Cauchy's integral formula for derivatives

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 integer $n \ge 0$ and each $a \in \mathbb{C}$ with $|a - c| < r$, $$f^{(n)}(a) = \frac{n!}{2 \pi i} \oint_{|z - c| = r} \frac{f(z)}{(z - a)^{n+1}} \, dz,$$ where $f^{(n)}$ is the $n$-th complex derivative of $f$. This derivative exists by holomorphic functions are infinitely differentiable.

Remarks

For $n = 0$, this is Cauchy's integral formula. Formally, the formula comes from differentiating $\frac{1}{z - a}$ $n$ times under the integral sign with respect to $a$. For example, with $f(z) = e^z$, $c = a = 0$ and $r = 1$: $\oint_{|z| = 1} \frac{e^z}{z^{n+1}} \, dz = \frac{2 \pi i}{n!}$. It gives Cauchy's estimates for the derivatives.

Explanations

L1 human

The values of a holomorphic function on a circle determine not only its values inside, but also all its derivatives inside. To get the $n$-th derivative, you put a higher power of $z - a$ in the denominator and multiply by $n!$.