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latentcorpse
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This may be because I am tired but i can't find the residue of
e^{-\frac{1}{z^2}} at z=0
help!
e^{-\frac{1}{z^2}} at z=0
help!
Finding the residue of e^(-1/z^2) at z=0
- Thread starter latentcorpse
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- Residue
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SUMMARY
The residue of the function e^{-\frac{1}{z^2}} at z=0 is confirmed to be 0. This conclusion is reached by deriving the Laurent series from the Taylor series expansion of e^z around z=0, substituting -1/z^2 into the series. The resulting series is 1 - \frac{1}{z^2} + \frac{1}{2z^4} + ..., which contains no terms of the form \frac{1}{z} that would contribute to a non-zero residue. Concerns regarding the validity of this approach are addressed, affirming that the derived series is acceptable despite originating from the Taylor series.
PREREQUISITES- Understanding of Laurent series and their properties
- Familiarity with Taylor series expansions
- Knowledge of complex analysis, particularly residues
- Basic proficiency in mathematical notation and manipulation
- Study the derivation of Laurent series from Taylor series in complex analysis
- Explore examples of calculating residues for various complex functions
- Learn about the implications of residues in contour integration
- Investigate the behavior of e^{-\frac{1}{z^2}} in different regions of the complex plane
Students and professionals in mathematics, particularly those focusing on complex analysis, as well as anyone interested in understanding residue calculations and series expansions.
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