How to prove $\frac{1}{2\pi} \int_0^{2\pi} u(z_0+re^{i\theta}) d\theta = u(z_0)$ for harmonic polynomial $u$

Question

Suppose $U\subseteq \mathbb{C}$ be an open set, Let $z_0 \in U$ and $r>0$ and assume $\{ z : |z-z_0|\leq r\}\subseteq U$, and $u$ is a harmonic polynomial on $U$. Then textbook says \begin{align} \frac{1}{2\pi} \int_0^{2\pi} u(z_0+re^{i\theta}) d\theta = u(z_0) \end{align} but i wonder how it holds.

Answer 1

This result is the famous mean-value equality for harmonic functions, see for example the section "Properties of harmonic functions -> The mean value property" in the article https://en.wikipedia.org/wiki/Harmonic_function (it is formulated in the general n-dimensional case there!). Moreover, this result holds for all harmonic functions on $U$, not only for polynomials. A very good reference on this topic is "Functions of One Complex Variable I" by J. Conway (published with Springer), see the Mean Value Theorem 1.4 on page 253 in chapter X on Harmonic Functions and the following pages. Another standard reference is "Complex Analysis" by Lars V. Ahlfors, section 4.6 on Harmonic Functions, see in particular Theorem 20 and following.