The value of a holomorphic function at the center of a closed disc in its domain equals its average over the boundary circle and its average over the disc.Suggest a slogan
Let \(f\) be holomorphic on an open set containing the closed disc \(\overline{D_r(z_0)}\). Then \[ f(z_0)=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+re^{i\theta})\dtheta=\frac{1}{\pi r^2}\iint_{D_r(z_0)}f(z)\dA. \] The same identities hold for \(u=\Re f\) in place of \(f\), by taking real parts.
The Cauchy integral formula on the circle \(z=z_0+\rho e^{i\theta}\), \(0<\rho\le r\), gives \[ f(z_0)=\frac{1}{2\pi i}\int_0^{2\pi}\frac{f(z_0+\rho e^{i\theta})}{\rho e^{i\theta}}\,i\rho e^{i\theta}\dtheta=\frac{1}{2\pi}\int_0^{2\pi}f(z_0+\rho e^{i\theta})\dtheta. \] Multiplying by \(\rho\) and integrating over \(0\le\rho\le r\) in polar coordinates gives \(\frac{r^2}{2}f(z_0)=\frac{1}{2\pi}\iint_{D_r(z_0)}f\dA\).