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.

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.
