# How to find the Laurent series of $f(z) = \frac{\sin z - z}{z^2 \cos z}$?

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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I would evaluate the integrals by residue theorem. – Christopher A. Wong Sep 24 '12 at 16:42