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Homework Help: Complex valued line integrals

  1. Apr 25, 2007 #1
    Hi there I got a couple of question regarding the topic above

    1. The problem statement, all variables and given/known data

    (a) Given the integrals

    [tex]\int \limit_{0}^{i} \frac{dz}{(1-z)^2}[/tex]
    [tex]\int_{i}^{2i} (cos(z)) dz [/tex]
    [tex]\int_{0}^{i\pi} e^{z} dz[/tex]

    (1)write this as a Line integral on the form [tex]\int_{\gamma} f(\gamma(t)) \cdot \gamma'(t) dt[/tex] and
    (2)Next find sum the of integrals using anti-derivatives.

    Here am I unsure. How do I approach to calculate ?

    [tex]\int_{\gamma_{n}} \frac{dz}{z}[/tex] where is [tex]\gamma_{n}:[0,2\pi] \rightarrow \mathbb{C} [/tex] is a parameter presentation of the unit circle, where [tex]n \in \mathbb{Z} - \{0\}[/tex] and which runs through

    [tex]\gamma_{n}(t) = e^{itn}[/tex]

    2. Relevant equations

    Have I understood and solved (a) correctly?

    What about (b) could somebody please be so kind give me a hint/(some help) :) ?

    3. The attempt at a solution

    attempted Solution A:

    I choose [tex]\gamma(t) = t \cdot i[/tex] where [tex]t \in [0,1][/tex]

    Since [tex]\gamma(t)' = i[/tex] then the solution is

    [tex]\int_{\gamma} \frac{dt \cdot i}{(1-(it))^2} = \int_{0}^{1} \frac{dt \cdot i}{(1-(it))^2} = -1/2 + 1/2 \cdot i[/tex]

    the finding the sum of the original integral

    [tex]\int_{0}^{i} \frac{dz}{(1-(z))^2} = -1/2 + 1/2 \cdot i [/tex]

    Second integral:

    [tex]\int_{0}^{i\pi} e^{z} dz = \int_{\gamma} (e^{it} \cdot i) dt = \int_{0}^{\pi} (e^{it} \cdot i) dt = -2 [/tex]

    finding the sum of the integral:

    [tex]\int_{0}^{i\pi} e^{z} dz = -2[/tex]

    (B) Attempted solution

    If [tex]\int_{\gamma_{n}} \frac{dz}{z}[/tex], then to solve this integral using the definition of the line integral

    then I take [tex]\int_{\gamma_{n}} \frac{dz}{z} = \int_{0}^{2\pi} \frac{1}{e^{t \cdot n \cdot i}} \cdot \frac{d}{dt}(e^{t \cdot n \cdot i}) dt = 2 \cdot n \cdot \pi \cdot i[/tex]

    where [tex]n \in \mathbb{Z} - \{0\}[/tex]

    Could this be the solution?

    Best Regards
    Last edited: Apr 25, 2007
  2. jcsd
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