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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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    I would evaluate the integrals by residue theorem.2012-09-24
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    There is a problem at $\pm\pi/2$,\pm3\pi/2$.2012-09-24
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    @AD. what does the original question imply then? Do I need to evaluate Laurent series at different points on singularities then?2012-09-24
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    @ChristopherA.Wong am I going in right direction? still the problem looks quite complicated to me to evaluate it even with Wolframalpha2012-09-24
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    @saurs thanks for the comment. could you please [check this](http://www.wolframalpha.com/input/?i=taylor+expansion+%28sin%28z%29+-+z%29%2F%28z^2+cos%28z%29%29+at+z%3D0). Why is Taylor expansion same as Laurent expansion at $z=0$?2012-09-24
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    Man, _my_ friends never ask me to calculate Laurent series for them...2012-09-24
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    @StevenStadnicki he didn't demanded either, I'm doing this entirely for sake of my interest :D2012-09-24
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    @Monkey D. Luffy see my comment to M. Strochyk's answer2012-09-24
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    @saurs thanks for your response.2012-09-24
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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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    [seems so](http://www.wolframalpha.com/input/?i=taylor+expansion+%28sin%28z%29+-+z%29%2F%28z^2*cos+%28z%29%29+at+z%3D0) is Taylor series equivalent to [Laurent series](http://www.wolframalpha.com/input/?i=laurent+expansion+%28sin%28z%29+-+z%29%2F%28z^2*cos+%28z%29%29+at+z%3D0)?2012-09-24
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    @Monkey D. Luffy Obviously, not equivalent. I mean expand $\sin{z}$ and $\cos{z}$ in Taylor series and then select main part2012-09-24
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    could you please give me some relevant link (solved similar problems)? I'm still unable to figure this out.2012-09-24
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    @Monkey D. Luffy You know what the Laurent series of $\sin z$ and $\cos z$ are (about the appropriate singularities). So to find the Laurent series of $f(z)$ you just divide the Laurent series of $\sin(z) - z$ by that of $z^2 \cos(z)$ (again about the appropriate singularities). So you are dividing an infinite sum by another infinite sum. But you only want the first four terms, so you can truncate the sums and do polynomial long division.2012-09-24
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    @Monkey D. Luffy $\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots;$ $\cos{z}=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\ldots;$ $f(z) = \frac{\sin z - z}{z^2 \cos z}=\frac{-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots}{z^2(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\ldots)}=\frac{-\frac{z}{3!}+\frac{z^3}{5!}-\ldots}{(1-\frac{z^2}{2!}+\frac{z^4}{4!}-\ldots)}$2012-09-24
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    all right ... thank both of you!! this looks less rigorous.2012-09-24