If \(f\) is holomorphic on \(D_r(z_0)\) \begin{align*} f(z_0) = {1\over 2\pi} \int_0^{2\pi} f(z_0 + re^{i\theta}) \,d\theta = {1\over \pi r^2} \iint_{D_r(z_0)} f(z)\, dA .\end{align*} Taking the real part of both sides, one can replace \(f=u+iv\) with \(u\).
\begin{align*} f\left(z_{0}\right)=\frac{1}{2 \pi i} \int_{\left|z-z_{0}\right|=r} \frac{f(z)}{z-z_{0}} d z=\frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{f\left(z_{0}+r e^{i \theta}\right)}{r e^{i \theta}} r i e^{i \theta} d \theta=\frac{1}{2 \pi} \int_{0}^{2 \pi} f\left(r e^{z_{0}+r i \theta}\right) d \theta, .\end{align*}