Residue Theorem: Finding Residue at ##|z|=2##

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    Residue Theorem
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Homework Help Overview

The discussion revolves around finding the residue of the integral ##\oint { \frac { \sin z }{ 2z-\pi } } dz## along the contour where ##|z|=2##. Participants are analyzing the calculation of the residue and comparing their results with a reference solution.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss their individual calculations of the residue, with one expressing confidence in their answer while another suggests a possible oversight regarding a factor of half. Questions arise about the origin of this factor and the necessary form for the calculation.

Discussion Status

The discussion is active, with participants sharing their results and questioning specific aspects of the calculations. There is no explicit consensus yet, but some guidance has been provided regarding the factor that may need to be considered in the residue calculation.

Contextual Notes

Participants are working under the constraints of a homework problem, which may limit the information available for their calculations. The original poster expresses uncertainty about their approach compared to the textbook solution.

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Homework Statement



Find the residue of ##\oint { \frac { sinz }{ 2z-\pi } } dz## where ##\left| z \right| =2##[/B]

Homework Equations



##f\left( z_{ o } \right) =\frac { 1 }{ 2\pi i } \oint { \frac { f\left( w \right) }{ w-z_{ o } } } dw##

The Attempt at a Solution



It seems to me that the answer is ##2\pi i##, but the book gives ##\pi i##.

Not sure what I did wrong, I am pretty confident in my answer.

Chris
 
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For your integral, f(z_0)=\frac 1 2 \sin \frac \pi 2 and z-z_0=z-\frac \pi 2.
 
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I got \pi i as well. You probably forgot a factor of half when calculating the residue.
 
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Where does the 1/2 factor come from?

Thanks,
Chris
 
Never mind, I see that it the bottom needs to have a two factored out to fit the form.

Thanks,
Chris
 

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