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    \displaystyle\int_{\gamma} z^2\ dz

    where \gamma is the parabolic segment given by \phi (t) = t^2 + it, \  0 \leq t \leq 1

    Might be a silly question, but I thought that this is 0, simply because z^2 is holomorphic on \gamma since \gamma traces out a simple path. Or am I wrong here? Can someone give me some hints as to how to use Cauchy's Theorem to do this?
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    It's not a closed path so it won't necessarily evaluate to zero; I don't think Cauchy's theorem really comes into this.
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    (Original post by nuodai)
    It's not a closed path so it won't necessarily evaluate to zero; I don't think Cauchy's theorem really comes into this.
    That's weird, since the question actually mentions to 'use Cauchy's Theorem or otherwise', and normally this lecturer doesn't do 'trick' questions (per se). My first instinct was to evaluate it normally, and I got something along the lines of -\frac{5}{12} -\frac{4}{15}i or something along those lines... I'd assume that sounds about right?

    Also, if I were evaluating say \displaystyle \int_\gamma |z|^2 dz over the straight line segment [0,1] then why would my answer not be inconsistent with Cauchy's theorem? Is it because I'm integrating over an open path and so Cauchy's Theorem can't be applied? Seems like too simple an answer.
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    Thinking about it, if you join \gamma with the line segment [1+i,0] = \{ (1+i)(1-t)\, :\, 0 \le t \le 1 \} then you have a closed curve, so you can use Cauchy's theorem by evaluating the integral of z^2 along [1+i,0] and then use the fact that \displaystyle \int_{\gamma} \cdots + \int_{[1+i,0]} \cdots = 0 (lazy notation because I'm lazy).
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    Now that's a fantastic idea.
 
 
 
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