Complex integral

  • #1
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Homework Statement


[tex]\oint_{L} \frac{ \mbox{d} z}{ z(z+3) }[/tex] and [tex]L:|z|=4[/tex]

The Attempt at a Solution


what is assumption, is it oriented positive or negative? and Cauchy formula, can it be done like this?
[tex]\frac{ 1 }{ 3 } \left( \oint_{L} \frac{ \mbox{d} z}{ z } - \oint_{L} \frac{ \mbox{d} z}{ z+3 } \right)[/tex]
 

Answers and Replies

  • #2
HallsofIvy
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If nothing else is said, the integral is assumed to be in the positive orientation, counter-clockwise.
Yes, your partial fraction reduction is correct and the integral can be done in that way. Letting [itex]z= e^{i\theta}[/itex] in the first integral and [itex]z= 3+ e^{i\theta}[/itex] in the second will give very simple integrals, giving the residues at z= 0 and z= 3.
 
  • #3
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0
thanks for answer, I got 0, is it possible? and please tell me why residue in 3, not -3?
 
Last edited:
  • #5
Dick
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thanks for answer, I got 0, is it possible? and please tell me why residue in 3, not -3?

Sure the pole is at -3. And 0 is right. In fact, it's generally true that if f(z) is a polynomial of degree greater than 1 then the integral is zero around a contour that encloses all of the poles.
 

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