# Finding Laurent series

Find the Laurent series for $\frac{(z+1)}{z(z-4)^3}$ in $0<|z-4|<4$.

How can I do this series and what is the most efficent way of doing it? Do I find the partial fractions or since we are given the region, which is defined by $0<|z-4|<4$, then I know its form must correlate accordingly by $\frac{1}{1+(\frac{1}{a})}$, where $a$ is the series where we must manipulate so that we can apply the geometric series in the specified region?

Let $\frac{z-4}{4}=t\implies z=4t+4$
then $|t|<1$ and $$\frac{(1+z)}{z(z-4)^3}=\frac{4t+5}{64(4t+4)(t^3)}=\frac{1}{64t^3}+\frac{1}{256t^3(1+t)}\big(=\frac{1}{256t^3}(1-t+t^2-t^3+\cdots)\big)$$ $$=\frac{1}{64t^3}+\frac{1}{256t^3}-\frac{1}{256t^2}+\frac{1}{256t}-\frac{1}{256}+\frac{t}{256}+\cdots$$ $$=\frac{5}{256t^3}-\frac{1}{256t^2}+\frac{1}{256t}-\frac{1}{256}+\frac{t}{256}+\cdots$$
Now, put back $t=\frac{(z-4)}{4}$
• That is a neat trick, thanks! One question , I dont understand when you did $\frac{1}{256t^3(1-(-t))}\big(=\frac{1}{256t^3}(1-t+t^2-t^3+\cdots)\big)$? – Q.matin Mar 6 '13 at 8:02
• $\frac{1}{1+t}=1-t+t^2-t^3+\cdots$ for $|t|<1$. Multiply bith sides by $\frac{1}{256t^3}$ – Aang Mar 6 '13 at 8:05