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Complex integral question

  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Assuming a counterclockwise orientation for the unit circle, calculate
    ∫ [itex]\frac{z+i}{z^3+2z^2}[/itex] dz

    2. Relevant equations

    3. The attempt at a solution

    I don't understand these types of questions. What does the |z| have to do with the integral? It's written on the bottom limit of the integral in case that wasn't clear.

    From the answers, it writes:

    Edit: Nevermind. I found out how the formula works. I still don't understand what the |z| does. What would happen if it was |z|=10 for example?
    Last edited: Oct 25, 2011
  2. jcsd
  3. Oct 25, 2011 #2


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    |z|=1 is represents the closed curve (unit circle) around which the contour integration should be performed.

    do you know the residue theorem, could be pretty useful here
  4. Oct 25, 2011 #3
    I somewhat understand the residue theorem and how singularities work. I'm not sure how that explains what the circle does. As long as the singularity is located in the circle, then I can do the integral. If I had something like |z-2|=1, then a problem would occur since the singularity is outside the circle. How do I adjust it then?
  5. Oct 25, 2011 #4


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    you don't the residue theorem relates the integral around a closed curve to the poles located inside the curve - so you need to find which poles are located with the unit circle

    start with finding the poles of the function
  6. Oct 26, 2011 #5
    So if I had something like the expression below, where C is the square with vertices ([itex]\pm2+2i[/itex],[itex]\pm2-2i[/itex])

    ∫ [itex]\frac{cos(z)}{z(z^2+8)}[/itex] dz

    Would I ignore the poles at +/-isqrt(8) and proceed to use the CIF?

    Last edited: Oct 26, 2011
  7. Oct 26, 2011 #6


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    Yup, you only care about the poles inside the contour.

    To answer your question in post 3: If there are no singularities inside the contour, the integral is equal to 0.
    Last edited: Oct 26, 2011
  8. Oct 26, 2011 #7
    omg I feel so stupid for not seeing something this simple...

    Thanks for the help so far. I have one last question. What happens when you have multiple singularities within the contour. For example:

    ∫ [itex]\frac{sin(z)cos(z)}{((z-\frac{\pi}{4}}[/itex])(z+[itex]\frac{\pi}{4}[/itex]) dz

    Edit: blah I can't fix the fractions in the integral. The z+pi/4 is on the denominator

    Do we treat each one individually and then add them up?



    Last edited: Oct 26, 2011
  9. Oct 26, 2011 #8


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    Yes, that's what you do. The integral is equal to [itex]2\pi i\sum~\mathrm{residues}[/itex].
  10. Oct 26, 2011 #9
    Awesome. I finally understand this topic! Thanks Vela and Lanedance.
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