Existence of Laurent expansions on an annulus

Statement

Let $c \in \mathbb{C}$, let $0 \le r < R \le \infty$, and let $A = \{z : r < |z - c| < R\}$ be an annulus. Let $f : A \to \mathbb{C}$ be holomorphic on $A$. Then there are unique complex numbers $(a_n)_{n \in \mathbb{Z}}$ such that the Laurent series $\sum_{n \in \mathbb{Z}} a_n (z - c)^n$ converges to $f(z)$ for each $z \in A$. For each $\rho$ with $r < \rho < R$ and each $n \in \mathbb{Z}$, $$a_n = \frac{1}{2 \pi i} \oint_{|z - c| = \rho} \frac{f(z)}{(z - c)^{n+1}} \, dz.$$ The series converges absolutely on $A$ and uniformly on each closed annulus $\rho_1 \le |z - c| \le \rho_2$ with $r < \rho_1 \le \rho_2 < R$.

Remarks

Uniqueness means: each Laurent series that converges to $f$ on $A$ has these coefficients. So we can find them by any method, for example with known power series. For example, on $0 < |z| < 1$, $\frac{1}{z(1 - z)} = \frac{1}{z} + 1 + z + z^2 + \cdots$. On $|z| > 1$, the same function is $-\frac{1}{z^2} - \frac{1}{z^3} - \cdots$. If $f$ is holomorphic on the whole disc $|z - c| < R$, then $a_n = 0$ for $n < 0$, and the Laurent series is the Taylor series.

Explanations

L1 human

On a ring around $c$, each holomorphic function is a sum of powers of $z - c$, with negative and positive exponents. The coefficients come from circle integrals, as for Taylor series. The same function can have different Laurent series on different rings.