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Math Help - clockwise and counterclockwise integration

  1. #1
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    clockwise and counterclockwise integration

    This is one of the problems that I have to do for Monday, and I really don't get it. I didn't understand the explainations in class. If someone could help me with these, I would really appreciate it. I think something like this could be on the final as well, so if you could really explain how to do every step that would be fantastic!


    1. Furnishing a sketch of the curve you are integrating over, integrate the following:

    a. (cos 4z)/(z^3(4z - (pi)) counterclockwise around the circle |z - 1| = 1/2

    b. Integral (with c at the bottom) of 1/ (z^2 - 1) dz where c is the curve with the given orientations. HINT: C = C1 union C2 where C1: |z - 1| = 1, counterclockwise, C2: |z + 1| = 1 clockwise.

    c. e^z / (ze^z - 2iz) where C is the circle |z| = 0.5, counterclockwise.

    d. 4(z + 2i)^(-1) + 2(z + 4i)^(-1) clockwise around the circle |z - 1| = 2.5

    e. (z^3)(e^z) / (2z - 1)^3 counterclockwise around the unit circle.
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  2. #2
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    I am a noob when it comes to complex analysis. Maybe there is a nicer way to do #1. But here is my unfinish solution.
    Attached Thumbnails Attached Thumbnails clockwise and counterclockwise integration-picture10.gif  
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  3. #3
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    I really didn't follow that at all.
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  4. #4
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    I can at least tell you how to find the singular points.

    Quote Originally Posted by Matt View Post
    c. e^z / (ze^z - 2iz) where C is the circle |z| = 0.5, counterclockwise.
    Look when denominator is zero.
    That is, ze^z - 2iz=0
    Thus,
    z(e^z - 2i)=0
    When can have, z=0 or e^z-2i =0
    The second one says that,
    e^z = 2i
    It turns out there are infinitely many solutions to this equation,
    z = ln |2i| + i*(Arg(2i)+2pi*k)
    z = ln 2 + i (pi+2*pi*k)
    Luckily for us no integers k make z lie inside |z|=.5
    Thus none of them are singular points.
    And hence only the origin is the problem.
    d. 4(z + 2i)^(-1) + 2(z + 4i)^(-1) clockwise around the circle |z - 1| = 2.5
    The contour integral of this is the sum of these contour integrals. The second one is easy. The only singular point is z=-4i, but luckily it is not inside the |z-1|=2.5. Thus the second summand is analytic and by the Cauchy closed curve theorem its contour integral is zero (remember all poles are in Eastern Europe). Thus, you can just calculate the first integral only. The singular point is on z=-2i.
    e. (z^3)(e^z) / (2z - 1)^3 counterclockwise around the unit circle.
    Simple 2z-1=0 thus, z=1/2. Which is pole of order 3.
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