Let $f$ a holomorphic function in $\mathbb{D}(0,1)$ such that $f(z)=\sum_{n\ge 0}a_n z^n$. Then apparentely we can deduce that for all $r\in [0,1[$, $n\ge1$ : $\int \limits_{0}^{2\pi} \overline{f(r\exp(i\theta))}\exp(-in\theta)\mathrm{d}\theta=0$.
I thought that it was a Cauchy formula but $\bar{f}$ is not necessarily holomorphic.
Thanks in advance !
Conjugation is linear and continuous, so $$ \int _{0}^{2\pi} \overline{f(r\exp(i\theta))}\exp(-in\theta)\mathrm{d}\theta =\overline{\int _{0}^{2\pi} {f(r\exp(i\theta))}\exp(in\theta)\mathrm{d}\theta} $$