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I understand how to use long division to find the reciprocal of a power series.

Are there any theorems that allow you to divide Laurent series that have a finite number of terms with negative exponents? That is, if I have a Laurent series around $0$, such as

$f(z) = \frac{1}{z^2} + 3z + 5z^4 + \ldots$,

can I use long division to come up with a series for $\frac{1}{f(z)}$ and know anything about its convergence?

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    $\begingroup$ To know something about it convergence (orwhatever)... where ? $\endgroup$ – DonAntonio May 13 '13 at 15:21
  • $\begingroup$ Since you say you know how to find the reciprocal of a power series, why not calculate $$z^2\frac{1}{z^2f(x)}?$$ The denominator is a power series. $\endgroup$ – MJD May 13 '13 at 15:22
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If you want to find out how the function behaves around a tiny neighbourhood of $\,z=0\,$, for example, then you can do

$$\frac1{f(z)}=\frac1{\frac1{z^2}+3z+5z^4+\ldots}=\frac{z^2}{1+3z^3+5z^6+\ldots}=$$

$$=z^2\cdot\frac1{1+3z^3+\mathcal O(z^6)}=z^2\left(1-3z^3+9z^6-27z^9+\ldots\right)=z^2-3z^5+\ldots$$

And the above is valid whenever

$$|3z^3|<1\implies|z|<\frac1{\sqrt[3]3}$$

Thus, our function, which is undefined at $\,z=0\,$, has a removable singularity there and that point behaves as a double zero.

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