Let $|z| < 1$ and consider:
$$\overline{f(0)} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi$$
Expand f as a power series centered at zero.
Then $\overline{f(0)} = \overline{a_0}$ and we obtain:
$$\overline{a_0} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{a_0 + a_1e^{i\phi} + a_2e^{2i\phi} + ...}\;\;d\phi$$
$$\;\;\;\;\;\;\;\;\;\; =\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\bar{a_0} + \bar{a_1}e^{-i\phi} + \bar{a_2}e^{-2i\phi} + ...\;\;d\phi$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =\frac{1}{2\pi}\int_0^{2\pi}\frac{1}{e^{i \phi}-z}\bar{a_0}e^{i\phi} + \bar{a_1} + \bar{a_2}e^{-i\phi} + \bar{a_3}e^{-2i\phi}+...\;\;d\phi$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{\bar{a_0}}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}d\phi + \frac{1}{2\pi}\sum_{n=0}^{\infty}\int_0^{2\pi}\frac{\overline{a_{n+1}}}{e^{i \phi}-z}e^{-ni\phi}d\phi$$
$$\;\;\;\;\;\;\;\;\;=\frac{\bar{a_0}}{2\pi i}\int_{\alpha}\frac{d\zeta}{\zeta-z}d\zeta + \frac{1}{2\pi}\sum_{n=0}^{\infty}\int_0^{2\pi}\frac{\overline{a_{n+1}}}{e^{i \phi}-z}e^{-ni\phi}d\phi$$
$$=\bar{a_0} + \frac{1}{2\pi}\sum_{n=0}^{\infty}\overline{a_{n+1}}\int_0^{2\pi}\frac{1}{e^{i \phi}-z}e^{-ni\phi}d\phi.$$
Now we need to show that for all $n\geq 0$
$$\int_0^{2\pi}\frac{1}{e^{i \phi}-z}e^{-ni\phi}d\phi = 0.$$
Making the substitution $u = e^{i\phi}$ we obtain:
$$-i\int_{\alpha}\frac{1}{u^{n+1}(u-z)}du.$$
From here one can either perform partial fraction decomposition or use the residue theorem to prove that:
$$-i\int_{\alpha}\frac{1}{u^{n+1}(u-z)}du = 0.$$
Note that $\alpha$ is the parametrization of the unit circle.