Ahlfors complex integration. 
Suppose $f(z)$ is analytic on a closed curve $\gamma$ (i.e $f$ is analytic in a region contains $\gamma$ ).  Prove that $\int\limits_{\gamma}\overline{f(z)}f'(z)dz$ is purely imaginary.

How  can  we  prove  this  statement? Is  there  any  special  theorem  to  apply? 
 A: 
Is there any trick without Stokes theorem?

Yes. We could for example use the fact that integrals of exact differentials over closed curves always vanish, so
$$\int_\gamma d\lvert f(z)\rvert^2 = 0,$$
and
$$d\lvert f(z)\rvert^2 = \frac{\partial}{\partial z} \lvert f(z)\rvert^2 \,dz + \frac{\partial}{\partial\overline{z}}\lvert f(z)\rvert^2\,d\overline{z} = \overline{f(z)}f'(z)\,dz + f(z)\overline{f'(z)}\,d\overline{z},$$
so
$$d\lvert f(z)\rvert^2 = 2\operatorname{Re}\left(\overline{f(z)}f'(z)\,dz\right).$$
A: Stokes' theorem would not apply because no one promised that $f$ is defined in the entire domain bounded by $\gamma$. But yes, there is a trick:
$$
\int \bar f f'dz=\int \bar f df=\int |f|^2\frac {df}f=\int |f|^2d(\ln f)=\int |f^2|d\ln|f|+i\int|f|^2d\operatorname{arg}f.
$$
It remains to note that
$$
 \int |f^2|d\ln|f|=\int |f|d|f|=\int d(|f|^2/2)=0.
$$
Of course this only works well if $|f|\ne0$ on $\gamma$. But if there are zeros of $f$ on $\gamma$, you can make an arbitrarily small deformation of the contour so that there will be no zeros on the deformed contour. Then the integral will be imaginary, and letting the deformed contour tend to the original one you get what you need.
