Calculating the residues of $f(z)=\frac{ze^{iaz}}{z^4+4}$ I have $f(z)=\frac{ze^{iaz}}{z^4+4}$. The poles of $f$ are {$\pm1\pm i$}.
How do I calculate $2\pi i \sum \operatorname{Res}\limits_{z = \pm 1 + i} f(z)$ ? (I'm only looking at positive $i$'s.)
I've tried using the property $\operatorname{Res}f(z) = \frac1{g'(p)}$ and that gives me $\frac1 4 e^{ia(\pm1+i)}$. But then I realised the property only works for one pole. How do I do this?
By l'Hôpital's rule, the residue is$$\begin{align}\lim_{z\to\pm1+i}\frac{z-(\pm1+i)}{(z^3+4z^{-1})e^{-iaz}}&amp;=\lim_{z\to\pm1+i}\frac{1}{(-iaz^3-4iaz^{-1}+3z^2-4z^{-2})e^{-iaz}}\\&amp;=\frac{e^{-a\pm ia}}{-ia(\pm1+i)^3-4ia(\pm1+i)^{-1}+3(\pm1+i)^2-4(\pm1+i)^{-2}},\end{align}$$which I leave you to simplify further using e.g. $(\pm1+i)^{-1}=(\pm1-i)/2$.
Consider, for instance, the residue at $1+i$. Since$$z^4+4=(z-1-i)(1-1+i)(z+1-i)(z+1+i),$$define$$g(z)=\frac{ze^{iaz}}{(z-1+i)(z+1-i)(z+1+i)},$$and you have $f(z)=\frac{g(z)}{z-1-i}$. Therefore,\begin{align}\operatorname{res}_{z=1+i}\bigl(f(z)\bigr)&amp;=\operatorname{res}_{z=1+i}\left(\frac{g(z)}{z-1-i}\right)\\&amp;=g(1+i)\\&amp;=-\frac{ie^{(-1+i)a}}8.\end{align}