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Let $a\in \mathbb{C}, |a|\ne 3$ and $\gamma$ is a circle with center at $0 $and radius $3$. How to compute the following integral $$\int \limits_\gamma \! \dfrac{\bar{z}}{z-a} \, \mathrm{d} z$$ ?

$\bar{z}$ is not analytic, that is why I do not what should I do

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on the circle you have $\bar z = 3/z$. That's the trick used in this kind of problems. – user8268 Nov 6 '12 at 12:26
up vote 1 down vote accepted

Fleshing out (and slightly correcting) the trick mentioned in the comments:

$$z\in\gamma\Longrightarrow 9=|z|^2=z\overline z\Longrightarrow \overline z=\frac{9}{z}\Longrightarrow$$

$$\oint_\gamma\frac{\overline z}{z-a}dz=9\oint_\gamma \frac{dz}{z(z-a)}dz =:9 I$$

(1) If $\,|a|>3\,$ , then using Cauchy's Theorem:

$$9I=9\oint_\gamma\frac{\frac{1}{z-a}}{z}dz=9\cdot 2\pi i\left.\frac{1}{z-a}\right|_{z=0}=-\frac{18\pi i}{a}$$

(2) If $\,|a|<3\,$ , then we can integrate over little circles $\,\gamma_0\,,\,\gamma_a\,$ around zero and $\,a\,$ resp. that do not pass through the other point and are completely contained within $\,\gamma\,$ , and get:

$$9I=9\oint_{\gamma_0}\frac{\frac{1}{z-a}}{z}dz+\oint_{\gamma_a}\frac{\frac{1}{z}}{z-a}dz=18\pi i\left(\left.\frac{1}{z-a}\right|_{z=0}+\left.\frac{1}{z}\right|_{z=a}\right)=$$

$$=18\pi i\left(-\frac{1}{a}+\frac{1}{a}\right)=0$$

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$$\int_\gamma\frac{\bar{z}}{z-z}dz=\int_\gamma\frac{9}{z(z-a)dz}=\frac{9}{a}\int_\gamma\left(\frac{1}{z-a}-\frac{1}{z}\right)dz=\frac{18\pi i}{a}(n(\gamma ,a)-n(\gamma , 0))$$ Where $n(\gamma ,a )$ is the winding number ( of $\gamma$ around $a$. So the given integral is equal to $0$ if $|a|<3$ and is $-18\pi i/a$ if $|a|>3$.

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In the final result I think you must multiply, and not divide, by $\,2\pi i\,$ – DonAntonio Nov 6 '12 at 13:29
@DonAntonio: Yes correct,thanks. – pritam Nov 6 '12 at 13:36
Thanks, to all, we should also consider the case a=0, but it is easy by Cauchy formula – StudentMath Nov 6 '12 at 17:09

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