Cauchy's integral formula

Statement

Let $U \subseteq \mathbb{C}$ be an open set and let $f : U \to \mathbb{C}$ be holomorphic on $U$. Let $c \in \mathbb{C}$ and $r > 0$ be such that the closed disc $\{z : |z - c| \le r\}$ is contained in $U$. Then for each $a \in \mathbb{C}$ with $|a - c| < r$, $$f(a) = \frac{1}{2 \pi i} \oint_{|z - c| = r} \frac{f(z)}{z - a} \, dz,$$ where the integral is the circle integral over the circle $|z - c| = r$.

Remarks

The values of $f$ on the circle determine the values of $f$ at each point inside the circle. With $a = c$ and $z = c + r e^{i\theta}$, the formula gives $f(c) = \frac{1}{2\pi} \int_0^{2\pi} f(c + r e^{i\theta}) \, d\theta$. So $f(c)$ is the average of $f$ on the circle. For $f = 1$, the formula is the integral of 1/(z - a). The formula gives the formula for the derivatives, and it shows that holomorphic functions are analytic.

Explanations

L1 human

If you know a holomorphic function on a circle, you know it everywhere inside the circle. The formula computes the value at an inside point $a$ as a weighted average of the values on the circle, with the weight $1/(z - a)$.