Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$i^4 = (i^2)^2 = 1$$
$$ \frac{(1+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
$$ \frac{4(-1-i) + (-i+1+1+i)}{(-i+1+1+i)}$$
$$ \frac{-4-4i + 2}{2}$$
$$ \frac{-2-4i}{2} = -1-2i$$
Where is the error?

By the way, since $i^4=1$ the fraction simplifies as
$$
\frac{i^4+3}{i-1}=\frac{4}{i-1}=\frac{4(i+1)}{-2}=-2(i+1).
$$
