Compute the integral $\int_{|z|=\rho}|z-a|^{-4}|dz|$ with $|a|\neq \rho$ I need help in computing the integral indicated above. What I've tried so far:
Parametrize the curve indicated by $|z|=\rho$ with $\gamma = z(t) = \rho \cos t + i\sin t$. Then by definition
$$
\int_\gamma f(z)|dz|=\int_\gamma f(z(t))|z'(t)| dt
$$
gives the following
\begin{align}
\int_\gamma |z-a|^{-4} |dz| & = \int_0^{2\pi} |\rho \cos t+i\rho \sin t-a_1-ia_2|^{-4}\rho dt\\
& = \rho\int_0^{2\pi}\frac{1}{(\rho^2-2a_1\cos t-2a_2\sin t + a_1^2+a_2^2)^2} dt
\end{align}
Where $a=a_1+ia_2$. It's not hard to see how this becomes complicated very easily. I want to know if there is some sort of 'trick' I'm not aware of or something I might be missing.
 A: I think there's not much way to avoid some tedious calculations. I won't provide the full calculations, but I think this roadmap should be sufficient.
The cases of $a=0$ and $\rho = 0$ are pretty straightforward. For the other cases, I think this identity is a useful hint
\begin{align*}
\left| z-a \right|^{-4} = (z-a)^{-2} \overline{(z-a)^{-2}}
\end{align*}
From here, after replacing $z$ with $\rho e^{2\pi i t}$, you have to consider the cases of $|a| > \rho$ and $|a| < \rho$ and use Cauchy's Integration formula.
A: Here is a solution with the aid of the technical subsititution: $|dz|=-i\rho\frac{dz}{z}$ and the fact that $z\bar{z}=\rho^{2}$ on the contour.
\begin{align}
\int_{C}|z-a|^{-4}|dz|&=-i\rho\int_{C}(z-a)^{-2}(\bar{z}-\bar{a})^{-2}z^{-1}dz\\
&=-\frac{i\rho}{\bar{a}^{2}}\int_{C}\frac{zdz}{(z-a)^{2}(z-a')^{2}}\\
\text{where }a'=\frac{\rho^{2}}{\bar{a}}
\end{align}
Observe that $a$ and $a'$ lie in two different regions determined by the circle. Suppose $a$ is in the circle, then $f=\frac{z}{(z-a')^{2}}$ is analytic in the circle. Thus $\int_{C}\frac{zdz}{(z-a)^{2}(z-a')^{2}}=2\pi if'(a)$ by CIF.
