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If $P(z)$ is a polynomial and $C$ denotes the circle $|z-a|=R$ what is the value of $\int_{C}^{} P(z)d\overline{z} $ ? The answer in Ahlfors is $-2\pi i R^2 P'(a)$ I don't know if I'm doing it right but I made a substitution $d\overline{z} = -R^2 \frac{dz}{(z-a)^2} $

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    i'm confused about why we get $R^2$ and why use (z-a)^2 ??2013-04-23

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$z=a+Re^{it}\Rightarrow \overline z=a+Re^{-it}\Rightarrow d\overline z=-iRe^{-it}dt$ so $\int_{C} P(z)d\overline z=-i\int_{0}^{2\pi}P(a+Re^{it})Re^{-it}dt$

$=-iR^2\int_{0}^{2\pi}{P(a+Re^{it})\over R^2e^{2it}}Re^{it}dt$ $=-iR^2\int_{C}{P(z) dz\over (z-a)^2}=-2i\pi R^2 P'(a)$ because from cauchy integral formula we get $f^{(n)} (a)={n!\over 2\pi i}\int_{C}{f(z) dz\over (z-a)^{n+1}}$

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    thanks a lot for the answer although it came a little to late. But i appreciate that you found time for this forgotten question.2013-05-04