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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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If $P$ is polynomial, i.e. $P(z) = \sum_{k=0}^n a_k \cdot z^k$ for some $a_k \in \mathbb{C}$, $n \in \mathbb{N}_0$, then $P$ is holomorphic and therefore the integral $\int_C P(z) \, dz$ is equal to 0 for all closed curve $C$. –  saz Dec 11 '12 at 16:53
    
If the integral was wrt dz then yes it would be trivial but it's wrt d(conjugate of z). –  Dan Dec 11 '12 at 18:41
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Okay, sorry, I have overlooked that. –  saz Dec 11 '12 at 18:50
    
I still don't get how to solve this. –  d13 Apr 23 '13 at 16:42
    
i'm confused about why we get $R^2$ and why use (z-a)^2 ?? –  d13 Apr 23 '13 at 17:11

1 Answer 1

$z=a+Re^{it}\Rightarrow \overline z=a+Re^{-it}\Rightarrow d\overline z=-iRe^{-it}dt$ 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. –  d13 May 4 '13 at 6:01

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