Finding Laurent series for $f(z)=\frac{4z^2+2z-4}{z^3-4z}$ around $z=2$ Having
$$f(z)=\frac{4z^2+2z-4}{z^3-4z}$$
find the Laurent series in $z=2$
the scope of $z$ is  $0<|z-2|<2$
here is my approach:
$f(z)=\frac{4z^2+2z-4}{z^3-4z}=\frac{1}{z}+\frac{2}{z-2}+\frac{1}{z+2}=\frac{1}{z-2+2}+\frac{2}{z-2}+\frac{1}{z-2+4}$

but now I can't figure out a way to transform the denumenators to something to make use of maclaurin series like  $\frac{1}{1-(z-2)}$

for the first and last fraction I wrote something like $\frac{1}{2(1-(-\frac{z-2}{2}))}$ and $\frac{1}{4(1-(-\frac{z-2}{4}))}$ I'm not sure about these and for the middle one I don't know what to write I appreciate any help
 A: Consider first the function$$g(z)=\frac{4z^2+2z-4}{z^2+2z}=4-\frac2z-\frac4{2+z}.$$Then, since\begin{align}\frac1z&=\frac1{2-(2-z)}\\&=\frac12\frac1{1-\frac{2-z}2}\\&=\frac12\sum_{n=0}^\infty\left(\frac{2-z}2\right)^n\\&=\sum_{n=0}^\infty\frac{(-1)^n}{2^{n+1}}(z-2)^n\end{align}and since\begin{align}\frac1{2+z}&=\frac1{4+(z-2)}\\&=\frac14\frac1{1+\frac{z-2}4}\\&=\sum_{n=0}^\infty\frac{(-1)^n}{4^{n+1}}(z-2)^n,\end{align}you have\begin{align}g(z)&=4-\sum_{n=0}^\infty(-1)^n\left(\frac1{2^n}+\frac1{4^n}\right)(z-2)^n\end{align}and therefore\begin{align}f(z)&=\frac{g(z)}{z-2}\\&=\frac4{z-2}-\sum_{n=-1}^\infty(-1)^{n+1}\left(\frac1{2^{n+1}}+\frac1{4^{n+1}}\right)(z-2)^n\end{align}
A: Just another way to do it $$f(z)=\frac{4z^2+2z-4}{z^3-4z}$$ Let $z=x+2$ and consider now
$$g(x)=\frac{2 \left(2 x^2+9 x+8\right)}{x \left(x^2+6 x+8\right)}=\frac{1}{x+2}+\frac{1}{x+4}+\frac{2}{x}$$
Now use the McLaurin series around $x=0$
$$\frac 1{x+a}=\sum_{n=0}^\infty (-1)^n a^{-(n+1)}\,x^n$$ Which, by the end, makes
$$g(x)=\frac 2x+\sum_{n=0}^\infty (-1)^n \,2^{-2 (n+1)} \left(2^{n+1}+1\right) \,x^n$$ Replace $x$ by $z-2$ and you are done.
