real part is five. On the left, the imaginary part is three 𝑏. Remember, that’s the coefficient of 𝑖. And we could say that on the right, the imaginary part is zero.

By equating the real parts, we find 𝑎 is equal to five. Then, by equating the imaginary parts, we get three 𝑏 is equal to zero. But this, of course, means that 𝑏 itself is equal to zero. So, going back to that general form of a complex number, 𝑎 plus 𝑏𝑖, we can say that 𝑧 must be equal to five plus zero 𝑖, which is simply five.