Average of a holomorphic function on an annulus

Question

Let $A = \{z \in \mathbb{C}: 1 < |z| < 2\}$ and $f: A \rightarrow \mathbb{C}$ be holomorphic. Why is $\frac{1}{2\pi}\int_{0}^{2\pi}(\textrm{re} f)(re^{i\theta})\, d\theta$ constant on $\{r: 1 < r < 2\}$?

Answer 1

Observe that $I=\frac1{2\pi} \int_0^{2\pi} u(r e^{i\theta}) d\theta = \mathcal{R} [\frac1 {2\pi i} \int_\gamma \frac {f(z)}{z} dz ]=\mathcal{R} [a_0] $