Prove the integral is always imaginary

Question

Show that if f is analytic on D and γ is a closed curve in the region then the integral $$\int \overline{f(z)}f'(z)$$ is purely imaginary. I think this problem would use some extension of cauchy integral formula. I know that f'(z) is analytic but don't know much about $$\overline{f(z)} $$. Someone suggested to use Green's theorem here but I can't figure out how to do that. Thank you for reading.

Answer 1

If $f = u + iv$ then $\bar{f}f' = (u - iv)(u_x + iv_x) = (uu_x + vv_x) + i(uv_x - vu_x)$. So

\begin{align}\bar{f}f'\, dz &= [(uu_x + vv_x) + i(uv_x - vu_x)](dx + i\, dy)\\ & = [(uu_x + vv_x)\,dx - (uv_x - vu_x)\, dy] + i[(uu_x + vv_x)\, dy + (uv_x - vu_x)\, dx] \end{align}

Since $$\Re\int \overline{f(z)}f'(z)\, dz = \int \Re\{\overline{f(z)}f'(z)\, dz\}$$ the result will be proved if we show that $\Re\{\bar{f}f'\, dz\}$ is an exact differential. As $f$ is analytic, $u_x = v_y$ and $u_y = -v_x$. Thus $uv_x - vu_x = -uu_y - vv_y = -(uu_y + vv_y)$ and

$$\Re\{\bar{f}f'\, dz\} = (uu_x + vv_x)\, dx + (uu_y + vv_y)\, dy = d\left(\frac{u^2 + v^2}{2}\right)$$

So $\Re\{\bar{f}f'\, dz\}$ is exact.