# Calculating residues of multiple poles?

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!

• first webiste popping up when asking google about "residue calculation". I can't believe that people are even too lazy to ask a searchengine... :( en.wikipedia.org/wiki/Residue_%28complex_analysis%29 – tired May 26 '15 at 13:39
• If $f$ has a pole of order $n$ at $z=z_0$ then $\text{Res}(f,z_0) = \lim_{z\to z_0}\frac{1}{(n-1)!}\frac{d^{n-1}}{dz^{n-1}}[(z - z_0)^{n}f(z)]$. Plug in $n=3$ and $z_0 = -1/2$. – Winther May 26 '15 at 13:43
• @Winter Do poles other than 1 have any contribution to a contour integral? I thought they all came out to be 0. For example $e^{1/z}$, only $1/z$ in the expansion has a contribution. – grdgfgr May 26 '15 at 13:59
• @grdgfgr No only the $1/z$ term in the Laurent expansion matter for countour integration and that is exactly what the formula I gave above is trying to calculate. To see this lets take the example of $f$ having a pole of order $2$ at $z=0$. This means that $f(z) = \frac{a_{-2}}{z^2} + \frac{a_{-1}}{z^1} + a_0 + \ldots$ so $\frac{d}{dz}[f(z) z^2] = \frac{d}{dz}[a_{-2} + a_{-1}z + a_0 z^2 + \ldots] = a_{-1} + 2a_0 z + \ldots$. Taking the limit $z\to 0$ gives us the residue $a_{-1}$. – Winther May 26 '15 at 18:12

$$A=\text{Res}\left(\frac{\pi}{\sin(\pi z)(2z+1)^3},z=-\frac{1}{2}\right)=-\frac{\pi}{8}\cdot\text{Res}\left(\frac{\sec(\pi z)}{z^3},z=0\right)$$ 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.