Laurent series of $f(z) = \frac{z - 4}{(z+1)^2(z - 2)}$

Question

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?

Answer 1

Hint:

1. Compute the Laurent expansion of $\frac1{1+z}$
2. Notice that $\frac{d}{dz}\frac{1}{1+z}=-\frac{1}{(1+z)^2}$

Answer 2

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 $t$.