<s>[INST] Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stuck. The general idea I know is to get each of the 3 terms above into the form $D\cdot \frac{1}{1-z}$, where $D$ is a constant. But for the first term, it would be $\frac{1}{4}\cdot\frac{1}{(0-z)}$. I can't get rid of the zero. 
 [/INST] Let $\zeta = z - 2$. We are trying to find the Laurent series for $1/(\zeta+2)\zeta^2$ centred at $\zeta = 0$, so
$$ f = \frac{1}{(\zeta+2)\zeta^2} = \zeta^{-2} \frac{1}{\zeta + 2}
$$
The latter is holomorphic in a neighbourhood of $0$, so let $(a_j)_{j=0}^\infty$ such that
$$ \frac{1}{\zeta + 2} = \sum_{j=0}^\infty a_j \zeta^j, 1 = 2\sum_{j=0}^\infty a_j \zeta^j + \sum_{j=0}^\infty a_j \zeta^{j+1}
$$
Therefore,
$$ 1 = 2a_0 + \sum_{j=1}^\infty 2a_j \zeta^j + \sum_{j=1}^\infty a_{j-1} \zeta^j = 2a_0 + \sum_{j=1}^\infty (2a_j + a_{j-1}) \zeta^j 
$$
so $a_0 = \frac{1}{2}$, and we establish the recurrence relation $a_{j+1} = -\frac{1}{2}a_j$. Hence $a_j = (-1/2)^j$.
The original series becomes
$$ \zeta^{-2} \frac{1}{\zeta + 2} = \zeta^{-2} \sum_{j=0}^\infty (-\frac{1}{2})^j \zeta^j
$$
Then substitute $\zeta = z-2$.
To find the region of convergence, notice that $f$ is holomorphic except at the poles $z = 0, z = 2$. The result follows easily from Laurent's theorem.
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