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How can I expand $\dfrac{\pi \csc(z\pi)}{(2z+1)^3}$? so then I can find the residue ? thanks

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You do not need any particular expansion to find the residues of: $$f(z)=\frac{\pi}{(2z+1)^3\sin(\pi z)}.$$ The zeroes of $\sin(\pi z)$ occurs at $z\in\mathbb{Z}$ and they are all simple zeroes, hence $f(z)$ has simple poles at $z\in\mathbb{Z}$ and a double pole at $z=-\frac{1}{2}$. We trivially have: $$\forall n\in\mathbb{Z},\quad\text{Res}(f(z),z=n) = \frac{(-1)^n}{(2n+1)^3}$$ and: $$\text{Res}\left(f(z),z=-\frac{1}{2}\right) = -\text{Res}\left(\frac{\pi}{8z^3\cos(\pi z)},z=0\right)=-\frac{\pi^3}{16} $$ by considering the Taylor series of $\sec(\pi z)$.

Related (classical) question: Infinite Series $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}$.

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