1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Another complex integral, TOUGH!

  1. Jun 22, 2007 #1

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

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

    [tex] \int _{c} \dfrac{1- Log z}{z^{2}} dz [/tex]

    where C is the curve:

    [tex] C : z(t) = 2 + e^{it} ; - \pi / 2 \leq t \leq \pi / 2 [/tex]


    2. Relevant equations
    I know the independance of path in a domain where f(z) is analytical, but I tried the standard parametrization just to beging with someting.


    3. The attempt at a solution

    [tex] z^{2} = 4 + 4e^{it} + e^{2it} [/tex]

    [tex] Log(2 + e^{it} ) = \frac{1}{2} \ln (5 + \cos t) +it [/tex]

    [tex] dz = ie^{it} dt [/tex]

    [tex] i \int _{- \pi / 2} ^{\pi / 2} \dfrac{1 -\frac{1}{2} \ln (5 + \cos t) -it }{4 + 4e^{it} + e^{2it}}e^{it} dt [/tex]

    lol iam lost
     
  2. jcsd
  3. Jun 22, 2007 #2

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    do you think I shall use independance of path?
     
  4. Jun 22, 2007 #3

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    never mind, I solved it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Another complex integral, TOUGH!
  1. Tough Integral (Replies: 4)

  2. A tough integral (Replies: 1)

  3. Tough integral (Replies: 1)

  4. Tough integral (Replies: 2)

  5. Tough Integral (Replies: 3)

Loading...