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Homework Help: 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
    |z|=1

    2. Relevant equations

    f'(a)=[itex]\frac{n!}{2i\pi}[/itex]=∫[itex]\frac{f(z)}{{z-a}^(n+1)}[/itex]
    ???
    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:
    f(z)=[itex]\frac{z+i}{z+2}[/itex]
    2[itex]\pi[/itex]f'(0)=[itex]\frac{\pi}{2}[/itex]+[itex]\pi[/itex]i


    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

    lanedance

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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

    lanedance

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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
    C

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

    2[itex]\pi[/itex]i*f(0)=2[itex]\pi[/itex]i*[itex]\frac{cos(0)}{(0^2+8)}[/itex]=[itex]\frac{i\pi}{4}[/itex]
     
    Last edited: Oct 26, 2011
  7. Oct 26, 2011 #6

    vela

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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
    |z|=[itex]\pi[/itex]


    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?

    2[itex]\pi[/itex]i*f([itex]\frac{\pi}{4}[/itex])=2[itex]\pi[/itex]i*[itex]\frac{1}{2*(\pi/2}[/itex])=2i

    2[itex]\pi[/itex]i*f([itex]\frac{-\pi}{4}[/itex])=2[itex]\pi[/itex]i*[itex]\frac{-1}{2*(-\pi/2)}[/itex]=2i

    2i+2i=4i
     
    Last edited: Oct 26, 2011
  9. Oct 26, 2011 #8

    vela

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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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