A integral similar to Cauchy's integral formula Let $f$ be holomorphic on an open set containing the closed unit disc, and $\gamma$ be the unit circle, parametrized counterclockwise. Prove that for any $a \in \mathbb{C},$ $|a| \neq 1$,
$$
\frac{1}{2 \pi i} \int_{\gamma} \overline{f(z)}/(z-a) dz= 
\begin{cases} \overline{f(0)} &\mbox{if} \,\, |a|<1, \\
              \overline{f(0)}-\overline{f(1/ \bar{a})} &\mbox{if}\,\, |a|>1.
\end{cases}
$$
I don't know how to deal with the integral since $\overline{f}$ is not holomorphic. I was trying to use Mean Value Theorem of harmonic function but still not working. Any thoughts are appreciated.
 A: I suggest you think about the following approach. The problem with conjugation, as you have observed already, is that it is not holomorphic, but even though $\overline{f(z)}$ is not a holomorphic function the "double" conjugate $f^*(z) := \overline{f(\overline{z})}$ is holomorphic.  (Indeed, if the power series of $f(z)$ around a point has coefficients $c_n$, the function $f^*(z)$ has a power series around the same point with coefficients $\overline{c_n}$; this was an exercise when I took my first complex analysis course.) So you'd be better off if somehow you could work with $\overline{f(\overline{z})}$ instead of $\overline{f(z)}$. But you don't want to make the change of variables $w = \overline{z}$ because that is not holomorphic. 
Key idea: on the unit circle, which is the path of integration, we have $\overline{z} = 1/z$ and the function $1/z$ is holomorphic.  So make the change of variables $$w = 1/z$$ in the integral. Be careful about the path of integration, which remains the unit circle but gets reversed orientation, so you'll have some sign issues to deal with. In the integrand the numerator will be $\overline{f(1/w)} = \overline{f(\overline{w})} = f^*(w)$ (we have $1/w = \overline{w}$ because the path of integration is the unit circle) and the denominator will be a quadratic polynomial in $w$ that you can factor and split up into a sum of two terms using partial fractions. You can  apply the usual Cauchy integral formula to the two integrals, taking cases on one of them if $|a| < 1$ or $|a| > 1$. I tried this myself and everything worked, but I will leave the detailed calculations to you because it is a homework problem and I think it will be instructive to do it on your own.
A: This is from Remmert's book. Define $h(w)= \dfrac{\bar z f(w)}{s^2-\bar z w}$. This is holomorphic in a neighborhood of the origin, and $f(\zeta)\zeta^{-1}+h(\zeta)=\frac{f(\zeta)}{\zeta} \dfrac{ \bar\zeta}{\bar\zeta-z}$ when $|\zeta|=s$. Now integrate this over such circle and use Cauchy's formula. You should get that $$2\pi i f(0)=\int_{|\zeta|=s} \frac{f(\zeta)}{\zeta} \dfrac{ \bar\zeta}{\bar\zeta-z}d\zeta$$
Now we use that $\displaystyle \overline{\int g(z)dz}=\int\overline{g}{(z)}d\overline z$. Moreover, since $\zeta\overline \zeta=s^2$, we get $ \zeta d\overline\zeta+\overline\zeta d\zeta =0$, so the conjugate of this last equality gives what we want for $|z|<s$. Can you deal with the case $|z|>s$?

Spoiler To light  up Remmert's proof, I will try to adapt what was mentioned in the comments for any general circle $|z|=s$. It is better to read this after solving the problem by one's own, however. Note that $z\mapsto s^{2}z^{-1}$ sends this circle to itself and reverses orientation, but this gets cancelled by $d(z^{-1})=-z^{-2}dz$. Thus, we have (remember $\bar z=z^{-1}s^2$!) $$\int_{|z|=s}\frac{\overline{f(z)}}{z-w}dz=\int_{|z|=s}\frac{s^2\overline{f(\overline z)}}{z(s^2-zw)}dz$$
Note that $d(s^2z^{-1})=-s^2z^{-2}dz$ entails when $z\bar z=s^2$ the equality $zd\bar z+\bar z  dz=0$ just as in the proof of Remmert. Morover, if in the last integral we write things in terms of $u=\bar z$, since $\bar z=s^2z^{-1}$, we get the integral $$\int_{|u|=s}\frac{u\overline{f(u)}}{s^2-\bar u w}d\bar u=\overline{\int_{|u|=s}\frac{\overline u{f(u)}}{s^2- u \bar w}du}$$
This is almost the mysterious function in Remmert's proof. At any rate, since $$\frac{s^2}{z(s^2-zw)}=\left(\frac{1}{z}-\frac{1}{zw-s^2}\right)$$ we can decompose our integral into two summands. One will give the term with $\bar f(0)$ and the other will give the more extraneous term. 
