Complex integration, any ideas? I need to solve the following integral of a function analytic on an open environmemt of the unit circle, but i dont know how to get that answers
$$\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a} dz$$
The answer is $\overline{f(0)}$ if $|a| < 1$
and $\overline{f(0)}- \overline{f(\frac{1}{\overline{a}})}$ if $|a|>1$
 A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
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 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
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$\ds{}$
\begin{align}&\color{#66f}{\large{1 \over 2\pi\ic}\int_{\verts{z}\ =\ 1}{\ol{\fermi\pars{z}} \over z - a} \,\dd z}
=\ol{\vphantom{\Huge A^{a}}\bracks{-\,{1 \over 2\pi\ic}
\int_{\verts{\color{#c00000}{\LARGE\ol{z}}}\ =\ 1}
{\fermi\pars{z} \over \ol{z} - \ol{a}}\,\dd\ol{z}}}
\qquad\qquad\qquad\qquad\qquad\quad\pars{1}
\\[3mm]&=\ol{\vphantom{\Huge A^{a}}\bracks{-\,{1 \over 2\pi\ic}
\int_{\verts{z}\ =\ 1}{%
\fermi\pars{z} \over 1/z - \ol{a}}\,\pars{-\,{\dd z \over z^{2}}}}}
=\ol{\vphantom{\Huge A^{a}}\bracks{-\,{1 \over 2\pi\ic}
\int_{\verts{z}\ =\ 1}{%
\fermi\pars{z} \over \ol{a}z\pars{z - 1/\ol{a}}}\,\dd z}}\qquad\quad\pars{2}
\\[3mm]&=\left\lbrace\begin{array}{rclcl}
\ol{\left.-\,{1 \over 2\pi\ic}\,2\pi\ic\,
{\fermi\pars{z} \over \ol{a}\pars{z - 1/\ol{a}}}
\right\vert_{z\ =\ 0}} \!\!\!\!\!& = & \!\!\!\!\!\ol{\fermi\pars{0}} & \mbox{if} & \verts{a} < 1
\\[3mm]
\ol{\left.-\,{1 \over 2\pi\ic}\,2\pi\ic\,
{\fermi\pars{z} \over \ol{a}\pars{z - 1/\ol{a}}}
\right\vert_{z\ =\ 0}}\
-\
\ol{\left.\,{1 \over 2\pi\ic}\,2\pi\ic\,{\fermi\pars{z} \over \ol{a}z}
\right\vert_{z\ =\ 1/\ol{a}}} \!\!\!\!\!
& = & \!\!\!\!\!
\ol{\fermi\pars{0}} - \ol{\fermi\pars{1 \over \ol{a}}} & \mbox{if} & \verts{a} > 1
\end{array}\right.
\end{align}

Note that it's a change of sign between the end of line $\pars{1}$ and the
  beginning of line $\pars{2}$ since conjugation change the rotation sense which we 'loosely' indicates with $\ds{\color{#c00000}{\LARGE\ol{z}}}$ in line $\pars{1}$.

\begin{align}&\color{#66f}{\large{1 \over 2\pi\ic}\int_{\verts{z}\ =\ 1}{\ol{\fermi\pars{z}} \over z - a} \,\dd z
=
\left\lbrace\begin{array}{lcl}
\ol{\fermi\pars{0}} & \quad\mbox{if}\quad & \verts{a} < 1
\\[3mm]
\ol{\fermi\pars{0}} - \ol{\fermi\pars{1 \over \ol{a}}}
& \quad\mbox{if}\quad & \verts{a} > 1
\end{array}\right.}
\end{align}
A: First assume $f(0)=0$. Since $f$ is defined around $0$, extend up to the unit circle, $f$ have a Taylor series of the form $\sum\limits_{k=1}^{\infty}a_{k}z^{k}$ that have radius of convergence at least $1$. Thus $f(\overline{z})=\overline{f(z)}$ on the unit circle boundary. Also, on the boundary, then $\frac{1}{z}=\overline{z}$. Hence our integral reduce to $\frac{1}{2\pi i}\int_{|z|=1}\frac{f(\frac{1}{z})}{z-a}$. Now we don't even know if $\frac{f(\frac{1}{z})}{z-a}$ is defined inside the circle, in fact $\frac{f(\frac{1}{z})}{z-a}$ is different from our original function, but its value on the boundary match so that's good enough. And even better $\frac{f(\frac{1}{z})}{z-a}$ is now analytic now that we got rid of the conjugate. We know that $f(\frac{1}{z})$ is also defined outside the boundary all the way to infinity due to property of $f$. Hence we will calculate this using residue outside the boundary.
To get residue at infinity, consider $-\frac{1}{z^{2}}\frac{f(z)}{\frac{1}{z}-a}=\frac{f(z)}{z(az-1)}$ at $z=0$. This is clearly a removable singularity (remember we assume $f(0)=0$). Hence the residue at infinity of $\frac{f(\frac{1}{z})}{z-a}$ is $0$.
If $|a|>1$ we got another singularity at $z=a$. This is a simple pole so residue is just $f(\frac{1}{a})$. Using the Taylor expansion argument above we get $\overline{f(\frac{1}{\overline{a}})}$.
Hence $\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}=\frac{1}{2\pi i}\int_{|z|=1}\frac{f(\frac{1}{z})}{z-a}=0$ if $|a|<1$ and $-\overline{f(\frac{1}{\overline{a}})}$ if $|a|>1$.
To complete the argument we remove the assumption that $f(0)=0$. Define $g(z)=f(z)-f(0)$. Then our integral is $\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{f(z)}}{z-a}=\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{g(z)}}{z-a}+\overline{f(0)}\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z-a}$. Apply the above we get $\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{g(z)}}{z-a}=0$ if $|a|<1$ and $-\overline{g(\frac{1}{\overline{a}})}=\overline{f(0)}-\overline{f(\frac{1}{\overline{a}})}$ if $|a|>1$. And easily calculate: $\overline{f(0)}\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z-a}=\overline{f(0)}$ if $|a|<1$ and $0$ if $|a|>1$. So as the whole we get $\frac{1}{2\pi i}\int_{|z|=1}\frac{\overline{g(z)}}{z-a}=\overline{f(0)}$ if $|a|<1$ and $\overline{f(0)}-\overline{f(\frac{1}{\overline{a}})}$ if $|a|>1$.
