Computing $\int_{|z|=\rho} {1 \over |z-a|^{4}}\ |dz|$ when $0 > \rho \ne |a|$ Goal: In what follows we let $\gamma = \rho e^{it}$ on $0 \le t \le 2\pi$ serve as the paramaterization of the curve $|z| = \rho$.  We also assume that $\rho > 0$ (else the computation is trivially equal to zero).
We wish to compute
$$
\int_{|z|=\rho} {1 \over |z-a|^{4}}\ |dz| 
$$
when $0 < \rho \ne |a|$.
Attempt, with questions in step (3) and (4):


*

*We have that 
\begin{align*}
= & \int_{|z|=\rho} {1 \over |z-a|^{4}}\ |dz| \\
= & \int_{|z| = \rho} {1 \over (z-a)^2 \overline{(z - a)^2}}\ |dz| \\
= & \int_{|z| = \rho} {1 \over (z-a)^2 (\overline{z} - \overline{a})^2}\ |dz| \\
= & \int_0^{2 \pi} {1 \over (\rho e^{i t} - a)^2 (\rho e^{-i t} - \overline{a})^2} |i\rho e^{i t}|\ dt\\
= & \rho \int_0^{2 \pi} {1 \over (\rho e^{i t} - a)^2 (\rho e^{-i t} - \overline{a})^2} \ dt\\
\end{align*}

*As a consequence, we can say either of the following:
\begin{equation*}
\tag{*}
\int_{|z|=\rho} {1 \over |z-a|^{4}}\ |dz| = \rho \int_0^{2 \pi} {{1 \over (\rho e^{i t} - a)^2} \over (\rho e^{-i t} - \overline{a})^2} \ dt
\end{equation*}
or also
\begin{equation*}
\tag{**}
\int_{|z|=\rho} {1 \over |z-a|^{4}}\ |dz| = \rho \int_0^{2 \pi} {{1 \over (\rho e^{-i t} - \overline{a})^2} \over (\rho e^{i t} - a)^2} \ dt
\end{equation*}

*Now if $\rho < |a|$, then doesn't the fact that ${1 \over (z-a)^4}$ is analytic on the disk of size $\rho$ imply that the integral is zero (via Cauchy's integral theorem on a disk)?

*On the other hand, if $\rho > |a|$, can we apply Cauchy's Integral Formula to the numerator of, say, $(*)$  to yield the computation?


EDIT: Substituting $|dz| = {-i\rho \over z}\  dz$ and $\overline{z} = {\rho^2 \over z}$ yields:

\begin{align*}
= & \int_{|z|=\rho} {1 \over |z-a|^{4}}\ |dz| \\
= & \int_{|z| = \rho} {1 \over (z-a)^2 \overline{(z - a)^2}}\ |dz| \\
= & \int_{|z| = \rho} {1 \over (z-a)^2 (\overline{z} - \overline{a})^2}\ |dz| \\
= & \int_{|z| = \rho} {1 \over (z-a)^2 ({\rho^2 \over z} - \overline{a})^2}\ {-i\rho \over z}\  dz \\
= & \int_{|z| = \rho} {{-i\rho \over z({\rho^2 \over z} - \overline{a})^2} \over (z-a)^2 }\ \  dz \\
\end{align*}
But it seems like here Cauchy's Integral Formula doesn't apply since ${-i\rho \over z({\rho^2 \over z} - \overline{a})^2}$ is undefined at $0$ and hence not analytic inside $\gamma$.
 A: I'm in the process of working through this exact problem, and had the exact same point of confusion as the OP. Since this is so far unanswered, I'll put my answer here for the future benefit of others like myself. Please check it thoroughly. I have posted it in good faith, but everyone makes mistakes!
Beginning from the last step in the OP's work:
$$\int_{|z|=\rho}\dfrac{\left(\frac{-i\rho}{z\left(\dfrac{\rho^2}{z}-\overline{a}\right)^2}\right)}{(z-a)^2}dz=-i\rho\int_{|z|=\rho}\dfrac{1}{z\left(\dfrac{\rho^2}{z}-\overline{a}\right)^2(z-a)^2}dz$$
$$=-i\rho\int_{|z|=\rho}\dfrac{z}{(\rho^2-\overline{a}z)^2(z-a)^2}dz,$$
and we can (finally) employ the Residue Theorem. If $\rho<|a|$, then there is no pole at $z=a$, and regardless, there is a pole of order $2$ at $z=\rho^2/\overline{a}$. Calculating the residues is basic (and exceedingly tedious to type out), so we leave that to the reader. The residue of the integrand at $z=\rho^2/\overline{a}$ is $-\dfrac{a\overline{a}^3+\rho^2\overline{a}^2}{(\rho^2-a\overline{a})^3}$ so:
For $\rho<|a|$:
Our result is $-2\pi\rho\dfrac{a\overline{a}^3+\rho^2\overline{a}^2}{(\rho^2-a\overline{a})^3}$.
For $\rho>|a|$: In this case there is also a pole of order $2$ at $z=a$ and the residue there is $\dfrac{\overline{a}a+\rho^2}{(\rho^2-a\overline{a})^3}$. So, our result in this case is the previous residue plus this one, and of course we must multiply by $-i\rho(2\pi i)$. After a little simplification, we arrive at $\dfrac{2\pi\rho}{(\rho^2-a\overline{a})^3}(a\overline{a}-a\overline{a}^3+\rho^2-\rho^2\overline{a}^2)$.
