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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?

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  • $\begingroup$ You may appreciate following technique : just integrate the expansion of $\ \tan'(x)=1+\tan(x)^2\,$ repetitively. $\endgroup$ Nov 14, 2013 at 12:58

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You can do this quickly if you know the Maclurin expansion of $\tan{z}$ off the top of your head:

$$\tan{z} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} B_{2 n} 2^{2 n} \left (2^{2 n}-1\right )}{(2 n)!} z^{2 n-1}$$

where $B_{2 n}$ is a Bernoulli number.

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