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

  • #1
3
0
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 teh domain {Rez>0}∖{2}. Is that correct?
 
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Answers and Replies

  • #2
35,518
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I agree.
Where is the point in excluding that point, by the way? It is not special.
 
  • #3
Svein
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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...
 
  • #4
35,518
11,989
Why?
The Laurent series is easy to construct here.
 
  • #5
Svein
Science Advisor
Insights Author
2,176
711
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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