Suppose that $f$ is analytic on a close curve γ. Prove or disprove
$$\int_\gamma \overline{f(z)}f'(z)dz$$
is purely imagine.
I know that $f$ is analytic on a close curve, then
$$\int_\gamma f(z) dz=0$$
I tried an example with $\gamma =e^{it}$ with $0\leq t\leq 2\pi$, I often get the real part of the integral equal zero.
I tried let $f=u+iv$ so $f'=u_x+iv_x$. Since $f=u+iv$, $\overline{f}=u-iv$
$$\overline{f(z)}f'(z)=uu_x+vv_x+i(uv_x-vu_x)$$
But this doesn't get me anywhere. Any help would be greatly appreciated.
Best Answer
$$ 0 = \int_{\gamma} d\, |f|^2 = \int_{\gamma} \left(\overline{f} \cdot f' dz + f\cdot \overline{f'} d\overline{z}\right) = 2 \textrm{Re} \int_{\gamma}\overline{f} \cdot f' dz. $$