# line integral versus complex integral

Let $a\in \mathbb C, r>0$ and $\gamma_r=\partial D(0,r)$. I want to evaluate the following line integral

$$I=\int_{\gamma_r}\frac{1}{|z-a|^2}ds.$$

I'm looking for a complex function $g(z)$ such that

$$I=\int_{\gamma_r}g(z)dz.$$

I put $g(z)=\dfrac{1}{iz|z-a|^2}$. Then

$$\int_{\gamma_r}g(z)dz=\int_0^{2\pi}\frac{rie^{it}}{rie^{it}|re^{it}a|^2}dt=\frac{1}{r}I.$$

If $f(z)=\dfrac{1}{i|z-a|^2}$, I can apply Cauchy's integral formula to conclude

$$f(0)=\frac{1}{2\pi i}\int_{\gamma_r}\frac{f(z)dz}{z}dz=\frac{1}{2\pi i}\int_{\gamma_r}g(z)dz.$$

I find

$$I=r\int g(z)dz=2\pi r i f(0)=\frac{2\pi r}{|a|^2}.$$

Unfortunately, the correct solution should be

$$\dfrac{2\pi r}{| |a|^2-r^2|}.$$

Where did I go wrong?

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$g(z)$ is not a holomorphic function, thus Cauchy's integration formula does not apply to $g$. – Sangchul Lee Nov 14 '12 at 11:26
but if i take r<|a| can i apply? – Federica Maggioni Nov 14 '12 at 11:33
Ii is not a matter of existence of singularity. It is because $g(z)$ is not complex differentiable! A clever modification, however, would work. – Sangchul Lee Nov 14 '12 at 11:36

Let $z = re^{i\theta}$ on $\partial D(0, r)$. Then the length element is given in terms of $\theta$ by $ds = r d\theta$ an also we have $dz = iz \, d\theta$. Thus

$$ds = \frac{r}{iz} dz.$$

This allows us to rewrite $I$ as

$$I = \oint_{|z|=r} \frac{r}{iz \left| z - a \right|^2} \, dz.$$

To apply Cauchy integration formula, we must represent $I$ as an integral of meromorphic function. Note that

$$r^2 = \left| z \right|^2 = z \bar{z}.$$

Thus

$$\left|z - a \right|^2 = (z - a)(\bar{z} - \bar{a}) = (z - a)(r^2 z^{-1} - \bar{a})$$

and hence $I$ can be written as

$$I = \oint_{|z|=r} \frac{r}{iz (z - a)(r^2 z^{-1} - \bar{a})} \, dz = \frac{2\pi r}{2\pi i}\oint_{|z|=r} \frac{1}{(z - a)(r^2 - \bar{a} z)} \, dz.$$

To evaluate this integral, we divide into two cases according to the pole's position.

Case 1. Assume $|a| < r$, then $z = a$ is the unique pole of the integrand inside $D(0, r)$. Thus by Cauchy integration formula, we have

$$I = 2\pi r \operatorname{Res}_{z = a} \frac{1}{(z - a)(r^2 - \bar{a} z)} = \frac{2\pi r}{r^2 - |a|}.$$

Case 2. Assume $|a| > r$, then $z = r^2 \bar{a}^{-1}$ is the unique pole of the integrand inside $D(0, r)$. Thus similarly we have

$$I = 2\pi r \operatorname{Res}_{z = r^2 \bar{a}^{-1}} \frac{1}{(z - a)(r^2 - \bar{a} z)} = \frac{2\pi r}{(r^2 \bar{a}^{-1} - a)(-\bar{a})} = \frac{2\pi r}{|a|^2 - r^2}.$$

Combining two cases gives the desired answer.

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than you for solution. I can't understand $dz=izd\theta$ – Federica Maggioni Nov 14 '12 at 12:04
sorry, i've understood right now – Federica Maggioni Nov 14 '12 at 12:06
i think i can definitely say i've understood your solution. However i did some changes, since i (officially) shouldn't know the notion of residue at a point. So i distinguished the two cases and i've choosen suitable integrands so to apply cauchy's integral formula to compute integrals. The result is the same, so i think it works. Thank you for your help – Federica Maggioni Nov 14 '12 at 12:59