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If $P(z)$ is a polynomial and $C$ denotes the circle $|z-a|=R$, what is the value of $\int_CP(z)d\overline{z}$?

I parametrize the circle as $z(t)=a+Re^{it}$ where $t\in[0,2\pi]$. Then

$$\int_CP(z)d\overline{z}=\int_0^{2\pi}P(z(t))\overline{z'(t)}dt=\int_{0}^{2\pi}P(a+Re^{it})\cdot(-iRe^{-it}) dt$$ This equals $$-iR\int_{0}^{2\pi}P(a+Re^{it})\cdot(e^{-it})dt$$

I'm not sure what to do from here.

PJ Miller
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2 Answers2

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Expand $P(a+Re^{it}) = P(a) + P'(a)Re^{it} + \dots$ using the Taylor expansion (this is a finite expansion since $P$ is a polynomial).

The integral of $e^{inx}$ over $[0,2\pi]$ vanishes by periodicity (or by a direct calculation), for any nonzero $n\in \mathbf Z$. Therefore, the only term in the integral which doesn't integrate to $0$ is the term $P'(a)Re^{it}$ in the expansion above (the coefficient $e^{it}$ cancels out with the $e^{-it}$ factor). The integral is therefore

$$-iR \int_0^{2\pi} P'(a)R dt = -2\pi iR^2P'(a).$$

Bruno Joyal
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For $z(t)=a+R e^{it}$, $e^{-it}=\frac{R}{z(t)-a}$, so $$d\bar{z}=-iR e^{-it}dt=-e^{-2it}dz=-\frac{R^2}{(z(t)-a)^2}dz.$$ Then by Cauchy's integral formula,

$$\int_C P(z)d\bar{z}=-R^2\int_C\frac{P(z)}{(z-a)^2}dz=-2\pi i R^2P'(a).$$

Hu Zhengtang
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  • This is the better solution. I figured OP didn't want to use Cauchy's formula, since s/he specified $P$ was a polynomial, and it is possible to avoid it in that case. – Bruno Joyal Oct 02 '13 at 04:43
  • @Marie: Thank you. I didn't realize that the OP might want to avoid using Cauchy's formula. By the way, I think our approaches are equivalent, because in my approach, I have to suppose that $P$ is holomorphic in the open disk and continuous on the closed disk. Then the Taylor expansion of $P$ at $a$ has radius of convergence no less than $R$, so your approach is still valid; just in case that the radius of convergence is exactly $R$, we may replace $C$ with a smaller circle $C_r: |z-a|=r$, $r<R$ and let $r\to R$. – Hu Zhengtang Oct 02 '13 at 05:13
  • @Giraffe I'll give the bounty to Marie's answer, which uses a more elementary method. I'll upvote yours nevertheless. Thanks! – PJ Miller Oct 02 '13 at 18:22
  • @PJMiller: You are welcome. – Hu Zhengtang Oct 03 '13 at 10:34