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 $t=\frac{(z-4)}{4}$

  • $\begingroup$ That is a neat trick, thanks! One question , I dont understand when you did $\frac{1}{256t^3(1-(-t))}\big(=\frac{1}{256t^3}(1-t+t^2-t^3+\cdots)\big)$? $\endgroup$ – Q.matin Mar 6 '13 at 8:02
  • $\begingroup$ $\frac{1}{1+t}=1-t+t^2-t^3+\cdots$ for $|t|<1$. Multiply bith sides by $\frac{1}{256t^3}$ $\endgroup$ – Aang Mar 6 '13 at 8:05
  • $\begingroup$ Ohh, I see what you did. I thought it was part of the problem. Thanks a lot for the help! $\endgroup$ – Q.matin Mar 6 '13 at 8:08
  • 1
    $\begingroup$ You're welcome. Have a Good Day :) $\endgroup$ – Aang Mar 6 '13 at 8:09

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