Certain inverse problem related to moments Suppose $D\subset \mathbb C$ is a smoothly bounded domain and it contains the origin. Let $ds$ denote the arc length measure on $\partial D.$ I am interested in the following two inverse problems (perhaps related to certain moment problem):
A) Suppose for some polynomial $p$ there exists $n_0\in\mathbb N$ so that the following identity holds  $$\int_{\partial D}\frac{p(z)}{z^n}ds_z=0$$ for all positive integer  $n\geq n_0.$
Does it imply that $\partial D$ must be a disk?
A')  With the same assumption as in (A) we have  $$\int_{\partial D}\frac{p(z)}{(z-t)^n}ds_z=0 \,\,\,\,\text{for all}\,\,\, t\in\ D$$
holding for all $n\geq n_0$. Does this force $\partial D$ to be a disk?
In here $ds$ denotes the arc length measure.
 A: (A') is equivalent to (A).
Note that (for any given $p$ and $n$) 
$$F_n(t) = \int_{\partial D} \dfrac{p(z)}{(z-t)^n} ds_z$$ is an analytic function of $t \in D$, and its Maclaurin series is
$$  F_n(t) = \sum_{k=0}^\infty {{k+n-1}\choose k} F_{n+k}(0) t^k $$
Thus if $F_n(0) = 0$ for all $n \ge n_0$, we also have $F_n(t) = 0$ for all $n \ge n_0$ and all $t \in D$.
EDIT: 
It's convenient to invert the domain: let $C = \{0\} \cup \{1/z: z \notin \overline{D}\}$, another smoothly bounded domain.
There is a positive measure $d\mu$ on $\partial C$ (not, in general, arc length, but absolutely continuous with respect to it) corresponding to $ds$ on $\partial D$, and $\int_{\partial C} p(1/w) w^n\; d\mu(w) = 0$ for $n \ge n_0$.  Now
harmonic measure $d\omega_q(w)$ on $\partial C$ for the pole $q \in C$ would have this property for all $n > \deg(p)$ if either $q = 0$ or $p(1/q) = 0$.
Is every positive measure on $\partial C$, absolutely continuous wrt arc length, that annihilates $p(1/w) w^n$ for sufficiently large $n$ a linear combination of these?  Is it possible to find $D$ (other than a disk) such that $d\mu$ is a 
linear combination of harmonic measures?
A: [EDIT]: as pointed out in the comments below, I misunderstood the question (missing the "arc length measure" bit) so that the answer is silly.
Maybe I'm misinterpreting the question, please let me know if this is the case; but I think that the answer is "No". Indeed, by usual analysis in one complex variable, for every $C^1$ curve $\gamma$ that parametrizes your boundary $\partial D$ and every holomorphic function $f$ on $C^*$ you have
$$
\int_\gamma f = 2\pi i \text{Res}(f,0).
$$
If you take $f(z) = \frac{p(z)}{z^n}$ where $p(z) = \sum a_j z^j$ is your polynomial, then $\text{Res}(f,0) = a_{n-1}$ regardless of the form of your boundary. So I would say that for every boundary and every $p(z)$ you always get 0 provided $n_0 \geq \text{deg}(p)+2$.
The answer for the second one is the same, simply by a change of variables.
(and since you are only interested in very regular functions, namely meromorphic functions, if I remember correctly the $C^1$ condition on the boundary may be dropped, too)
