Let $r>0$, $c \in \mathbb{C}$, and suppose $f$ is analytic in an open neighborhood of $\overline{B_r(c)}$. I would like to show that this implies that ${\int_{B_r(c)}} \overline{f(z)}f’(z)dz$ is pure imaginary.
My initial idea was to use the fact that $\overline{f(z)}=\frac{|f(z)|^2}{f(z)}$ and parameterize using $\gamma :[0,2 \pi] \rightarrow \mathbb{C}$ defined by $\gamma (t)=c+re^{it}$. Then we’d have
${\int_{B_r(c)}} \overline{f(z)}f’(z)dz={\int_{0}^{2\pi}} \frac{|f(\gamma (t)|^2}{f(\gamma (t))}f'(\gamma (t))\gamma'(t)dt$.
My next thought was then to make the substitution $u=f(\gamma (t))$ so that the integral would then become
${\int_{0}^{2\pi}} \frac{|u|^2}{u}du=0$.
But this seems to mean that the whole integral vanished, not just the real part. Am I interpreting this result incorrectly and this does imply that the real part vanishes? Or is my above work incorrect? Any hints or advice would be appreciated!