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Homework Help: Contour Integration

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    ∫[itex]_{\gamma}[/itex](x-y)dz where [itex]\gamma[/itex] has parametrization: z(t) = e[itex]^{it}[/itex] for [itex]\pi[/itex]/2 [itex]\leq[/itex] t [itex]\leq[/itex] 3[itex]\pi[/itex]/2
    2. Relevant equations
    the integral of the sum is the sum of the integral

    3. The attempt at a solution
    I tried to break it up and see if I could evaluate it as I normally would but it started to get really messy and I think I was going about it wrong.

    z=exp(it) and dz=iexp(it)
    ∫(x)dz - i∫(y)dz

    can someone please push me into the right direction? thank you
  2. jcsd
  3. Mar 20, 2012 #2


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    Homework Helper

    as you mean to evaluate directly you need to use a substitute for the parameterisation variable

    \cint_{\gamma} f(z) dz = \int_{t_a}^T_b f(z(t)) z'(t) dt[/tex]
  4. Mar 20, 2012 #3
    could you elaborate just a little more? I really appreciate your help
  5. Mar 20, 2012 #4


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    Homework Helper

    what don't you understand? I won't do the problem for you, but am happy to help
  6. Mar 20, 2012 #5
    I dont understand how to set up a substitution so I can get it into a form that I know how to work with
  7. Mar 20, 2012 #6


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    You intend that z= x+ iy, right? So if [itex]z= e^{it}= cos(t)+ i sin(t)[/itex] what are x and y in terms of t?
  8. Mar 20, 2012 #7
    Okay this is what I have so far:

    ∫[itex]_{\gamma}[/itex](x-y)dz where [itex]\gamma[/itex] has parametrization: z(t)=e[itex]^{it}[/itex] for [itex]\pi[/itex]/2[itex]\leq[/itex]t[itex]\leq[/itex]3[itex]\pi[/itex]/2

    ∫[itex]_{\gamma}[/itex](x-y)dz = ∫[itex]_{\gamma}[/itex](cos(t)-isin(t))dz

    which we can break up like,

    ∫[itex]_{\gamma}[/itex](cos(t))dz - i∫[itex]_{\gamma}[/itex](sin(t))dz

    since z(t)=e[itex]^{it}[/itex] then dz=ie[itex]^{it}[/itex]

    so we have

    ∫[itex]_{\gamma}[/itex](cos(t))(ie[itex]^{it}[/itex]) - i∫[itex]_{\gamma}[/itex](sin(t))(ie[itex]^{it}[/itex] )

    is this correct so far?

    or would my dz be -sin(t)+icos(t)?
    Last edited: Mar 20, 2012
  9. Mar 20, 2012 #8


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    you may want to check your "y" value, remember z= x+ iy,
  10. Mar 20, 2012 #9
    so instead of isin(t) just simply sin(t)

    ∫[itex]_{\gamma}[/itex](x-y)dz = ∫[itex]_{\gamma}[/itex](cos(t)-sin(t))dz


    every time I try evaluating the above expression I never can seem to get the correct answer. I'm using ie^(it) for my dz

    AHHH im getting so frustrated. I feel like I am missing an important concept, this problem should not take me this much time
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