I want to find the Laurent expansion of $\tan z$ about $a = 0$ in the punctured ball $0 < |z| < \frac{\pi}{2}$. Because $\tan z$ is holomorphic on this punctured ball, its Laurent expansion has no negative coefficients and so $\tan z = \sum_{k \geq 0} a_k z^k$, with $$a_k = \frac{1}{2\pi i} \int_{|z| = \frac{\pi}{4}} \frac{\tan z}{z^{n+1}} dz = \operatorname{Res} \left( \frac{\tan z}{z^{n+1}},0\right).$$ Now I can evaluate the residue on the right because the function in question has a pole of order $z^{n+1}$; I do this using the formula
$$\operatorname{Res} \left( \frac{\tan z}{z^{n+1}},0\right) = n! \frac{d^n}{dz^n} \tan z\Big|_{z=0}.$$
My question is: Apart from calculating this explicit, under time pressure is there a quicker way to evaluate the Laurent series of $\tan z$ or the residue in question above?
