Calculating Residium of Complex Integral |z|=1

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Homework Help Overview

The discussion revolves around calculating the residue of a complex integral defined along the unit circle in the complex plane, specifically the integral of the function \(\frac{1}{z}e^{\frac{1}{z}}\). Participants are exploring the implications of the integral's lack of an upper bound and the nature of singularities present.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the interpretation of the integral as a path integral around the unit circle, questioning the significance of the absence of an upper bound. There are suggestions to expand the exponential function as a power series and to consider a change of variables to facilitate the calculation of the residue.

Discussion Status

The discussion is active, with various approaches being proposed, including power series expansion and variable substitution. Participants are engaging with the problem's complexities and exploring different interpretations without reaching a consensus.

Contextual Notes

Some participants note the absence of poles in the integral and emphasize the presence of singularities, raising questions about how to handle these in the context of calculating the residue.

nhrock3
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[tex]\int_{|z|=1}^{nothing } \frac{1}{z}e^{\frac{1}{z}}[/tex]
in this integral there is no upper bound
its around |z|=1

there are no poles here
only singular significant
what to do here
when calclating the residium
??
 
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I believe, and don't trust me on this, that it's asking you to calculate a path integral around the unit circle on the complex plane. There's no "upper bound" because the integral is describing a path, not just a starting point and an ending point. Incidentally, I believe that starting and ending points are the same.
 
Start by expanding [tex]\exp (1/z)[/tex] as a power series, multiply by 1/z and look for the z^{-1} term. That will be your residue.

Mat
 
A change of variables [itex]w = 1/z[/itex] will also do the trick.
 

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