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I am practicing with the residue calculation and I am having a bit of misunderstanding.

Let $f(z) = \frac{1}{(z^3+1)^2}$. I want to calculate the Res of $f$ in all its singularities which are the poles $-1 , e^{\frac{i \pi}{3}}$ and $ e^{\frac{-i \pi}{3}}$ with order $n =2$ if I am not wrong. I thought of using the formula of residue for poles with order higher than 1, which consists in multiplying $f(z)$ with $(z-z_0)^m$ where $ m \geq n$ and then taking the derivative of the result $ m-1 $ times and dividing it by $(m-1)!$ and calculating the limit as $z$ approaches to $z_0$. Somehow that calculation is not getting simplified or booting to a result. am I doing something wrong here?

David G. Stork
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MWSC1996
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  • Could you share what exactly you did? That would help in pointing out your mistake if there is one. – user170231 Jul 26 '22 at 21:36
  • ok so starting with the pole $-1$ we will get after the applying the method above, $1/(z^2-z+1)^2$ . taking the limit to $-1$ . we should have 1/9 as a result. is that correct? – MWSC1996 Jul 26 '22 at 21:42
  • For a pole $z_0$ of order $n$, the residue there is given by$$\frac1{(n-1)!} \lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}(z-z_0)^n f(z)$$You seem to have just evaluated$$\lim_{z\to-1}(z+1)^2f(z)$$ – user170231 Jul 26 '22 at 21:45
  • Yes I think this solves my question. Thanks a lot! – MWSC1996 Jul 26 '22 at 21:49

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