Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $C$ be a circle in the complex plane with center $a$ and radius $R$. I am trying to evaluate $\oint_{C} P(z) d \bar{z}$.

If I set $z=\bar{u}$ then we have $\bar{z}=u$ and $d\bar{z}=du$. Thus we may write $\oint_{C} P(z) d \bar{z} = \oint_{\bar{C}} P(\bar{u}) du$ where $\bar{C}$ is circle with radius $R$ and center $\bar{a}$. If $P(\bar{u})$ was an analytic function of $u$ I could apply Cauchy's Theorem and get 0. Does it matter that the conjugation function is not analytic?

Thank you for any help or hints.

share|cite|improve this question
That really does matter, since Cauchy Integration Theorem only applies to analytic functions. How about considering the relation $d\bar{z} = R^2 \, d\left( \frac{\bar{z} - \bar{a}}{(z - a)(\bar{z} - \bar{a})} \right) = - \frac{R^2}{(z - a)^2} \, dz$? – Sangchul Lee May 11 '11 at 15:50
sos440: That relation is very helpful since we could then apply Cauchy Integral formula and be done. How exactly did you get that first equality? Thank you so much for the help! – pel May 11 '11 at 15:59
It's just a simple trick. First, $d\bar{z} = d(\bar{z} - \bar{a})$. Then note that on $C$, we must have $R^2 = |z - a|^2 = (z - a)(\bar{z} - \bar{a})$. Combining these observations gives the first equality. – Sangchul Lee May 11 '11 at 16:03
@sos440: That's perfect! If you copy paste your 1st comment as an answer I would be more than happy to select it as the answer and vote it up. Thanks again. – pel May 11 '11 at 16:09

Expand $P$ about $a$ as $P(z) = a_0 + a_1 (z-a) + \ldots$, and parametrize the integral with $z=a+Re^{it}$, $0\le t \le 2\pi$, so that $\bar{z} = \bar{a} + R e^{-it}$, $d\bar{z} = -iRe^{-it} dt$. Then the integral becomes $$ \begin{split}\int_0^{2\pi} & P(a+Re^{it}) (-iRe^{-it}) \, dt = -i\int_0^{2\pi} \sum_{k=0}^\infty a_k (Re^{it})^k Re^{-it}dt \\& = -i \sum_{k=0}^\infty R^{k+1} a_k \int_0^{2\pi} e^{it(k-1)}dt = -iR^2 a_1 (2\pi) = -2\pi i R^2 P'(a)\end{split},$$ since all the integrals vanish except for $k=1$. Interchanging integration and series is justified by uniform convergence. Note that the infinite series in your case is really a finite sum, and that this result with the same proof applies to the case where $P$ is analytic in a neighborhood of the closed disk $|z-a| \le R$.

share|cite|improve this answer
$+1$ for the nice and easy solution – Un Chien Andalou Apr 28 '13 at 8:25

Your Answer


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.