Closed complex integral that does not vanish My girlfriend has the following math problem in complex analysis:

Let $p(z)$ be a complex polynomial, $w \in \mathbb C$ and $r > 0$. Show that
  $$ \intop_{|w - \xi| = r} \overline{p(\xi)} \, \mathrm d\xi = 2 \pi \mathrm i r^2 \overline{p'(\omega)} $$
  holds, where the circle is traversed in a mathematically positive sense.

Having the residue theorem in mind, I would think that the closed integral of any function that is holomorphic except for some isolated singularies would just be the sum of the residues (times winding number and so on). Since $p$ is a polynomial, which should be holomophic on $\mathbb C$, I would think that any integral would have to vanish.
I assume that the overline means complex conjugation, which is not a holomorphic operation in itself. However, since the complex conjugate of the whole function is taken, it will be just mirrored and the integration path will be in a negative direction compared to the function.
While writing this, I just thought that the previous paragraph would only hold if it was $\overline{p(\xi) \, \mathrm d\xi}$ and not $\overline{p(\xi)} \, \mathrm d\xi$. Multiplying something complex conjugated with something not conjugated looks like an absolute value.
Could you please give me a hint where to start with this problem?
 A: Let $D$ be the closed disc $\big\{ z \in \mathbb{C} : |z-\omega| \le r \big\}$.
For any function $p(z)$ holomorphic over $D$, we can apply Stroke's theorem
in complex coordinates $\xi$ and $\bar{\xi}$ to get
$$\int_{|\xi-\omega|=r} \overline{p(\xi)} d\xi
= \int_{\partial D} \overline{p(\xi)} d\xi
= \int_D d\overline{p(\xi)} \wedge d\xi
= \int_D \overline{p'(\xi)} d\bar{\xi} \wedge d\xi
= 2i\int_D \overline{p'(\xi)} \frac{d\bar{\xi} \wedge d\xi}{2i}
$$
In real coordinate $x, y$ where $\xi = x + iy$,
$\;\displaystyle\frac{d\bar{\xi} \wedge d\xi}{2i} = dx \wedge dy\;$ is the area element. Since $p(\xi)$ is holomorphic, so does $p'(\xi)$ and hence its real and imaginary parts
are harmonic functions. Apply the mean value property of harmonic functions, the integral
becomes
$$2i\int_D \left(\Re(p'(\xi)) - i \Im(p'(\xi))\right) dx dy
=2i ( \pi r^2 )\left(\Re(p'(\omega)) - i \Im(p'(\omega))\right)
=i 2\pi r^2 \overline{p'(\omega)}
$$
For polynomials, we have a more elementary proof. Let $n$ be the degree of $p(\xi)$. If one
expand $p(\xi)$ with respect to $\omega$, for $\xi \in \partial D$, we have
$$\overline{p(\xi)}
= \overline{ \sum_{k=0}^n \frac{p^{(k)}(\omega)}{n!} (\xi-\omega)^k }
= \sum_{k=0}^n \frac{\overline{p^{(k)}(\omega)}}{n!} (\overline{\xi-\omega} )^k
= \sum_{k=0}^n \frac{\overline{p^{(k)}(\omega)}}{n!} \left(\frac{r^2}{\xi-\omega}\right)^k
$$
We can then evaluate $\int_{|\xi-\omega|=r} \overline{p(\xi)} d\xi$ using regular residue theory to get desired result.
A: You may assume $\omega=0$, $\>p(z)=c z^n$ with $n\geq0$. Then $\overline{p(z)}=\bar c\>r^{2n} z^{-n}$ for all $z\in\partial D_r$ and therefore
$$\int\nolimits_{\partial D_r}\overline{p(z)}\ dz=\bar c \>r^{2n}\int\nolimits_{\partial D_r} z^{-n}\ dz=\cases{0&$(n\ne1)$\cr 2\pi i \ \bar c r^{2n}\quad&$(n=1)$\cr}\quad .$$
It is easy to verify that the right side is $=2\pi i r^2\overline{p'(0)}$ for all $n\geq0$.
