# complex integration

• Apr 12th 2009, 08:45 PM
Stiger
complex integration
Suppose that $f(z)$ is analytic on a closed curve $\gamma$(i.e., $f$ is analytic in a region that contains $\gamma$). Show that
$\int_\gamma \overline{f(z)}f'(z)dz$ is purely imaginary.
• Apr 13th 2009, 11:22 AM
Opalg
Quote:

Originally Posted by Stiger
Suppose that $f(z)$ is analytic on a closed curve $\gamma$(i.e., $f$ is analytic in a region that contains $\gamma$). Show that
$\int_\gamma \overline{f(z)}f'(z)dz$ 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 $\gamma$ is given by a differentiable path $z = \gamma(t)\;(0\leqslant t\leqslant1)$, with $\gamma(1) = \gamma(0)$. Then
\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.