<s>[INST] 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?
 [/INST] $$
\frac{(i^4+3)(-i-1)}{(i-1)(-i-1)}\neq \frac{(i^4+3)(-i-1)+(i-1)(-i-1)}{(i-1)(-i-1)}
$$
</s>