### Video Transcript

Solve two π§ minus π§ bar equals
five in β.

Here we have a complex number. And we can say that that could be
of the form π plus ππ, where π and π are real numbers. π§ bar is its conjugate. Thatβs π minus ππ. And β is used to denote the set of
complex numbers. Letβs substitute π§ and π§ bar into
our equation.

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

Now, this number is purely
real. Or we can say itβs a complex number
whose imaginary part is zero. And once weβve identified that, we
can equate the real and imaginary parts. We see that π must be equal to
five. And three π must be equal to
zero. In fact, if three π is equal to
zero, π must also be equal to zero. Weβre solving for π§. And weβve established that π β its
real part is equal to five. And π β its imaginary part is
equal to zero. So we could say that π§ is equal to
five plus zero π though we donβt need to write the imaginary part. So we say that π§ is simply equal
to five.