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 $$d\overline{z} = -R^2 \frac{dz}{(z-a)^2} $$
Solving a complex integral 1
4
$\begingroup$
complex-analysis
-
0If $P$ 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 $C$. – 2012-12-11
-
0If the integral was wrt dz then yes it would be trivial but it's wrt d(conjugate of z). – 2012-12-11
-
1Okay, sorry, I have overlooked that. – 2012-12-11
-
0I still don't get how to solve this. – 2013-04-23
-
0i'm confused about why we get $R^2$ and why use (z-a)^2 ?? – 2013-04-23