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

Solve 2π§ β π§ bar = 5 in β.

01:47

### Video Transcript

Solve two π§ minus π§ bar equals five in the set of complex numbers.

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

Letβs substitute these complex numbers into our original equation. When we do, we get two times π plus ππ minus π minus ππ equals five. Weβre now going to distribute our parentheses. Two times π is two π, and two times ππ is two ππ. Negative one times π is negative π, and negative one times negative ππ is positive ππ. So, our equation is two π plus two ππ minus π plus ππ equals five.

Letβs collect like terms. We get two π minus π, which is π, and two ππ plus ππ, which is three ππ. So, we get π plus three ππ equals five. What weβ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 π. And on the right, the real part is five. On the left, the imaginary part is three π. Remember, thatβs the coefficient of π. And we could say that on the right, the imaginary part is zero.

By equating the real parts, we find π is equal to five. Then, by equating the imaginary parts, we get three π is equal to zero. But this, of course, means that π itself is equal to zero. So, going back to that general form of a complex number, π plus ππ, we can say that π§ must be equal to five plus zero π, which is simply five.