How would I calculate
$$\mathrm{Res}\left(\frac{\pi}{\sin(\pi z)(2z+1)^3}\right)?$$ I understand it has singularities at $z=n$ and $z=-1/2$, I'm interested in the residue when $z=-1/2$. I know that calculating the residue at a simple pole is $\lim\limits_{z\to n} (z-n) f(z)$ but this is not valid for a function with multiple poles?
Any help would be much appreciated. Thank you!
Best Answer
When dealing with residues at multiple poles the standard approach is to consider derivatives as stated by Winther in the comments. In this case, a translation is enough, since:
$$A=\text{Res}\left(\frac{\pi}{\sin(\pi z)(2z+1)^3},z=-\frac{1}{2}\right)=-\frac{\pi}{8}\cdot\text{Res}\left(\frac{\sec(\pi z)}{z^3},z=0\right)$$ gives: $$ A = -\frac{\pi}{8}\cdot[z^2]\sec(\pi z) = -\frac{\pi}{8}\cdot\frac{\pi^2}{2}=\color{red}{-\frac{\pi^3}{16}}.$$ See also this related question.