<s>[INST] Laurent Series of $\frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$ Find the Laurent Series of $\displaystyle \frac{z+1}{z(z-4)^3}$ in $0<|z-4|<4$
I thought about doing partial fraction decomposition first, so I'd have 
$\displaystyle \frac{A}{z}+\frac{B}{z-4}+\frac{C}{(z-4)^2}+\frac{D}{(z-4)^3}$, but that would take awhile...so I'm not sure if this is the right method..?
 [/INST] Setting $\xi=\frac{z-4}{4}$, i.e. $z=4\xi+4$ we have $0<|\xi|<1$ and therefore
\begin{eqnarray}
\frac{z+1}{z(z-4)^3}&=&\frac{4\xi+5}{(4\xi+4)(4\xi)^3}=\frac{4\xi+5}{4^4\xi^3}\cdot\frac{1}{\xi+1}=\left(\frac{1}{4^3\xi^2}+\frac{5}{4^4\xi^3}\right)\cdot\frac{1}{1+\xi}\\
&=&\left(\frac{1}{4^3\xi^2}+\frac{5}{4^4\xi^3}\right)\sum_{k=0}^\infty(-1)^k\xi^k=\frac{1}{4^3}\sum_{k=0}^\infty(-1)^k\xi^{k-2}+\frac{5}{4^4}\sum_{k=0}^\infty(-1)^k\xi^{k-3}\\
&=&\frac{1}{4^3}\sum_{k=-2}^\infty(-1)^k\xi^k-\frac{5}{4^4}\sum_{k=-3}^\infty(-1)^k\xi^k\\
&=&\frac{5}{4^4\xi^3}-\frac{1}{4^4}\sum_{k=-2}^\infty(-1)^k\xi^k\\
&=&\frac{5}{4(z-4)^3}+\sum_{k=-2}^\infty\frac{(-1)^{k+1}}{4^{k+4}}(z-4)^k.
\end{eqnarray}
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