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Math Help - Complex Line Integration

  1. #1
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    Complex Line Integration

    I'm stuck on a question about complex line integration; I've managed to do some simpler questions, but integrating \int \exp(it) dt is giving me problems. Here's the full question:

    Calculate \int_\gamma |z| dz when \gamma = \{ z : z = \exp(it), -\pi \leq t \leq 0 \}

    So my first step is
    \int_{-\pi}^0 |\exp(it)| * \gamma ' (t) dt

    With \gamma ' (t) = i \exp(it) and |\exp(it)| = 1
    that becomes
    i \int_{-\pi}^0 \exp(it) dt

    I put that into Wolframm's online integrator and it said that's
    \sin(t) - i \cos(t)
    but I don't get how it works that out, or if it's even correct. TIA
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  2. #2
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    Quote Originally Posted by InfernoZeus View Post
    I'm stuck on a question about complex line integration; I've managed to do some simpler questions, but integrating \int \exp(it) dt is giving me problems. Here's the full question:

    Calculate \int_\gamma |z| dz when \gamma = \{ z : z = \exp(it), -\pi \leq t \leq 0 \}

    So my first step is
    \int_{-\pi}^0 |\exp(it)| * \gamma ' (t) dt

    With \gamma ' (t) = i \exp(it) and |\exp(it)| = 1
    that becomes
    i \int_{-\pi}^0 \exp(it) dt

    I put that into Wolframm's online integrator and it said that's
    \sin(t) - i \cos(t)
    but I don't get how it works that out, or if it's even correct. TIA
    \displaystyle \int e^{it} \, dt = \frac{1}{i} e^{it} (I have omitted the arbitrary constant since you will use this result to evaluate a definite integral).
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    \displaystyle \int e^{it} \, dt = \frac{1}{i} e^{it} (I have omitted the arbitrary constant since you will use this result to evaluate a definite integral).
    Cool, I'd already worked that out, but assumed that it was wrong because of what Wolframm told me Thanks for the help
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  4. #4
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    What wolfram told you and what you derived are both correct since they are the same by euler's formula!
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  5. #5
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    Quote Originally Posted by Vlasev View Post
    What wolfram told you and what you derived are both correct since they are the same by euler's formula!
    Duh! I should have been able to work that out Especially as I should know Euler's relations for my upcoming exam...
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  6. #6
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    Specifically, e^{it}= cos(t)+ i sin(t) so that i \int_{-\pi}^0 e^{it}dt= i\int_{-\pi}^0 cos(t)dt- \int_{-\pi}^0 sin(t)dt
    Last edited by mr fantastic; August 14th 2010 at 05:00 AM. Reason: Fixed some latex tags.
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