Computing $\int_\gamma { |dz| \over |z-a|^2}$ Goal: Compute
$$
\int_{|z|= \rho} {|\mathrm{d}z| \over |z-a|^2}
$$
under the condition $|a| \ne \rho$.  
Ahlfors' Hint: make use of the equations $z \bar{z} = \rho^2$ and 
$$
|\mathrm{d}z| = -i \rho {\mathrm{d}z \over z}.
$$
Old Attempt (see new one below):


*

*First let $\gamma$ be the arc of $|z| = \rho$ paramaterized by
$$
\gamma(t) = \rho e^{it} \text{ s.t. } 0 \le t \le 2\pi
$$
and whereby
$$
\gamma'(t) = \rho i e^{it}
$$

*At this point, I can only grab and present what I think might be useful facts to solve this problem.

*First, we have that
$$
\int_\gamma {|\mathrm{d}z| \over |z-a|^2} \ge \left|\int_\gamma {\mathrm{d}z \over (z-a)^2} \right|
$$

*Second, we have the fundamental fact that
$$
\int_\gamma {\mathrm{d}z \over z-a} = k 2 \pi i \text{ for some }k \in \mathbb{N}
$$
Question: How does one proceed in showing the desired result?

New Attempt Using T.A.E.'s Hint:


*

*We have first that
$$
       \frac{1}{|z-a|^{2}}=\frac{1}{(z-a)(\overline{z}-\overline{a})}=
      \frac{1}{(z-a)(\rho^{2}/z-\overline{a})}=\frac{z}{(z-a)(\rho^{2}-\overline{a}z)}
$$

*Our problem then reduces to the computation of
$$
\int_{|z|= \rho} {|dz| \over |z-a|^2} = \int_{|z|=\rho} \frac{z}{(z-a)(\rho^{2}-\overline{a}z)}\ |dz|
$$

*Now let $\gamma$ be the circle about the origin of radius $\rho$ with parameterization of 
$$
\gamma(t) = \rho e^{{it}} \text{ s.t. } 0 \le t \le 2 \pi
$$
which then yields us
$$
\gamma'(t) = \rho i e^{it}
$$

*Making use of the above parameterization along with Ahlfors' hint that
$$
|dz| = -i \rho {dz \over z}.
$$
we have that
$$
\int_{\gamma} \frac{z}{(z-a)(\rho^{2}-\overline{a}z)}\ |dz| = \int_{\gamma} \frac{-i \rho z}{(z-a)(\rho^{2}-\overline{a}z)z}\ dz = \int_{\gamma} \frac{-i \rho }{(z-a)(\rho^{2}-\overline{a}z)}\ dz
$$
so that then
$$
\int_{\gamma} \frac{-i \rho }{(z-a)(\rho^{2}-\overline{a}z)}\ dz = \int_0^{2 \pi} {-i \rho \gamma'(t) \over (\gamma(t) - a)(\rho^2 - \overline{a}\gamma(t))}\ dz = \int_0^{2 \pi} {\rho^2 e^{it} \over (\rho e^{{it}} - a)(\rho^2 - \overline{a}\rho e^{{it}})}\ dz
$$
Question: It's not clear to me how considering cases where $\rho < |a|$ and $\rho > |a|$ is going to be of help here.
 A: Try applying the formula another time:
$$
       \frac{1}{|z-a|^{2}}=\frac{1}{(z-a)(\overline{z}-\overline{a})}=
      \frac{1}{(z-a)(\rho^{2}/z-\overline{a})}=\frac{z}{(z-a)(\rho^{2}-\overline{a}z)}.
$$
Then consider the cases where $\rho < |a|$ and $\rho > |a|$. For example, the integral can now be written as
$$
    \oint_{|z|=\rho}\frac{d|z|}{|z-a|^{2}}=\oint_{|z|=\rho}\frac{z}{(z-a)(\rho^{2}-\overline{a}z)}\frac{-i\rho}{z}dz=2\pi\frac{1}{2\pi i}\oint_{|z|=\rho}\frac{dz}{(z-a)(\rho^{2}-\overline{a}z)}.
$$
Assuming $\rho \ne |a|$, partial fractions gives
$$
   \frac{1}{(z-a)(\rho^{2}-\overline{a}z)}
      =\frac{1}{\rho^{2}-|a|^{2}}\left[\frac{1}{z-a}+\frac{\overline{a}}{\rho^{2}-\overline{a}z}\right].
$$
You should be able to take it from there if you know basic complex analysis. If you're integrating on $|z|=\rho > |a|$, then $1/(z-a)$ has a singularity inside the circle $|z|=\rho$, but $1/(\rho^{2}-\overline{a}z)$ does not. If $|z|=\rho < |a|$, then $1/(z-a)$ does not have a singularity inside $|z|=\rho$, but $1/(\rho^{2}-\overline{a}z)$ does.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\color{#c00000}{\int_{\verts{z} = \rho}
     {\verts{\dd z} \over \verts{z - a}^{2}}}:\ {\large ?}.
     \quad a \equiv \alpha + \beta\,\ic\,,\quad \alpha, \beta \in {\mathbb R}\,,
     \quad \root{\alpha^{2} + \beta^{2}} \not= \rho}$.

With $\ds{z = \rho\expo{\ic\theta}\,,\qquad -\pi \leq \theta < \pi}$:
  \begin{align}
\color{#c00000}{\int_{\verts{z} = \rho} {\verts{\dd z} \over \verts{z - a}^{2}}}&=
\int_{-\pi}^{\pi}{\verts{d\pars{\rho\expo{\ic\theta}}} \over \verts{\rho\expo{\ic\theta} - a}^{2}}
=\int_{-\pi}^{\pi}{\rho\,\dd\theta \over
\bracks{\rho\cos\pars{\theta} - \alpha}^{2}
+\bracks{\rho\sin\pars{\theta} - \beta}^{2}}
\\[3mm]&=\rho\int_{-\pi}^{\pi}{\dd\theta \over \rho^{2} -2\rho\bracks{\alpha\cos\pars{\theta} + \beta\sin\pars{\theta}} + \verts{a}^{2}}
\end{align}

With $\ds{t \equiv \tan\pars{\theta \over 2}}$:
\begin{align}
\color{#c00000}{\int_{\verts{z} = \rho} {\verts{\dd z} \over \verts{z - a}^{2}}}&=
\rho\int_{-\infty}^{\infty}{2\,\dd t/\pars{1 + t^{2}} \over \rho^{2} + \verts{a}^{2}-2\rho\braces{\alpha\bracks{1 - t^{2}}/\bracks{1 + t^{2}}
+ \beta\bracks{2t}/\bracks{1 + t^{2}}}}
\\[3mm]&=2\rho\int_{-\infty}^{\infty}{\dd t \over
\pars{\rho^{2} + \verts{a}^{2}}\pars{1 + t^{2}} - 2\rho\alpha\pars{1 - t^{2}}
-2\beta t}
\\[3mm]&=2\rho\int_{-\infty}^{\infty}{\dd t \over
\pars{\rho^{2} + 2\alpha\rho + \verts{a}^{2}}t^{2} - 2\beta t
+\rho^{2} - 2\alpha\rho + \verts{a}^{2}}
\\[3mm]&={2\rho \over \pars{\rho + \alpha}^{2} + \beta^{2}}
\int_{-\infty}^{\infty}{\dd t \over t^{2} - 2\mu t + \nu^{2}}
\end{align}
where
\begin{align}
\mu &\equiv {\beta \over \pars{\rho + \alpha}^{2} + \beta^{2}}
={\beta \over \verts{\rho + a}^{2}}
\\[3mm]
\nu &\equiv
\root{%
\pars{\rho - \alpha}^{2} + \beta^{2} \over \pars{\rho + \alpha}^{2} + \beta^{2}}
={\verts{\rho - a} \over \verts{\rho + a}}
\end{align}

\begin{align}
\color{#c00000}{\int_{\verts{z} = \rho} {\verts{\dd z} \over \verts{z - a}^{2}}}&=
{2\rho \over \verts{\rho + a}^{2}}
\int_{-\infty}^{\infty}{\dd t \over \pars{t - \mu}^{2} - \mu^{2} + \nu^{2}}
={2\rho \over \verts{\rho + a}^{2}}
\int_{-\infty}^{\infty}{\dd t \over t^{2} - \mu^{2} + \nu^{2}}
\\[3mm]&\mbox{and}\quad
\nu^{2} - \mu^{2}=
{\pars{\rho - \alpha}^{2} \over \verts{\rho + a}^{2}}
\end{align}

When $\ds{\rho = \alpha = \Re\pars{a}}$, the integral
$\ds{\color{#f00}{\large\tt\mbox{diverges !!!}}}$. So, we consider the case
$\ds{\rho \not= \alpha = \Re\pars{a}}$:
\begin{align}
\color{#c00000}{\int_{\verts{z} = \rho} {\verts{\dd z} \over \verts{z - a}^{2}}}&=
{2\rho \over \verts{\rho + a}^{2}}\,{1 \over \root{\nu^{2} - \mu^{2}}}
\int_{-\infty}^{\infty}{\dd t \over t^{2} + 1}
={2\pi\rho \over \verts{\rho + a}^{2}}\,{1 \over \root{\nu^{2} - \mu^{2}}}
\end{align}

$$\color{#00f}{\large%
\int_{\verts{z} = \rho} {\verts{\dd z} \over \verts{z - a}^{2}}
=2\pi\,{\rho \over \verts{\rho -\Re\pars{a}}\ \verts{\rho + a}}}\,,
\qquad\rho \not= \Re\pars{a}\,,\quad \rho \not= a
$$

