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Can someone share how to find the Laurent series expansion of $$f(z)=\frac{1}{(z^2-1)(z^2-4)}$$ centered at $0$ on the annulus $1<|z|<2$?

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    $\begingroup$ You can use Apart to break the function to simpler rational expressions but I don't know if there is an automated command to give you the series for such cases. $\endgroup$ – Spawn1701D Apr 19 '13 at 3:04
  • $\begingroup$ It is not clear from this question as it is posed, whether it is a math question or a Mathematica question. Please clarify. $\endgroup$ – m_goldberg Apr 19 '13 at 4:51
  • $\begingroup$ Since (1/(z^2-4)-1/(z^2-1))/3 == 1/((z^2-1)(z^2-4)) you should be able to continue from there, assumiing you know the Laurent series for 1/(z-a). $\endgroup$ – Somos Oct 31 at 19:29
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One slick way to compute the coefficients $c_k$ in the Laurent series

$$f(z)=\sum_{k\in\mathbb Z} c_k (z-a)^k$$

is to recognize that the problem of computing them is equivalent to the problem of computing Fourier coefficients, if you take the contour $\gamma$ in the definition for Laurent coefficients,

$$c_k=\frac1{2\pi i} \oint_\gamma \frac{f(z)\,\mathrm dz}{(z-a)^{k+1}}$$

to be a circle of radius $r$ within the annulus of interest. In your case we can take $r=3/2$, so the computation of the coefficients can be done like so:

With[{r = 3/2, n = 8},
     Table[FourierCoefficient[With[{z = r Exp[I t]}, 1/((z^2 - 1) (z^2 - 4))], t, k]/r^k,
          {k, -n, n, 2}]]
   {-1/3, -1/3, -1/3, -1/3, -1/12, -1/48, -1/192, -1/768, -1/3072}

(Exercise: why did I skip the computation of the odd-indexed coefficients?)

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This Note http://courses.washington.edu/ph227814/228/W14/notes/Laurent.nb.pdf from Steve Sharpe Uni Washington was helpful for me.

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