Laurent series expansion

Question

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 $1<|z|<2$?

Answer 1

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?)

Answer 2

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

Answer 3

Another way is to use partial fraction decomposition:

Apart[1/((z^2 - 1) (z^2 - 4))]

$$\frac{1}{\left(z^2-1\right) \left(z^2-4\right)}=-\frac{1}{6 (z-1)}+\frac{1}{6 (z+1)}-\frac{1}{12 (z+2)}+\frac{1}{12 (z-2)}$$

You expand the terms with a pole on |z|=1 at infinity (laurent series), and the terms with a pole on |z|=2 at zero (taylor series) and add them up

Normal[Series[1/(12 (-2 + z)) - 1/(12 (2 + z)), {z, 0, 10}]] + 
 Normal[Series[-(1/(6 (-1 + z))) + 1/(6 (1 + z)), {z, \[Infinity], 
    10}]]