Question

8-i/2+3i
write complex # in standard form.
I know to FOIL, not sure if I did it right,
I got -13/4-9i then I'm not sure what to do next or if I'm done...so lost can anyone help me?

Answer 1

I will assume that you mean
(8 - i)/(2 + 3i) ....... the brackets are crucial here

= (8 - i)/(2 + 3i) * (2-3i)/(2-3i_
= (16 - 24i - 2i + 3i^2)/(4 + 9)

= (13 - 26i)/13

= 1 - 2i

Answer 2

Thanks for the help Reiny!