# How to integrate $\int\limits_{|z| = R} \frac{|dz|}{|z-a|^2}$

I need to integrate, $\int\limits_{|z| = R} \frac{|dz|}{|z-a|^2}$ where $a$ is a complex number such that $|a|\ne R$.

So first I tried polar coordinates, which gives something I cannot continue.

Then I tried to write $|dz| = rd\theta = dz/ie^{i\theta}$ and I have $$\int\limits_{|z| = R} \frac{dz}{ie^{i\theta}(z-a)(\overline{z}-\overline{a})}$$

which makes me want to use cauchy's integral formula, but I'm not sure if it has a pole at $z = a$ or not.

How to I calculate this integral?

First sub $z=R e^{i \phi}$, $dz = i R e^{i \phi} d\phi$, $|dz|=R d\phi$. Then realize that

$$|z-a|^2 = R^2 + |a|^2 - 2 R |a| \cos{\phi}$$

(I set an arbitrary phase to zero - it won't matter for the integration.)

The integral then becomes

$$R \int_0^{2 \pi} \frac{d\phi}{R^2 + |a|^2 - 2 |a| R \cos{\phi}}$$

Now - and this might seem weird - we go back to a complex representation so we may evaluate the integral using the residue theorem. That is, set $\zeta = e^{i \phi}$, $d\phi = -i d\zeta/\zeta$ and get that the integral is equal to

$$i R \oint_{|\zeta|=1} \frac{d\zeta}{|a| R \zeta^2 - (|a|^2+R^2) \zeta + |a| R}$$

To evaluate via the residue theorem, we find the poles of the integrand, which are at $\zeta=|a|/R$ and $\zeta=R/|a|$. Clearly, the analysis depends on whether $|a|$ is greater than or less than $R$. For example, when $|a| \lt R$, the integral is, by the residue theorem,

$$i 2 \pi (i R) \frac{1}{2 |a| R (|a|/R) - |a|^2-R^2} = \frac{2 \pi R}{R^2-|a|^2}$$

The analysis is similar for $R \lt |a|$. The end result is that

$$\oint_{|z|=R} \frac{|dz|}{|z-a|^2} = \frac{2 \pi R}{\left|R^2-|a|^2 \right|}$$

• How do you get this: $|z-a|^2 = R^2 + |a|^2 - 2 R |a| \cos{\phi}$? I only get $|z-a|^2 = R^2+|a|^2-z\overline{a}-\overline{z}a$ – mez Oct 15 '13 at 20:10
• $z \bar{a} + \bar{z} a = 2 \Re{(z \bar{a})}$. Then write $z$ and $a$ in polar form. – Ron Gordon Oct 15 '13 at 20:11
• Wow, what made you so fast in spotting that you should substitute $\zeta = e^{i\phi}$ back? – mez Oct 15 '13 at 20:39
• @mezhang: it is simply one approach to evaluating the integral. In this case, any rational function of sine and cosine integrated from $0$ to $2 \pi$ may be evaluated using that substitution and the residue theorem fairly easily, as you can see. – Ron Gordon Oct 15 '13 at 20:40

You may assume $a=|a|>0$. On $\gamma:=\partial D_R$ one has $\bar z={R^2\over z}$; furthermore the parametrization $$\gamma:\quad \phi\mapsto z=R e^{i\phi}\qquad(0\leq\phi\leq2\pi)$$ gives $dz=iR e^{i\phi}\ d\phi$ and therefore $$|dz|=Rd\phi=-i{R\over z}\ dz\ .$$ (Complexifying the real $|dz|$ in this way is a trick that will enable us to use the residue theorem later on.)

Altogether it follows that your integral ($=: J$) can be written as $$J=\int\nolimits_\gamma{-i R/z\over(z-a)\bigl({R^2\over z}-a\bigr)}\ dz={iR\over a}\int\nolimits_\gamma{dz\over(z-a)\bigl(z-{R^2\over a}\bigr)}\ .$$ The last integral can now be evaluated by means of the residue theorem. When $a<R$ we have a single pole in $D_R$ at $z=a$, and standard computation rules tell us that $$J=2\pi i\cdot {iR\over a}{1\over a-{R^2\over a}}={2\pi R\over R^2-a^2}\ .$$ I leave the case $a>R$ to you.