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Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point.

$${z^2 \over (z^2-1)}; \ \ z=1$$

I've broken the function into two parts ${1 \over 2}z({1 \over z+1} + {1 \over z-1})$, I compute the Laurent series but I can't get the answer

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Can anybody explain the question with details?

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  • $\begingroup$ I believe the residue is $\frac{1}{2}$, since the residue for a Laurent series $f(z)$ about the point $z=i$ is given by the coefficient of the $(z-i)^{-1}$ term. I have myself only studied this recently so I suggest you look this up for the details as to why. $\endgroup$ – AlexBowring Apr 11 '14 at 22:08
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$$\frac{z^2}{z^2-1}=1+\frac1{(z-1)(z+1)}=1+\frac12\left(\frac1{z-1}-\frac1{z+1}\right)=$$

$$1+\frac12\frac1{z-1}-\frac12\frac1{2+(z-1)}=1+\frac12\frac1{z-1}-\frac14\frac1{1+\left(\frac{z-1}2\right)}=$$

$$=1+\frac12\frac1{z-1}-\frac14\left(1-\frac{z-1}2+\frac{(z-1)^2}4-\ldots\right)$$

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  • $\begingroup$ This is as I was saying, just with the details. :) $\endgroup$ – Sam T Apr 11 '14 at 22:23
  • $\begingroup$ That is, it is the first method that I suggested. :) $\endgroup$ – Sam T Apr 11 '14 at 22:23
  • $\begingroup$ Wwll, no @SmileySam, as I didn't put $\;\omega:=z-1\;$ and didn't meant to develop around $\;z=0\;$ . That method is handy with harder functions, but really unnecessary with simple ones as in this case. $\endgroup$ – DonAntonio Apr 11 '14 at 22:25
  • $\begingroup$ But the method was identical. I just thought that if the original poster wasn't very used to finding Laurent series, then (s)he may find it helpful. :) $\endgroup$ – Sam T Apr 12 '14 at 7:05
  • $\begingroup$ As you correctly say though, when it is a simple function, not really necessary. :) $\endgroup$ – Sam T Apr 12 '14 at 7:05
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You could try letting $w = z-1$ and then expanding about $w=0$. Then the term with coefficient $w^{-1}$ is the required residue.

Alternatively you can use the fact that the residue is just the formula multiplied by $z - 1$, and then the limit taken as $z \rightarrow 1$, since the singularity is a simple pole.

If this has helped, then please remember to upvote and/or accept my answer! :)

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    $\begingroup$ I am upvoting your answer as I find it helpful. Sometimes newbies don't understand they can/should upvote all the helpful answers, though they can only accept one. +1 $\endgroup$ – DonAntonio Apr 11 '14 at 22:37

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