What is the residue of f(z) = e^(-2/z^2) using a laurent series?

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SUMMARY

The residue of the function f(z) = e^(-2/z^2) is confirmed to be 0 when analyzed using a Laurent series expansion. The series expansion begins with 1 - (2/z^2) + (2/z^4) + ..., indicating that there is no a_{-1} term present. This absence of the residue term directly leads to the conclusion that the residue is indeed 0, as stated in the referenced textbook.

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cragar
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Homework Statement


Use a laurent series to find the indicated residue
f(z)=e^{\frac{-2}{z^2}}

Homework Equations

The Attempt at a Solution


So I expand the series as
follows 1-\frac{2}{z^2}+\frac{2}{z^4} ...
my book says the residue is 0 , is this because there is no residue term ?
the a_{-1} term
 
Last edited:
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Yes
 
ok just wanted to make sure
 

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