Complex number - wrong result at the end

Question

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?

Answer 1

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

Answer 2

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

Answer 3

Following your line of algebraic manipulation, \begin{align} \frac{i^4+3}{i-1} &= \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i} \\ &= \frac{(i^4+3)(-1-i)}{(i-1)(-1-i)} \\ &= \frac{(1+3)(-1-i)}{-i-i^{2}+1+i} \\ \end{align} I'll leave to you to finish the details. I think the mistake was in your first line of algebra in the numerator.

Answer 4

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