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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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    That really does matter, since Cauchy Integration Theorem only applies to analytic functions. How about considering the relation $d\bar{z} = R^2 \, d\left( \frac{\bar{z} - \bar{a}}{(z - a)(\bar{z} - \bar{a})} \right) = - \frac{R^2}{(z - a)^2} \, dz$?2011-05-11
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    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!2011-05-11
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    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.2011-05-11
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    @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.2011-05-11
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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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