Help with equality of complex integrals I need to prove this equality of integrals...but i dont know how to begin, so if anyone can give an idea...
Let f a continuous function on $\overline{D}=\{z : |z|\leq 1\}$. Then:
$$\overline{\int_{|z|=1} f(z) dz} = - \int_{|z|=1} \frac{\overline{f(z)}}{z^2} dz$$
And then ihave to calculate that
$$\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}dz$$
for every a.
 A: Here is an approach. First note that

$$\overline{\int_{|z|=1} f(z) dz} =  \int_{|z|=1} {\overline{f(z)}}\,  d\bar{z}.$$

Then we use the fact that $|z|^2=z\bar{z}=1$, since $ |z|=1 $, as

$$  z\bar{z}=1 \implies \bar{z}=\frac{1}{{z}}\implies d\bar{z}= -\frac{1}{z^2} dz.  $$

Substituting back in the integral the desired result follows 

$$\overline{\int_{|z|=1} f(z) dz} =  \int_{|z|=1} {\overline{f(z)}}\,  d\bar{z}= -\int_{|z|=1} {\overline{f(z)}}\,  \frac{d{z}}{z^2} $$

A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\ol{\int_{\verts{z}\ =\ 1}\fermi\pars{z}\,\dd z}
     =-\int_{\verts{z}\ =\ 1}{\ol{f(z)}}\,{\dd z \over z^{2}}:\ {\large ?}}$

Note that
  \begin{align}&\ol{\int_{\verts{z}\ =\ 1}\fermi\pars{z}\,\dd z}
=\int_{\verts{z}\ =\ 1}\ol{\fermi\pars{z}}\,\dd\ol{z}
\end{align}

\begin{align}
&\mbox{However,}
\\[3mm]&1 = \verts{z}^{2} = z\,\ol{z}\quad\imp\quad
0 = \pars{\dd z}\ol{z} + z\,\dd\ol{z}\quad\imp\quad\dd\ol{z} 
=-\,{\ol{z}\,\dd z \over z}=-\,{\verts{z}^{2}\,\dd z \over z^{2}}
=-\,{\dd z \over z^{2}}
\end{align}

$$
\ol{\int_{\verts{z}\ =\ 1}\fermi\pars{z}\,\dd z}
=-\int_{\verts{z}\ =\ 1}\ol{\fermi\pars{z}}\,{\dd z \over z^{2}}
$$

A: We know that $\overline{e^z}=e^{\overline{z}}$ and use circle parametrisation $\gamma(t)=e^{it}$, where $t \in [0,2\pi)$
$\overline{\int_{|z|=1} f(z) dz}=\overline{\int_{0}^{2\pi}ie^{it}f(e^{it})} dt=\int_{0}^{2\pi}\Re{ie^{it}f(e^{it})}dt-i\int_{0}^{2\pi}\Im{ie^{it}f(e^{it})}dt$
Next $e^{it}=\cos t+i \sin t$, so:
$\int_{0}^{2\pi}\Re{ie^{it}f(e^{it})}dt=\int_{0}^{2\pi}=-\int_{0}^{2\pi}\sin t \Re{f(e^{it})}dt$
$\int_{0}^{2\pi}\Im{ie^{it}f(e^{it})}dt=\int_{0}^{2\pi}\cos t \Im{f(e^{it})}dt$
Now you can compute the second integral, it's(with the same parametrisation):
$\int_{0}^{2\pi}ie^{-it}\overline{f(e^{it})}dt=\int_{0}^{2\pi}\Re{ie^{-it}\overline{f(e^{it})}}dt+i\int_{0}^{2\pi}\Im{ie^{-it}\overline{f(e^{it})}}dt$
But $e^{-it}=\cos t-i \sin t$, so:
$\int_{0}^{2\pi}\Re{ie^{-it}\overline{f(e^{it})}}dt=\int_{0}^{2\pi}=\int_{0}^{2\pi}\sin t \Re{f(e^{it})}dy$
$\int_{0}^{2\pi}\Im{ie^{-it}\overline{f(e^{it})}}dt=\int_{0}^{2\pi}\cos t \Im{f(e^{it})}dt$
