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 $$d\overline{z} = -R^2 \frac{dz}{(z-a)^2} $$

  • $\begingroup$ 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 $C$. $\endgroup$ – saz Dec 11 '12 at 16:53
  • $\begingroup$ If the integral was wrt dz then yes it would be trivial but it's wrt d(conjugate of z). $\endgroup$ – A. Napster Dec 11 '12 at 18:41
  • 1
    $\begingroup$ Okay, sorry, I have overlooked that. $\endgroup$ – saz Dec 11 '12 at 18:50
  • $\begingroup$ I still don't get how to solve this. $\endgroup$ – d13 Apr 23 '13 at 16:42
  • $\begingroup$ i'm confused about why we get $R^2$ and why use (z-a)^2 ?? $\endgroup$ – d13 Apr 23 '13 at 17:11

$z=a+Re^{it}\Rightarrow \overline z=a+Re^{-it}\Rightarrow d\overline z=-iRe^{-it}dt$ 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 $$f^{(n)} (a)={n!\over 2\pi i}\int_{C}{f(z) dz\over (z-a)^{n+1}}$$

  • $\begingroup$ thanks a lot for the answer although it came a little to late. But i appreciate that you found time for this forgotten question. $\endgroup$ – d13 May 4 '13 at 6:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.