<s>[INST] How to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ I want to determine the Laurent expansion of $\tan{z}$ around $z=0$ that is convergent in $z=\pi$ (only the first couple of terms). Now I know that if $\sum_{n=-\infty}^{\infty}c_nz^n$ then
$$c_n=\frac{1}{2\pi i}\int_K \frac{\tan(z)}{z^{n+1}}$$
Where I need to choose $K$ in the annulus $\frac{\pi}{2}<|K(t)|<\frac{3\pi}{2}$. This means that
$$c_n=\text{Res}_{z=0}\frac{\tan(z)}{z^{n+1}}+\text{Res}_{z=-\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}}+\text{Res}_{z=\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}}$$
However I'm not sure how I can determine these residues, without already knowing the Laurent expansion of $\tan(z)$. Only the residue in $z=0$ is easy, since the 'standard' Taylor series of $\tan(z)$ is valid there. However, in $z=\pm\frac{\pi}{2}$ I have no clue of how I could approach this.
Thanks
Edit:
Using Mathematica I found out that
$$\text{Res}_{z=-\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}} = (-1)^{1-n}\left(\frac{2}{\pi}\right)^n$$
$$\text{Res}_{z=\frac{\pi}{2}}\frac{\tan(z)}{z^{n+1}} = -\left(\frac{2}{\pi}\right)^n$$
But I have no Idea how it finds this. Any help would be much appreciated.
 [/INST] Oke so I decided to just brute-force this for the residu in $z=\frac{\pi}{2}$ ($z=-\frac{\pi}{2}$ is similar)
$$\begin{align}
(z-\frac{\pi}{2})\tan{z-\frac{\pi}{2}}&=(z-\frac{\pi}{2})\frac{\sin{z-\frac{\pi}{2}}}{\cos{z-\frac{\pi}{2}}}\\
&=(z-\frac{\pi}{2})\frac{1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi}{2})^4\cdots}{-(z-\frac{\pi}{2})+\frac{1}{3!}(z-\frac{\pi}{2})^3-\frac{1}{5!}(z-\frac{\pi}{2})^5\cdots}\\
&=-\frac{1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi}{2})^4\cdots}{1-\lbrack\frac{1}{3!}(z-\frac{\pi}{2})^2-\frac{1}{5!}(z-\frac{\pi}{2})^4\cdots\rbrack}\\
&=-\lbrack 1-\frac{1}{2!}(z-\frac{\pi}{2})^2+\frac{1}{4!}(z-\frac{\pi}{2})^4\cdots\rbrack\lbrack 1+ \left(\frac{1}{3!}(z-\frac{\pi}{2})^2-\frac{1}{5!}(z-\frac{\pi}{2})^4\cdots\right)+\left(\frac{1}{3!}(z-\frac{\pi}{2})^2-\frac{1}{5!}(z-\frac{\pi}{2})^4\cdots\right)^2+\cdots\rbrack\\
&=-1+\left(\frac{1}{2!}-\frac{1}{3!}\right)(z-\frac{\pi}{2})^2\cdots
\end{align}$$
So we find that
$$\tan{z}=-\frac{1}{z-\frac{\pi}{2}}+\cdots$$
Which is all we really need to know, since it's easy to see that
$$\frac{1}{z^{n+1}}=\left(\frac{2}{\pi}\right)^{n+1}+\cdots$$
And so we can finally conclude that indeed
$$\text{Res}_{z=\frac{\pi}{2}}\frac{\tan{z}}{z^{n+1}}=-\frac{2^{n+1}}{\pi^{n+1}}$$
And from here is easily observed that
$$\text{Res}_{z=-\frac{\pi}{2}}\frac{\tan{z}}{z^{n+1}}=(-1)^n\frac{2^{n+1}}{\pi^{n+1}}$$
And 
So we find for the first few coefficients:
$$c_{-3}=-\frac{\pi^2}{2}$$
$$c_{-1}=-2$$
$$c_{1}=1-\frac{8}{\pi^2}$$
$$c_{3}=\frac{1}{3}-\frac{32}{\pi^4}$$
$$c_{5}=\frac{2}{15}-\frac{128}{\pi^6}$$
Where the positive fractions come from the $\text{Res}_{z=0}\frac{\tan{z}}{z^{n+1}}$ term (which are easier to determine, since we can just use the standard taylor series for $\tan$ those terms).
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