<s>[INST] Laurent series expansion of $\frac{1}{z(z-2)^2}$ at $z=2$ Im struggling a bit with Laurent series expansions, and I am stuck on the following exercise:
Find a Laurent series expansion with center $z=2$ of $f(z)=\frac{1}{z(z-2)^2}$
My solution so far is not a lot:
I've found that there will be one Laurent series at the punctured disk $D(2,2)$ and one $A=(2,2,\infty)$
Starting with the punctured disk $D(2,2)$ I know I want the polynomial to have the form $\frac{1}{1-c*z}$ since we are expanding in our "inner" area.
The part im stuck on is how I should get $\frac{1}{z(z-2)^2}$ = $\frac{1}{z}\frac{1} {(z-2)^2}$ to $\frac{1}{1-c*z}$ to able to write as a series.
Can anyone help me out please?
Thanks!
 [/INST] Using partial fraction expansion, we can write
$$\begin{align}
\frac1{z(z-2)^2}=\frac1{4z}-\frac1{4(z-2)}+\frac1{2(z-2)^2}
\end{align}$$


To obtain the Laurent series for $|z-2|<2$, we write
$$\begin{align}
\frac1{z(z-2)^2}&=\frac1{4(z-2+2)}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac18 \frac1{1+\frac{z-2}2}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac1{2(z-2)^2}-\frac1{4(z-2)}+\frac18\sum_{n=0}^\infty (-1)^n\left(\frac{z-2}{2}\right)^n
\end{align}$$


And to obtain the Laurent series for $|z-2|>2$ we write
$$\begin{align}
\frac1{z(z-2)^2}&=\frac1{4(z-2+2)}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac1{4(z-2)} \frac1{1+\frac2{z-2}}-\frac1{4(z-2)}+\frac1{2(z-2)^2}\\\\
&=\frac1{2(z-2)^2}-\frac1{4(z-2)}+\frac1{8}\sum_{n=1}^\infty (-1)^{n-1}\left(\frac{2}{z-2}\right)^n\\\\
&=\frac1{8}\sum_{n=3}^\infty (-1)^{n-1}\left(\frac{2}{z-2}\right)^n
\end{align}$$
</s>