Problem:Let $f(z)$ be analytic on some domain containing the closed unit disk $|z|\leq1$. Calculate $$\frac{1}{2 \pi i}\int_D\frac{\overline{f(z)}dz}{z-a}$$ if $|a|<1$, where D is unit circle.
Solution: Let$$f(z)=u(z)+iv(z)$$ where, $u,v$ are real functions. By Cauchy integral formula we have: $$\frac{1}{2 \pi i}\int_D\frac{f(z)dz}{z-a}=f(a)$$ so,
$$\int_D\frac{(u(z)+iv(z))dz}{z-a}=2 \pi if(a)$$ $\int_D \frac {u(z)}{z-a}dz=-2 \pi Im(f(a))$ and $\int_D \frac {u(z)}{z-a}dz=2 \pi Re(f(a))$. So, $$\frac{1}{2 \pi i}\int_D\frac{\overline{f(z)}dz}{z-a}=\frac{1}{2 \pi i}(-2 \pi Im(f(a))-i2 \pi Re(f(a))=- \overline{f(a)}$$ What is wrong with this solution( in book solution is $\overline{f(a)}$)?
If $g$ is holomorphic on $D(0,R), R>1,$ then
$$\tag 1 g(0)=\frac{1}{2\pi}\int_0^{2\pi} g(e^{it})\,dt.$$
This falls right out of Cauchy's theorem (or the mean value property of harmonic functions if you like).
Do a little work to see our expression equals
$$\tag 2 \frac{1}{2\pi}\int_0^{2\pi} \frac{ \overline{f(e^{it})}}{1-ae^{-it}}\,dt = \overline{ \frac{1}{2\pi}\int_0^{2\pi}\frac{f(e^{it})}{1-\bar{a}e^{it}} }\,dt.$$
If we let $g(z)=f(z)/(1-\bar{a}z)$ and use $(1),$ we see that $(2) $ equals $\overline{f(0)}.$