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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. – Christopher A. Wong Sep 24 '12 at 16:42
There is a problem at $\pm\pi/2$,\pm3\pi/2$. – AD. Sep 24 '12 at 16:46
@AD. what does the original question imply then? Do I need to evaluate Laurent series at different points on singularities then? – Monkey D. Luffy Sep 24 '12 at 16:48
@ChristopherA.Wong am I going in right direction? still the problem looks quite complicated to me to evaluate it even with Wolframalpha – Monkey D. Luffy Sep 24 '12 at 16:48
@saurs thanks for the comment. could you please check this. Why is Taylor expansion same as Laurent expansion at $z=0$? – Monkey D. Luffy Sep 24 '12 at 17:20
up vote 2 down vote accepted

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 is Taylor series equivalent to Laurent series? – Monkey D. Luffy Sep 24 '12 at 17:06
@Monkey D. Luffy Obviously, not equivalent. I mean expand $\sin{z}$ and $\cos{z}$ in Taylor series and then select main part – M. Strochyk Sep 24 '12 at 17:23
could you please give me some relevant link (solved similar problems)? I'm still unable to figure this out. – Monkey D. Luffy Sep 24 '12 at 17:26
@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. – saurs Sep 24 '12 at 17:47
@Monkey D. Luffy $\sin{z}=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots;$ $\cos{z}=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\ldots;$ $f(z) = \frac{\sin z - z}{z^2 \cos z}=\frac{-\frac{z^3}{3!}+\frac{z^5}{5!}-\ldots}{z^2(1-\frac{z^2}{2!}+\frac{z^4}{‌​4!}-\ldots)}=\frac{-\frac{z}{3!}+\frac{z^3}{5!}-\ldots}{(1-\frac{z^2}{2!}+\frac{z‌​^4}{4!}-\ldots)}$ – M. Strochyk Sep 24 '12 at 18:00

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