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I have the function $\frac{z+2}{z(z+1)}$ which I am trying to find the integral for a circle of radius 3 centered at the origin. I'm able to use Cauchy's Second Residue theorem to find it to be $2 \pi i$, but let's say I want to find it by just expanding the function into an appropriate Laurent Series and finding the residue through there. How would you go about expanding this into a Laurent series centered around $z=-1$? I can only find the residue for the simple pole $z=0$.

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  • $\begingroup$ Decompose into partial fractions to get two rational terms of the form $\frac az+\frac b{z+1}$. Write the first term as $-\frac a{1-(1+z)}$ so that it more closely resembles a geometric series. $\endgroup$
    – user170231
    Oct 3 '17 at 4:41
  • $\begingroup$ I think this helped me out. I expanded the equation into $\frac{-2}{1-(1+z)}$-$\frac{1}{z+1}$. The residue for the first fraction is 0 since there are no negative exponents in the series, and the second one is clearly -1. Thanks. $\endgroup$ Oct 3 '17 at 4:56
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First, consider $\frac{z+2}z=1+\frac2z$. If $|z+1|<1$, then\begin{align}1+\frac2z&=1-\frac2{-z}\\&=1-\frac2{1-(z+1)}\\&=1-2\sum_{n=0}^\infty(z+1)^n\\&=-1-2\sum_{n=1}^\infty(z+1)^n\end{align}and therefore$$\frac{z+2}{z(z+1)}=\frac{-1}{z+1}-2-2(z+1)+\cdots$$So, the residue at $-1$ is $-1$.

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  • $\begingroup$ Shouldn’t it be -1/(z+1) -2 -2(z+1) - ... ? $\endgroup$
    – Ryukyu
    Sep 14 '18 at 12:34
  • $\begingroup$ @Poujh I've edited my answer. Thank you. $\endgroup$ Sep 14 '18 at 12:38

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