Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$ I want to expand function $$f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$$ for $1 < |z| < 2$.
My work so far
By partial fraction decomposition we can get that:
$$\frac{z - 4}{(z + 1)^2(z - 2)} = \frac{-3(z - 4)}{(z + 1)^2} + \frac{\frac 1 9 (z - 4)}{z - 2}$$
Now deriving Laurent expansion for second term is not so hard:
$$\frac{1}{z - 2} = -\frac{1}{2} \frac{1}{1 - \frac z 2} = - \frac 1 2 \sum_{n = 0}^\infty (\frac z 2)^n = - \sum_{n = 0}^\infty \frac{z^2}{2^{n + 1}}$$
So we have that:
$$ \frac{\frac 1 9 (z - 4)}{z - 2} = \sum_{n = 0}^\infty - \frac 1 9 \cdot \frac{z^n}{2^{n + 1}}(z - 4)$$
But I'm quite struggling with expanding $\frac{1}{(z + 1)^2}$. I tried several possibilities but those all led me nowhere. Can I ask you for a hand in expanding form $\frac{1}{(z + 1)^2}$?
EDIT
We know that $\frac{1}{1 + z} = \sum_{n = 0}^\infty \frac{(-1)^n}{z^{n + 1}}$
As I understood the idea with derivative is to:
$$\frac{d(\frac{1}{1 + z})}{dz} = \sum_{n = 0}^\infty \frac{d(\frac{(-1)^n}{z^{n + 1}})}{dz}$$
$$- \frac{1}{(1 + z)^2} = \sum_{n = 0}^\infty - (n + 1)z^{-n-2}(-1)^n$$
$$\frac{1}{(1 + z)^2} = \sum_{n = 0}^\infty (n + 1)z^{-(n + 2)}(-1)^n$$
Does it make any sense to you?

You want a series for $\frac1{(1+z)^2}$ that’s good outside the unit circle, don’t you? Thus a series in $1/z$. So write
$$
\frac1{(1+z)^2}=\frac1{z^2}\frac1{(1+1/z)^2}\,.
$$
If you find that confusing, expand $t^2\frac1{(1+t)^2}$ in powers of $t$, and substitute $1/z$ for $t$.
