1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.