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    Say I had a circle defined as \displaystyle \{ z \in \mathbb{C}:\lvert z-2 \rvert= 3 \} , say. Then would this qualify as a positively oriented contour and would I be able to apply CIF directly? (I assumed it would, but I'm having doubts, hence the silly question)
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    Well that would be the image of a contour; the orientation depends upon what parametrisation you use. [But to answer your question, yes... provided you parametrise it right.]
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    (Original post by nuodai)
    Well that would be the image of a contour; the orientation depends upon what parametrisation you use. [But to answer your question, yes... provided you parametrise it right.]
    I had an attempt with \phi(t) = 1 + 2e^{it} , \ t \in [0,2\pi] but the integrand is \dfrac{ze^z}{(z-i)^2} and it soon gets rather convoluted... most notably the annoying e^{(1+2e^{it})}.

    I should clarify that the circle used here is \{ z \in \mathbb{C}: \lvert z - 1 \rvert = 2 \}
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    What's the actual question? I have an inkling that this isn't actually a CIF question. [But it might be.] Are you trying to find \displaystyle \int_C \dfrac{ze^z}{(z-i)^2}\, dz, where C is your circle?
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    (Original post by nuodai)
    What's the actual question? I have an inkling that this isn't actually a CIF question. [But it might be.] Are you trying to find \displaystyle \int_C \dfrac{ze^z}{(z-i)^2}\, dz, where C is your circle?
    Yep, exactly that.
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    So use the 2nd formula here: http://en.wikipedia.org/wiki/Cauchy%...tegral_formula

    you don't need to parameterize the contour at all.
 
 
 
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