Evaluating a complex integral I'm having trouble figuring out how to evaluate the integral $\int_{|z|=\rho} \frac{|dz|}{|z-a|^2}$ where $|a| \neq \rho$. This is a problem in Ahlfors in the section on Cauchy's Integral Formula, and I think by convention when he says $|z|=\rho$, he means the parametrization $z=\rho e^{it}, \; 0 \leq t \leq \pi$ (so that the winding number of a point inside this circle would be 1). 
I'm guessing there is some smart way to apply the integral formula (since it's in this section), and I naively tried to expand the integrand. However, you end up with $\frac{1}{(z-a)(\bar{z}-\bar{a})}$, and I don't believe $\bar{z}-\bar{a}$ is an analytic function, so I'm not sure how to proceed from here.
 A: Hint: Try to write everything in terms of $z$.
Since we're working on the circle, we have $|z|^2 = z\overline{z} = \rho^2$.  Also, $|dz| = \rho\,dt$ and $dz = i\rho e^{it}\,dt = iz\,dt$.
You might then want to use partial fractions.
A: To use Zarrax and Jesse's hints, it's possible to see that
$$\int_{|z|=\rho} \frac{|dz|}{|z-a|^2} = \int_{|z|=\rho} \frac{1}{(z-a)(\bar{z}-\bar{a})}(-i\rho) \frac{dz}{z}$$
$$ =\int_{|z|=\rho} \frac{-i \rho}{(z-a)(\frac{\rho^2}{z}-\bar{a})(z)} dz$$
$$ =\int_{|z|=\rho} \frac{-i \rho}{(z-a)(\rho^2 - \bar{a}z)} dz$$
$$ =\frac{i \rho}{\bar{a}} \int_{|z|=\rho} \frac{1}{(z-a)(z-\frac{\rho^2}{\bar{a}})} dz$$
$$ =\frac{i \rho}{\bar{a}} 2 \pi i n(\gamma,a) \frac{1}{a-\frac{\rho^2}{\bar{a}}}$$
where $\gamma = \{|z|=\rho\}$, $n(\gamma,a)$ is the winding number of $a$ along $\gamma$, and where we assume that $|a| < \rho$ such that $a$ lies inside the circle $|z|=\rho$.\
If $|a|>\rho$, then we can reverse the role of the two functions, since then $\frac{1}{z-a}$ will be analytic inside $\gamma$.
Does that look correct?
