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Find all Laurent series of the form $\sum_{-\infty} ^{\infty} a_n $ for the function $f(z)= \frac{z^2}{(1-z)^2(1+z)}$

There are a lot of problems similar to this. What are all the forms? I need to see this example to understand the idea.

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up vote 1 down vote accepted

Here is related problem. First, convert the $f(z)$ to the form

$$f(z) = \frac{1}{4}\, \frac{1}{\left( 1+z \right)}+\frac{3}{4}\, \frac{1}{\left( -1+z \right) }+\frac{1}{2}\,\frac{1}{\left( -1+z \right)}$$

using partial fraction. Factoring out $z$ gives

$$ f(z)= \frac{1}{4z}\frac{1}{(1+\frac{1}{z})}- \frac{3}{4z}\frac{1}{(1-\frac{1}{z})}-\frac{1}{2z^2}\frac{1}{(1-\frac{1}{z})^2}\,. $$

Now, using the series expansion of each term yields the Laurent series for $|z|>1$

$$ = \frac{1}{4z}(1-\frac{1}{z}+ \frac{1}{z^2}+\dots) -\frac{3}{4z}(1+\frac{1}{z}+ \frac{1}{z^2}+\dots)-\frac{1}{2z}\sum_{k=0}^{\infty}{-2\choose k}\frac{(-1)^k}{z^k} \,, $$

$$ \sum_{k=0}^{\infty}\left( \frac{(-1)^k}{4}-\frac{3}{4}-\frac{(-1)^k{-2\choose k}}{2} \right)\frac{1}{z^{k+1}}\,. $$

Note that,

$$(1+x)^{-1}=1-x+x^2-\dots \,,$$

$$ (1-x)^{-1}=1+x+x^2+\dots \,,$$

$$ (1-x)^{m}=\sum_{k=0}^{\infty}{m\choose k}(-1)^kx^k $$

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