By Cauchy's Theorem we have
$
f(a)=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{w-a}dw
$
where $\gamma$ is the path $\gamma(t)=b+re^{it},~t\in[0,2\pi],$ $a\in B_r(b)$ where $B_r(b)$ is the closed ball of radius $r$ with center $b,$ fully lying in the domain of $f.$
Now, in case of $f(z)=\bar{z},$ we can't apply Cauchy's Theorem as $f$ is not an analytic function. So my doubt is whether we can say
$
\bar{a}=\frac{1}{2\pi i}\int_{\gamma}\frac{\bar{z}}{z-a}dz
$ for the path $\gamma$ with center different from $a.$
Any counter example or suggestions or hints is fully appreciated. Thanks in advance.
EDIT: There was a mistake in the question from my side. I have edited the question.
This is what I tried so far.
$\frac{1}{2\pi i}\int_{\gamma}\frac{\bar{z}}{z-a}dz\\
=\frac{1}{2\pi i}\int_{0}^{2\pi}\frac{(\bar{b}+re^{-i\theta})ire^{i\theta}d\theta}{(b-a)+re^{i\theta}}\\
=\frac{\bar{b}}{2\pi i}\int_0^{2\pi}\frac{ire^{i\theta}d\theta}{b-a+re^{i\theta}} +\frac{1}{2\pi}\int_0^{2\pi}\frac{r^2d\theta}{b-a+re^{i\theta}}\\
=\frac{\bar{b}}{2\pi i}\int_{\gamma}\frac{zdz}{z+b-a} +\frac{r^2}{2\pi}\int_0^{2\pi}\frac{d\theta}{b-a+re^{i\theta}}\\
=\bar{a}(b-a)+\frac{r^2}{2\pi}\int_0^{2\pi}\frac{d\theta}{b-a+re^{i\theta}}
$
How to proceed further?
Best Answer
Let's compute $$\oint _{|z-a| = r} \frac{1}{2\pi i} \frac{\overline{z}}{z-a} dz$$
Since $|z-a| = r$ implies $(z-a) (\overline{z}-\overline{a}) = r^2$, we have $ z\overline{z} -a\overline{z} - \overline{a}z + a\overline{a} = r^2$, so $$\overline{z} = \frac{r^2-a\overline{a}+\overline{a}z} {z-a}$$
Taking it into the integral: $$\oint _{|z-a| = r} \frac{1}{2\pi i} \frac{\overline{z}}{z-a} dz = \oint _{|z-a| = r} \frac{1}{2\pi i} \frac{r^2-a\overline{a}+\overline{a}z}{(z-a)^2} dz$$
This is the Cauchy's integral formula for derivatives, which equals to $$\frac{d}{dz}(r^2-a\overline{a}+\overline{a}z)\big|_{z=a} = \overline{a}$$ Which is the desired result.
EDIT: I see that you edited your question after my answer. But that's still OK. In this way, we are computing
$$\oint _{|z-b| = r} \frac{1}{2\pi i} \frac{\overline{z}}{z-a} dz = \oint _{|z-b| = r} \frac{1}{2\pi i} \frac{r^2-b\overline{b}+\overline{b}z}{(z-a)(z-b)} dz$$
If $b = a$, just as I said above. Otherwise, we have $$\frac{1}{(z-a)(z-b)} = \frac{1}{a-b} \left(\frac{1}{z-a} - \frac{1}{z-b}\right)$$
So $$\oint _{|z-b| = r} \frac{1}{2\pi i} \frac{r^2-b\overline{b}+\overline{b}z}{(z-a)(z-b)} dz = \frac{1}{a-b} ((r^2-b\overline{b}+\overline{b}a) - (r^2-b\overline{b}+\overline{b}b)) = \overline{b}$$