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Complex integration on a given path

  1. Mar 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Calculate the following integrals on the given paths. Why does the choice of path change/not change each of the results?

    (a) f(z) = exp(z) on
    i. the upper half of the unit circle.
    ii. the line segment from − 1 to 1.

    2. Relevant equations
    γf(z) = ∫f(γ(t))γ'(t)dt, with the limits being the limits of the parametrization.

    3. The attempt at a solution
    i) γ(t) = eit, t ∈ [0, π]

    Integral = ∫ez dz = ∫eeitieitdt

    u substitution: u = eit, du = ieit

    => Integral =∫eudu, I leave the lower bound at 0 and upper bound at π because I'm going to substitute for u at the end.

    Integral = eu]0π

    = eeit]0π

    = ee - ee0

    = ee - e1 = ee - e

    γ(t) = t, t ∈ [-1, 1]

    Integral = ∫e2 (1) dt, with lower bound = -1, upper bound = 1.

    = et ]-11

    = e1 - e-1

    = e - 1/e

    So the path does matter because two different paths gave two different answers.

    Whats wrong with my answer?
  2. jcsd
  3. Mar 20, 2017 #2


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    Do you know the value of e?
  4. Mar 20, 2017 #3
    I used it because that would be the unit circle in the complex plane, and then i restricted it to [0, pi].

    im not sure...is that the wrong parametrization?
  5. Mar 20, 2017 #4


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    in addition to what FactChecker pointed out, observe that the integration limits are the wrong way around in the following:
    The integral is from -1 to 1, which is from ##e^\pi## to ##e^0##.
  6. Mar 20, 2017 #5
    I'm sorry i don't understand.

    I thought for that integral we have to use the boundaries of the upper half of a unit circle.
    My parametrization was eit, where t is the angle. So then why would I not use 0 and π
  7. Mar 20, 2017 #6


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    You should use them, but with reversed order. The path has to have the same start and end points as the straight line. Since the straight line goes from the point -1+0i to +1+0i, the semi-circle must do that too - ie it must be traversed in the clockwise direction. Your limits make the traversal happen anti-clockwise. Swap the order.

    PS can you answer FactChecker's question? What is ##e^{i\pi}##? Think about how ##e^{it}## can be expressed in terms of trig functions.
  8. Mar 21, 2017 #7
    to FactChecker's response:

    ee = ecos(π) + isin(π) = e1 + 0i = e

    dang it seems so simple now..thanks for the hints

    to andrewkirk:

    I switched the integral's upper and lower limit like you said,

    and got this I = e]π0 = e - e = 0
  9. Mar 21, 2017 #8


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    cos(π) = -1
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