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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.

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    Here are links to different answers of the same question: (1) https://math.stackexchange.com/q/509622/121988 (2) https://math.stackexchange.com/q/2135680/1219882018-02-14

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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 $|z-a| \le R$.

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    $+1$ for the nice and easy solution2013-04-28