# Thread: imaginary number question

1. ## imaginary number question

I feel like this is probably a simple problem, but i'm stumped.
the question is "Write the expression in standard form"
$\frac{(2+i)^2(-i)}{1+i}$

I know the standard form, it's just that when I divide my answer doesn't match the answer in the back of the book which is $\frac{1}{2}-\frac{7}{2i}$

2. All yu need to do is exapnd the numerator and then multiply both numerator and denominator by the conmplex conjugate

$\frac{(2+i)^2(-i)}{1+i}$

$\frac{(2+i)^2(-i)}{1+i}$

$= \frac{(4+4i+i^2)(-i)}{1+i}$

$= \frac{(4+4i-1)(-i)}{1+i}$

$= \frac{(3+4i)(-i)}{1+i}$

$= \frac{-3i-4i^2}{1+i}$

$= \frac{-3i+4}{1+i}$

then

$= \frac{-3i+4}{1+i}\times \frac{1-i}{1-i}$

can you finish?