Let $f(z)$ be analytic in the $\operatorname{disc}|z|<R$. Assume the formula

$$f^{\prime}\left(z_{0}\right)=\frac{1}{2 \pi i} \int_{|z|=r} \frac{f(z) d z}{\left(z-z_{0}\right)^{2}}, \quad 0 \leqslant\left|z_{0}\right|<r<R .$$

By combining this formula with a complex conjugate version of Cauchy's Theorem, namely

$$0=\int_{|z|=r} \overline{f(z)} d \bar{z},$$

prove that

$$f^{\prime}(0)=\frac{1}{\pi r} \int_{0}^{2 \pi} u(\theta) e^{-i \theta} d \theta$$

where $u(\theta)$ is the real part of $f\left(r e^{i \theta}\right)$.