I want to prove:
If $p(z)$ is a non-constant polynomial in complex field and m > degP is an integer, then the arithmetic mean of the values of $p$ at the vertices of any regular m-gon in the complex plane equals the value of $p$ at the center of this m-gon.
My approach is the following:
Let $p(z) =a_nz^n + ... + a_1z + a_0$, $a_i \neq 0 $, and $m>n$. Let $z_k$, where $k=1,...,m$ be the m many vertices. Then $\sum_{k=1}^{m}p(z_k) = a_n\sum_{k=1}^{m}(z_k)^n +...+a_n\sum_{k=1}^{m}z_k+ma_o $. I consider the vertices in polar form: $z_i = r(cos\theta+i\sin\theta)$. Also, consider the center as the origin. Then for each $z_i$ considering $z_j$ as the reflection of $z_i$ with respect to origin,we have $z_j= r(cos(\theta\pm \pi) i\sin(\theta\pm \pi))=-r(cos\theta+i\sin\theta)$. For each odd $n$, similarly we can arrive $z_i^n + z_j^n=0$. If we can group the vertices for $n:$even like this, the sum $\sum_{k=1}^{m}p(z_k)$ becomes equal to $ma_0$ and we can get the desired result.
My intuition is that instead of dealing with symmetries with respect to origin, I should be able to write the polar forms of vertices such that using DeMoivre's identity the sum is always $ma_0$. However, I could not find a way to express this in a compact form such as $r(\frac{cos(2k\pi+\theta)}{n}+i\frac{sin(2k\pi+\theta)}{n})$ for some factor $k$.I also feel that this is closely related to the roots of unity expressed in the inscribed unit circle but my knowledge is very limited.
Apologies, you do not need harmonic functions. Just find two complex numbers $a$ and $b$ such that the image of the regular polygon $z_1,\ldots,z_m$ is the set of the $m$ roots of unity $1,w=e^{2i\pi/m},\ldots w^{m-1}.$ Taking $0<k<m$ observe that $\sum_{j=0}^{m-1}w^{kj}=0.$ To see this, if $r=pgcd(k,m)$ then $w^k$ is root of $z^{m/r}-1$ whose sum of roots is zero. If $k=0$ then $\frac{1}{m}\sum_{j=0}^{m-1}w^{kj}=1.$ Therefore, the result is proved in the particular case where $p(z)=z^k$ and the polygon is the polygon of the $m$ roots of unity. Passing from $z^k$ to $p(z)$ is done by linearity. Passing to an arbitrary regular polygon with $m$ vertices is done by the transformation $z\mapsto az+b$ above. The condition $k<m$ is essential, the result for $p(z)=z^m$ is false.