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Thread: complex integration

  1. #1
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    complex integration

    Suppose that $\displaystyle f(z)$ is analytic on a closed curve $\displaystyle \gamma$(i.e., $\displaystyle f$ is analytic in a region that contains $\displaystyle \gamma$). Show that
    $\displaystyle \int_\gamma \overline{f(z)}f'(z)dz$ is purely imaginary.
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  2. #2
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    Quote Originally Posted by Stiger View Post
    Suppose that $\displaystyle f(z)$ is analytic on a closed curve $\displaystyle \gamma$(i.e., $\displaystyle f$ is analytic in a region that contains $\displaystyle \gamma$). Show that
    $\displaystyle \int_\gamma \overline{f(z)}f'(z)dz$ is purely imaginary.
    If the curve is smooth enough to have a differentiable parametrisation then this is just integration by parts. In fact, suppose that $\displaystyle \gamma$ is given by a differentiable path $\displaystyle z = \gamma(t)\;(0\leqslant t\leqslant1)$, with $\displaystyle \gamma(1) = \gamma(0)$. Then
    $\displaystyle \begin{aligned}\int_\gamma \overline{f(z)}f'(z)\,dz &= \int_0^1\overline{f(\gamma(t))}f'(\gamma(t))\,\gam ma'(t)dt\\ &= \Bigl[\overline{f(\gamma(t))}f(\gamma(t))\Bigr]_0^1 - \int_0^1\overline{f'(\gamma(t))\gamma'(t)}f(\gamma (t))\,dt\\ &= -\overline{\int_0^1\overline{f(\gamma(t))}f'(\gamma (t))\,\gamma'(t)\,dt} = -\overline{\int_\gamma \overline{f(z)}f'(z)\,dz}.\end{aligned}$

    Thus the integral is equal to the negative of its complex conjugate, so it is purely imaginary.
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