# Find the laurent series

find the laurent series centered at $z=0$ for the rational functions below. Determine the largest open set in $C$ for which each series converges
$$\frac{1}{(z^2-1)(z^2-4)}.$$

I have no idea how to do this.

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\frac{1}{((z^2)-1)((z^2)-4)} –  MathLover Dec 12 '12 at 22:09
Actually try going by the route given here: en.wikipedia.org/wiki/Laurent_series. Try doing some integrations, put some work in it. Tell us where the problem exactly lies, based on yourndifficulties. –  Golob Dec 12 '12 at 22:14
Perhaps you can decompose the rational function by partial fractions, and then rewrite each term as a Laurent series about $z=0$. Then combine everything. –  yunone Dec 12 '12 at 22:25
Hint: 1. Where singularities of the function $\dfrac{1}{(z^2-1)(z^2-4)}$ are located? What is the type of these singularities? 2. Try decomposition of rational function $\dfrac{1}{(z^2-1)(z^2-4)}$ by partial fractions –  M. Strochyk Dec 12 '12 at 22:33

$$\frac{1}{(z-1)(z+1)(z-2)(z+2)}=\frac{A}{z-1}+\frac{B}{z+1}+\frac{C}{z-2}+\frac{D}{z+2}\Longrightarrow$$

$$1=A(z+1)(z^2-4)+B(z-1)(z^2-4)+C(z+2)(z^2-1)+D(z-2)(z^2+1)$$

The last equality above is one between polynomials so it must be true for any values of $\,z\,$, so for example:

$$z=-2\Longrightarrow 1=-20D\Longrightarrow D=-\frac{1}{20}$$

$$z=2\Longrightarrow 1=12C\Longrightarrow C=\frac{1}{12}$$

and etc. Thus, you can develop around $\,z=0\,$ as, for instance:

$$\frac{2}{z+2}=\frac{1}{1+\frac{z}{2}}=1-\frac{z}{2}+\frac{z^2}{4\cdot2!}-\frac{z^3}{8\cdot 3!}+\ldots$$

The above is valid whenener $\displaystyle{\left|\frac{z}{2}\right|<1\Longleftrightarrow |z|<2}\,$ , and etc.

At the end "glue" everything together putting attention on the values of $\,z\,$ for which your development are valid.

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