Please check this complex integral

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Discussion Overview

The discussion revolves around evaluating the complex integral $ \oint\frac{dz}{z(2z+1)} $ using the residue theorem, specifically within the context of a unit circle contour. Participants explore different methods for solving the integral and discuss the implications of their approaches.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a solution using the residue theorem, identifying the poles at 0 and $-\frac{1}{2}$ and calculating the residues, concluding that the integral evaluates to 0.
  • Another participant suggests using partial fractions as a potentially simpler method for evaluating the integral.
  • A participant questions the suggestion of using partial fractions, arguing that it still requires finding two residues, indicating a lack of clarity on the simplification proposed.
  • Another participant provides a fact about evaluating integrals based on whether poles are inside or outside the contour, suggesting that this could simplify the evaluation process without needing to calculate residues.
  • One participant concludes that since both poles are inside the contour, the result is 0, echoing the earlier conclusion but without resolving the method debate.

Areas of Agreement / Disagreement

Participants express differing views on the best method to evaluate the integral, with some supporting the residue theorem and others advocating for partial fractions. The discussion remains unresolved regarding which method is preferable.

Contextual Notes

Participants have not reached a consensus on the most efficient approach to the integral, and there are varying interpretations of the necessity of calculating residues versus using contour properties.

ognik
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An old exam has: Evaluate $ \oint\frac{dz}{z(2z+1)} $, where the contour is a unit circle

This look good for the residue theorem, it has 2 simple poles at 0, $-\frac{1}{2}$
$ Res(f, 0)= \lim_{{z}\to{0}}z\frac{1}{z(2z+1)}=1$
$ Res(f, -\frac{1}{2})= \lim_{{z}\to{-\frac{1}{2}}}(z+\frac{1}{2})\frac{1}{z2(z+\frac{1}{2})}=-1$
$ \oint\frac{dz}{z(2z+1)} =2\pi i\sum Res(f) = 2 \pi [1 + (-1)] = 0$
Is that right?
 
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Here is an easier suggestion. Use partial fractions first.
 
ognik said:
An old exam has: Evaluate $ \oint\frac{dz}{z(2z+1)} $, where the contour is a unit circle

This look good for the residue theorem, it has 2 simple poles at 0, $-\frac{1}{2}$
$ Res(f, 0)= \lim_{{z}\to{0}}z\frac{1}{z(2z+1)}=1$
$ Res(f, -\frac{1}{2})= \lim_{{z}\to{-\frac{1}{2}}}(z+\frac{1}{2})\frac{1}{z2(z+\frac{1}{2})}=-1$
$ \oint\frac{dz}{z(2z+1)} =2\pi i\sum Res(f) = 2 \pi [1 + (-1)] = 0$
Is that right?

Looks good to me :)
 
ThePerfectHacker said:
Here is an easier suggestion. Use partial fractions first.

Hi TPH, I don't see how partial fractions are easier $ \implies \oint \left[ \frac{1}{z}-\frac{1}{z+\frac{1}{2}} \right] dz $ - I would still end up finding 2 residues? What am I missing?
 
ognik said:
Hi TPH, I don't see how partial fractions are easier $ \implies \oint \left[ \frac{1}{z}-\frac{1}{z+\frac{1}{2}} \right] dz $ - I would still end up finding 2 residues? What am I missing?

Use the following fact,
$$ \oint_{\Gamma} \frac{dz}{z-a} = \left\{ \begin{array}{ccc} 2\pi i & \text{ if } & a \text{ is inside the contour } \\
0 & \text{ if } & a \text{ is not inside } \end{array} \right. $$

You do not need to calculate anything. Only see if $a$ is inside or outside your contour.
 
Nice. So both are inside, therefore we get $2 \pi i - 2 \pi i$ = 0?
 

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