How to find this Laurent series?

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    Laurent series Series
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SUMMARY

The discussion focuses on calculating the Laurent series for the function f(z) = 1/(z²·sin(z)). The user expresses confusion regarding the singularity at z=0 and the appropriate method to derive the series. It is established that the Taylor series for sin(z) can be inverted, and the Laurent series can be constructed by multiplying the series for csc(z) by z^(-2). This approach effectively addresses the singularity and provides insight into the function's behavior near the pole.

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  • Familiarity with the sine function and its series expansion
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Noone1982
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I understand perfectly well how to do Taylor series, but I am foggy on these Laurent series. Say, we have something like,

[tex]f\left( z \right)\; =\; \frac{1}{z^{2}\cdot \sin \left( z \right)}[/tex]

I think I need to use the taylor series expressions for sin(z) but otherwise, I am not sure what to do about that z^2. If I use that in a taylor series with z=0, then I get a singularity.

Since its 1/sin(z), do I just inverse the taylor series for sin(z) Yes, you can see I am very unknowledgeable about this. I turn here because the explanations I found in my book are totally lacking.
 
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It's clear that z=0 is a pole for your function, that's why the Taylor series is not well defined there. The Laurent series will tell you how this singularity behaves (how it grows to infinity).

A Laurent series has the form

[tex]\sum_{n=1}^\infty \frac{b_n}{z^n}+\sum_{n=0}^\infty a_nz^n,[/tex]

so you are right, in order to calculate the series, you just have write down the series for [itex]\csc z[/itex] and multiply it by [itex]z^{-2}[/itex].
 

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