Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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

2 Answers 2

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:

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

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

share|improve this answer

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.