Help me prove the identity $\overline{f(0)} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi$ Let f be an analytic function defined in an open set containing the closed unit disk and let z in ℂ be fixed.  I've simplified a more complicated expression down to this identity, and as implausible as it looks, after some numerical checking it does in fact appear to be true, although if you think you can find a counterexample be my guest:
$$\overline{f(0)} = \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi$$
The functions I tried were: z, z+1, and z+i, so nothing transcendental.  We don't know that $\overline{f(z)}$ is analytic and we can't even push the conjugate inside the function, thus I feel like I don't have many tools at my disposal except for algebraic manipulation, and so far that hasn't gotten me anywhere.
 A: 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.
A: Take complex conjugates of both sides: you're saying
$$ f(0) = \frac{1}{2\pi} \int_0^{2\pi} \frac{e^{-i\phi}}{e^{-i\phi}-z} f(e^{i\phi})\ d\phi$$
Now take $\zeta = e^{i\phi}$, $d\zeta = i e^{i\phi}\ d\phi$, so if $C$ is the positively oriented unit circle the right side is 
$$ 
\frac{1}{2\pi i} \oint_C \frac{\zeta^{-2}}{\zeta^{-1} - z} f(\zeta)\ d\zeta  = \frac{1}{2\pi i} \oint_C \frac{f(\zeta)}{1-\zeta z} \frac{d\zeta}{\zeta}$$ 
If $|z| &lt 1$, $f(\zeta)/(1 - \zeta z)$ is analytic in a neighbourhood of the unit disk, so by the Cauchy integral theorem the result is $f(0)/(1 - 0 z) = f(0)$.
On the other hand, as others have remarked, if $|z| > 1$ the result is false: you'd have to take into account the residue at $\zeta = 1/z$, obtaining $f(0) - f(1/z)$. 
A: Here is a simple counterexample. 
Let $f(z) = 1$. 
Change variables, let $\zeta = e^{i\phi}$. 
Then your identity claims 
$$1 = \frac{1}{2\pi i} \int_\gamma \frac{d\zeta}{\zeta-z}$$
where $\gamma$ is the unit circle centered at the origin.  But this integral depends on whether $z$ is in the unit circle or not. 
If $z$ is not in the unit circle your identity becomes $1 = 0$.  
Addendum: Let's look at the integral more closely. 
The claim is $\overline{f(0)} = I$, where 
$$\begin{eqnarray*}
I &=& \frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi \\
&=& \frac{1}{2\pi i} \int_\gamma \frac{d \zeta}{\zeta-z} \overline{f(\zeta)}.
\end{eqnarray*}$$
We assume $z$ is not on the contour. 
Take the conjugate of $I$,
$$\begin{eqnarray*}
\overline I &=& -\frac{1}{2\pi i} \int_\gamma \frac{\overline{d\zeta}}{\overline\zeta-\overline z} f(\zeta) \\
&=& \frac{1}{2\pi i} \int_\gamma d\zeta \, \frac{\overline\zeta}{\overline \zeta - \overline z}\frac{f(\zeta)}{\zeta} \\
&=& \frac{1}{2\pi i} \int_\gamma d\zeta \,\left[
\frac{f(\zeta)}{\zeta} - \frac{\overline z f(\zeta)}{\zeta\overline{z}-1}
\right].
\end{eqnarray*}$$
where we have used $d (\overline \zeta\zeta) = \zeta d \overline \zeta + \overline \zeta d \zeta = 0$.
If $z = 0$ we find $\overline I = f(0)$. 
If $z\ne 0$, there are singularities at $\zeta = 0$ and $1/\overline z$. 
When $|z| < 1$ we don't pick up the residue at $1/\overline z$ 
(since this point is not in the disk) 
and so again, $\overline I = f(0)$.
If $|z| > 1$ we get the contribution from the other singularity, 
and so $\overline I = f(0) - f(1/\overline z)$.
Therefore, 
$$\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{i\phi}}{e^{i \phi}-z}\overline{f(e^{i\phi})}d\phi
 = \begin{cases}
\overline{f(0)} & |z| < 1 \\
\overline{f(0)} - \overline{f(1/\overline z)} & |z| > 1. 
\end{cases}$$
Going back to our counterexample for the original formula, 
notice for $f(z) = 1$ we find $1 = 1$ for $|z| < 1$ and for $|z|>1$ we get $0 = 1 - 1$, as we should. 
A: Changing variables from $\phi$ to $-\phi$ in your integral, you get
$$\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{-i\phi}}{e^{-i\phi} - z}\overline{f(e^{-i\phi})}\,d\phi$$
$$= \frac{1}{2\pi}\int_0^{2\pi}\frac{1}{1 - ze^{i\phi}}\overline{f(e^{-i\phi})}\,d\phi$$
$$= \frac{1}{2\pi i}\int_0^{2\pi}\frac{1}{e^{i\phi}(1 - ze^{i\phi})}\overline{f(e^{-i\phi})}ie^{i\phi}\,d\phi$$
In contour integral form this is
$$= \frac{1}{2\pi i}\int_{|\zeta| = 1}\frac{1}{\zeta(1 - z\zeta)}\overline{f(\bar{\zeta})}\,d\zeta$$
Note that $g(\zeta) = \overline{f(\bar{\zeta})}$ is analytic whenever $f$ is. 
If $|z| &lt 1$, the integrand has a single pole, at $\zeta = 0$, with residue $g(0)$, so by the Residue Theorem the integral is $g(0) = \overline{f(0)}$ which is probably the case your teacher had in mind.
If $|z| > 1$, there's a pole at $\zeta = {1 \over z}$ with residue $-g(1/z)$ if you do the calculation, as well as the pole from before with residue $g(0)$, so by the Residue Theorem the value of the integral is actually $ g(0)-g(1/z) = \overline{f(0)}-\overline{f(1/ \bar{z})} $.
A: In addition to the above comment by oenamen, z cannot be ON the unit circle. So I would suggest we look at the problem as if z is interior to the unit circle. I'm suspicious under this modification that it IS true. Indeed I'm staring at the reference in Reinhold Remmert's, Theory of Complex Functions which says: If f is analytic in a neighborhood of the closure of the disc B = Bs(0) (the OPEN disc of radius s centered about the origin), then If g=f conjugate, then
g(0) is the integral you wrote for all z in B. The proof is based on the fact that
h(w)=z*f(w)/(s^2-z*w) is holomorphic in a neighborhood of the closure of B.
z* = z conjugate in the preceding. I see you wanted a proof and not just to know it's true. Well, I'll furnish the 'it's true, it's true' salute, and see if that helps for now.
