Integrating Sin(1/z) and Z Sin (1/Z^3) Over C (Circle of Radius 1)

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Discussion Overview

The discussion revolves around the integration of the functions Sin(1/z) and Z Sin(1/Z^3) over a circular contour of radius 1 centered at the origin in the complex plane. Participants explore various methods for evaluating these integrals, including the Cauchy integral formula, Laurent series, and Taylor series expansions.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about the integration process and mentions that their attempts using the Cauchy integral formula yield infinite results.
  • Another participant inquires about the locations of poles and their residues, suggesting a focus on singularities in the functions.
  • A suggestion is made to use Laurent series to facilitate the integration process.
  • It is proposed that participants should utilize the Taylor series for sin(z), substituting z with 1/z and 1/z^3 to derive the necessary Laurent series for the integrals.

Areas of Agreement / Disagreement

Participants have not reached a consensus on the integration results or methods, and multiple approaches are being discussed without resolution.

Contextual Notes

There are unresolved aspects regarding the identification of poles and the calculation of residues, as well as the implications of using different series expansions for the integrals.

Who May Find This Useful

Students preparing for exams in complex analysis, particularly those interested in contour integration and series expansions.

fahd
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hi there
im confused with this question..
Integrate 1) Sin(1/z) dz
and 2) Z sin (1/Z^3)

where Z is any complex number., over C which is a circle of radius 1 centred at 0

i tried using the cauchy integral formula and stuff but somehow the answer always comes infinity...is that right ..please help..got exam tomorrow!
 
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Where are the poles and what are its residues at each pole?
 
Presumably, you know the Taylor's series for sin(z). Replace z in that series by 1/z and 1/z3 to get the Laurent Series for sin(1/z) and sin(1/z3) respectively. Then the integrals should be easy.
 

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