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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.

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Hint: $\bar{z} = {\rho^2 \over z}$ and $1 = {1 \over iz} dz$ – Zarrax Feb 24 '11 at 7:15
up vote 2 down vote accepted

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.

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Hm, why did you suggest to parametrize the curve if you said to just write everything in terms of partial fractions (presumably to use Cauchy's integral formula)?? – user1736 Feb 24 '11 at 18:11
Well, you can't use Cauchy's Integral Formula on integrals with respect to $|dz|$. Somehow you need to convert it into an integral with respect to $dz$. The parametrization was just an intermediate step. – Jesse Madnick Feb 24 '11 at 20:53

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?

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