Integrals of the function f(z) = e^(1/z) (complex analysis)

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Discussion Overview

The discussion centers on the integration of the function f(z) = e^(1/z) within the multiply connected domain defined as {Re(z) > 0} ∖ {2}. Participants explore the properties of integrals of this function, including path independence and the implications of excluding the point z = 2.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that integrals of f(z) exist everywhere in the domain {Re(z) > 0} ∖ {2}, implying path independence.
  • Another participant questions the significance of excluding the point z = 2, stating it does not appear to be special.
  • A participant recalls that a modified version of the function, defined as f(z) = 0 for Re(z) ≤ 0 and f(z) = e^{-\frac{1}{z}} for Re(z) > 0, is C∞ in the whole complex plane but not analytic.
  • Another participant prompts for clarification on the reasoning behind the previous claim, noting that constructing the Laurent series for the function is straightforward.
  • A participant corrects their earlier notation from ℂ∞ to C∞ regarding the function's properties.

Areas of Agreement / Disagreement

Participants express differing views on the significance of the excluded point z = 2 and the analytic properties of the function, indicating that multiple competing perspectives remain without resolution.

Contextual Notes

There are unresolved assumptions regarding the implications of path independence and the nature of the function's analyticity in the specified domain.

Matt100
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How do you integrate f(z) = e^(1/z) in the multiply connected domain {Rez>0}∖{2}

It seems like integrals of this function are path independent in this domain since integrals of e^(1/z) exist everywhere in the domain {Rez>0}∖{2}. Is that correct?
 
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I agree.
Where is the point in excluding that point, by the way? It is not special.
 
I seem to remember that the function given by f(z)=0 for Re(z)≤0 and [itex]f(z)=e^{-\frac{1}{z}}[/itex] for Re(z)>0 is ℂ in the whole complex plane, but not analytic...
 
Why?
The Laurent series is easy to construct here.
 
Svein said:
I seem to remember that the function given by f(z)=0 for Re(z)≤0 and [itex]f(z)=e^{-\frac{1}{z}}[/itex] for Re(z)>0 is ℂ in the whole complex plane, but not analytic...
Sorry, that should be C.
 

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