Complex integration, any ideas?

Question

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 $|a|>1$

Answer 1

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

Answer 2

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 $|a|>1$.