<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] 
The function
\begin{align*}
 f(z)&=\frac{1}{z(z-2)^2} = \frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
\end{align*}
   has a simple pole at $z=0$ and a pole of order two at $z=2$.
Since we want to find a Laurent expansion with center $2$, we look at the other pole $0$ and see we have to distinguish two regions.
\begin{align*}
 0<|z-2|<2\qquad\quad\text{and}\qquad\qquad
 2<|z-2|
 \end{align*} 
  
  
*
  
*The first region $ |z-2|<2$ is a disc with center $2$, radius $2$ and the pole $0$ at the boundary of the disc. In the interior of this disc the fraction with pole $0$ admits a representation as power series at $z=2$, while the fractions with pole $2$ admit  a representation  as principal part of a Laurent series at $z=2$.
  
*The second region $|z-2|>2$ containing all points outside the disc with center $2$ and radius $2$ admits for all fractions a representation as principal part of a Laurent series at $z=2$.

A power series expansion of $\frac{1}{z+a}$ at $z=2$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{(a+2)+(z-2)}=\frac{1}{a+2}\cdot\frac{1}{1+\frac{z-2}{a+2}}\\
&=\frac{1}{a+2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{z-2}{a+2}\right)^n\tag{1}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(a+2)^{n+1}}(z-2)^n\tag{2}
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=2$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{(z-2)+(a+2)}=\frac{1}{z-2}\cdot\frac{1}{1+\frac{a+2}{z-2}}\\
&=\frac{1}{z-2}\sum_{n=0}^{\infty}(-1)^n\left(\frac{a+2}{z-2}\right)^n\tag{1}\\
&=\sum_{n=0}^{\infty}(-1)^n(a+2)^n\frac{1}{(z-2)^{n+1}}\\
&=\sum_{n=1}^{\infty}(-1)^{n-1}(a+2)^{n-1}\frac{1}{(z-2)^{n}}\tag{3}
\end{align*}

We can now obtain the Laurent expansion of $f(z)$ at $z=2$ for both regions
  
  
*
  
*Region: $|z-2|<2$
  
  
  \begin{align*}
f(z)&= \frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\cdot\frac{1}{2+(z-2)}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(z-2)^n-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\tag{4}\\
&=\frac{1}{8}\sum_{n=-2}^{\infty}\frac{(-1)^n}{2^n}(z-2)^n
\end{align*}
  
  
*
  
*Region: $2<|z-2|$
  
  
  \begin{align*}
f(z)&=\frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\frac{1}{(z-2)+2}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
&=\frac{1}{4}\sum_{n=1}^{\infty}(-1)^{n-1}2^{n-1}\frac{1}{(z-2)^n}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\tag{5}\\
&=\frac{1}{8}\sum_{n=3}^{\infty}(-1)^{n-1}2^{n}\frac{1}{(z-2)^n}
\end{align*}

Comment:


*

*In (1) we use the geometric series
\begin{align*}
\frac{1}{1+z}=\frac{1}{1-(-z)}=\sum_{n=0}^\infty (-z)^n=\sum_{n=0}^\infty (-1)^nz^n\qquad\qquad |z|<1
\end{align*}

*In (4) we apply the series expansion of (2)

*In (5) we apply the principal part representation of (3)
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