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Math Help - Contour integral

  1. #1
    Junior Member CuriosityCabinet's Avatar
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    Contour integral

    Evaluate the contour integral \int_\gamma \frac{1}{z} dz where \gamma = {z \in \mathbb{C} : z=e^{i\theta}, \theta \in [-\pi/2, \pi/2]}.
    I keep getting i\pi but the answer should be -i\pi, what am i doing wrong?
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  2. #2
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    Re: Contour integral

    The contour is the half circle centred at the origin of radius 1 starting at \displaystyle \theta = -\frac{\pi}{2} and ending at \displaystyle \begin{align*} \theta = \frac{\pi}{2} \end{align*}, so a decent parameterisation is \displaystyle \begin{align*} z = e^{i\theta} \implies dz = i\,e^{i\theta} \end{align*}, therefore

    \displaystyle \begin{align*} \int_{\gamma} { \frac{1}{z}\,dz } &= \int_{ -\frac{\pi}{2} }^{ \frac{\pi}{2} }{ \frac{1}{ e^{i\theta} } \, i\, e^{i\theta} \,d\theta} \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{i\,d\theta} \\ &= i\left[ \theta \right] _{-\frac{\pi}{2}}^{\frac{\pi}{2}} \\ &= i \left[ \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right] \\ &= i\, \pi  \end{align*}

    I agree with your answer - unless your contour is oriented in the other direction...
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  3. #3
    Junior Member CuriosityCabinet's Avatar
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    Re: Contour integral

    The solutions talk about how Log z is not holomorphic on [0,\infty) and that \gamma intersects with this branch cut. Also in the solutions they integrated directly without parametrising. I wonder how it would work with a parametrisation?
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