Could someone give me some pointers how to calculate:
$$\operatorname*{res}_{z=0} \frac{z^{2n}+1}{z^n[iaz^2+(1+a^2)z-ia]}$$
I don't think it's possible using the limit formula, but I'm having problems providing a Laurent-expansion.
#EDIT Given $0<a<1$
Split the fraction into two parts.
$$\frac{z^{2n}}{z^n[iaz^2 + (1+a^2)z - ia]} = \frac{z^n}{iaz + (1+a^2)z - ia}$$
is holomorphic at $0$ for $n > 0$, so that part has residue $0$ at $0$. The other part is best treated by a partial fraction decomposition,
$$\frac{1}{iaz^2 + (1+a^2)z - ia} = \frac{1}{(iaz+1)(z-ia)} = \frac{c_1}{iaz+1} + \frac{c_2}{z-ia},$$
with constants $c_1,c_2$ that are to be determined. Then you can expand both summands into geometric series, and the residue of
$$\frac{1}{z^n[iaz + (1+a^2)z-ia]}$$
at $0$ is the sum of the coefficients of $z^{n-1}$ in the two geometric series.