# How to compute this contour integral with a modulus sign in the integrand,

Evaluate the integral $$∫_{∣z∣=ρ} \frac {1}{|z−a|^{2}}|dz|$$ where ρ is a positive number, a is a complex number, and |a|<ρ.

I welcome any hints on how to get started on this problem.

The modulus sign is tripping me up a bit -- Cauchy's integral formula doesn't seem applicable either.

I then expanded the modulus function in the denominator to get

$$\sqrt {(x-u)^2+(y-v)^2}$$

and expanding the terms inside the square root just makes things messier and doesn't lead to anything insightful.

Another idea would be to use the Residue Theorem, since the integrand is real-valued, but what can I do about the |dz| term? And also, the integration appears to be over a circle and not a subset of the real line. Usually, applications of the Residue Theorem is to compute real integrals over R or $R^+$ and then we complexify the integral and choose a suitable contour to give some easy estimates of the integrand over certain parts of the contour. Here, a complex contour is already given in the problem statement: |z|=$\rho$

Thanks,

• what do you mean by $|dz|$ as a measure of the integral? Commented Oct 10, 2015 at 1:53
• Why don't you simply parametrize the integral as $z=\rho e^{it}$. Then if it really is $|dz|$ we have $|dz| = \rho dt$, the limits of the integral from $0$ to $2\pi$ and the denominator $(a^2 + \rho^2 - 2a\rho\cos t)$. This is a doable real integral. You don't need any fancy complex analysis to do this integral actually. Commented Oct 10, 2015 at 1:58
• can you give me link for that online source you mentioned which uses absolute value? Commented Oct 10, 2015 at 2:17
• In this case, the line integral is over measure $ds$ which is already an absolute value, so you have $ds = |\vec{r}'|dt$. In general it depends on what type of measure you have. Commented Oct 10, 2015 at 2:45
• math.stackexchange.com/questions/1437103/… Commented Oct 10, 2015 at 4:38

Along the curve $$\gamma:\quad\phi\mapsto z(\phi):=\rho e^{i\phi}\qquad(0\leq\phi\leq2\pi)$$ one has $$|z-a|^2=(z-a)(\bar z-\bar a)=(z-a)\left({\rho^2\over z}-\bar a\right)$$ and $dz = i \rho e^{i\phi} d\phi$. The latter implies $$|dz|=\rho d\phi={\rho\over i}{dz\over z}\ .$$ In this way we obtain $$Q:=\int_\gamma{|dz|\over |z-a|^2}={\rho\over i}\int_\gamma{1\over(z-a)\left({\rho^2\over z}-\bar a\right)}\>{dz\over z}={\rho\over i}\int_\gamma{1\over(z-a)(\rho^2-\bar az)}\>dz\ .$$ Since $|a|<\rho$ we therefore get $$Q={\rho\over i}\>2\pi i\>{\rm res}\left({1\over(z-a)(\rho^2-\bar az)}\biggm| z:=a\right)={2\pi\rho\over \rho^2-|a|^2}\ .$$
• How this comes : $$|dz|=\rho d\phi={\rho\over i}{dz\over z}\$$
• Because, $$|dz|=|i \rho e^{i\phi} d\phi|=\rho|d\phi|$$