Question

Divide the complex numbers 8+i over 2-i

Answer 1

start with
\[\frac{8+i}{2-i}\times \frac{2+i}{2+i}\] i .e. multiply top and bottom by the conjugate of the denominator

Answer 2

the conjugate of \(a+bi\) is \(a-bi\) and this works because
\((a+bi)(a-bi)=a^2+b^2\) a real number

Answer 3

so what you are going to get is
\[\frac{8+i}{2-i}\times \frac{2+i}{2+i}=\frac{(8+i)(2+i)}{2^2+1^2}\]

Answer 4

or
\[\frac{(8+i)(2+i)}{5}\]
then your last job is to multiply out in the numerator