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The original question to me (from a friend) was stated as

Q:Find the first four Laurent series of $f(z) = \frac{\sin z - z}{z^2 \cos z}$ in the region $0 < |z| < 2 \pi$

I'm not sure how to do it, if possible I wish only to know this expansion about zero.

The coefficients are given by $ a_n = \frac1{2i\pi}\int _\gamma \frac{f(z)}{(z-0)^n} dz $ So I change $z = r e^{i \theta}$ and integrate from $0$ to $2\pi$ putting $r=1$ $ a_n = \frac1{2i\pi}\int _\gamma \frac{\sin (r {e^{i \theta}) - r {e^{i \theta}}}}{r^{n+2}e^{i\theta {(n+2)}} \cos (re^{i\theta})} r ie^{i \theta}d\theta $

Am I going in right direction?

EDIT:: Any similar solved problem link will be highly welcome as answer :D

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    If you are able somehow to find the Laurent series of $(z^2-\pi^2/4)(z^2-9\pi^2/4)f(z)$, then you would get a nice representation of $f$ in \{z:|z|<2\pi,z\ne\pm\pi/2,\pm3\pi/4\}, also note that $\sin z-z$ has a zero of order 3 at $z=0$, which cancels the singularity at $z=0$.2012-09-24

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Expand each term in the fraction in Taylor series in a neighbourhood of $z_0=0$, paying attention on the radius of convergence of those expansions, and then select powers of $z$ what you need. Supplement to previous answer: because $f(z)$ have single poles in $\pm \frac{\pi}{2}$ and $\pm \frac{3\pi}{2}$, we obtain different Laurent expansions in such annuli: $\{z\colon 0<|z|<\frac{\pi}{2}\};$ $\{z\colon \frac{\pi}{2}<|z|<\frac{3\pi}{2}\};$ $\{z\colon \frac{3\pi}{2}<|z|<{2\pi}\};$

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    all right ... thank both of you!! this looks less rigorous.2012-09-24