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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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    **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 fractions2012-12-12

1 Answers 1

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

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

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.