[proofplan]
We decompose the Cauchy integral formula over two concentric circles bounding the annulus and expand the Cauchy kernel as geometric series converging in complementary directions. The outer circle produces the non-negative powers (the analytic part $f_+$), while the inner circle produces the negative powers (the principal part $f_-$). Uniform convergence justifies interchanging summation and integration in both cases, and uniqueness of the Laurent coefficients follows from the coefficient integral formula.
[/proofplan]
[step:Decompose the Cauchy integral formula over two concentric circles]
Fix $z \in A$ with $r < |z - a| < R$. Choose $\rho_1, \rho_2$ with $r < \rho_1 < |z - a| < \rho_2 < R$. By the [Cauchy integral formula](/theorems/345) applied to the annular region between $|w - a| = \rho_1$ and $|w - a| = \rho_2$ (with $z$ in the interior):
\begin{align*}
f(z) = \frac{1}{2\pi i} \int_{|w-a|=\rho_2} \frac{f(w)}{w - z} \, dw - \frac{1}{2\pi i} \int_{|w-a|=\rho_1} \frac{f(w)}{w - z} \, dw.
\end{align*}
The minus sign arises because the inner circle is traversed clockwise (as the inner boundary of the annulus).
[guided]
Why does the Cauchy integral formula involve two circles?
The function $f$ is holomorphic on the annulus $A$ but may have a singularity at $a$ (or fail to extend inside the inner boundary).
The standard Cauchy formula on a disc does not apply directly because the annulus is not simply connected.
Instead, we deform the contour.
Homotopy invariance in the annulus shows that $\int_{\Gamma} \frac{f(w)}{w-z} \, dw$ is the same for any simple closed curve $\Gamma$ in $A$ winding once around $z$.
Choosing $\Gamma$ as the boundary of the sub-annulus $\rho_1 < |w-a| < \rho_2$ (traversed with the outer circle counter-clockwise and the inner circle clockwise) gives the two-circle decomposition.
The minus sign on the inner integral arises because the inner circle, as part of the annular boundary, is traversed clockwise (the standard positive orientation of an annular boundary requires the inner circle to go clockwise).
[/guided]
[/step]
[step:Expand the outer integral as a power series in $(z - a)$ (non-negative powers)]
On the outer circle $|w - a| = \rho_2$, expand the kernel exactly as in the Taylor series proof:
\begin{align*}
\frac{1}{w - z} = \frac{1}{(w-a) - (z-a)} = \sum_{n=0}^\infty \frac{(z-a)^n}{(w-a)^{n+1}},
\end{align*}
converging uniformly in $w$ on $|w-a| = \rho_2$ since $|z-a|/\rho_2 < 1$. Interchanging sum and integral:
\begin{align*}
\frac{1}{2\pi i} \int_{|w-a|=\rho_2} \frac{f(w)}{w-z} \, dw = \sum_{n=0}^\infty c_n (z-a)^n,
\end{align*}
where $c_n = \frac{1}{2\pi i} \int_{|w-a|=\rho_2} \frac{f(w)}{(w-a)^{n+1}} \, dw$. This series converges for $|z - a| < \rho_2$, and since $\rho_2 < R$ was arbitrary, it converges for $|z - a| < R$.
[/step]
[step:Expand the inner integral as a power series in $(z - a)^{-1}$ (negative powers)]
On the inner circle $|w - a| = \rho_1$, we have $|w - a| = \rho_1 < |z - a|$, so we factor $(z - a)$ from the denominator instead:
\begin{align*}
\frac{1}{w - z} = \frac{-1}{(z-a) - (w-a)} = \frac{-1}{z-a} \cdot \frac{1}{1 - \frac{w-a}{z-a}} = -\sum_{m=0}^\infty \frac{(w-a)^m}{(z-a)^{m+1}},
\end{align*}
converging uniformly in $w$ on $|w-a| = \rho_1$ since $\rho_1/|z-a| < 1$. Substituting into the inner integral (with the sign from the decomposition):
\begin{align*}
-\frac{1}{2\pi i} \int_{|w-a|=\rho_1} \frac{f(w)}{w-z} \, dw = \sum_{m=0}^\infty \left(\frac{1}{2\pi i} \int_{|w-a|=\rho_1} f(w)(w-a)^m \, dw\right) \frac{1}{(z-a)^{m+1}}.
\end{align*}
Setting $n = -(m+1)$ (so $m = -n - 1$ and the sum runs over $n = -1, -2, \ldots$):
\begin{align*}
\sum_{n=-\infty}^{-1} c_n (z-a)^n, \quad \text{where } c_n = \frac{1}{2\pi i} \int_{|w-a|=\rho_1} \frac{f(w)}{(w-a)^{n+1}} \, dw.
\end{align*}
This series converges for $|z - a| > \rho_1$, and since $\rho_1 > r$ was arbitrary, it converges for $|z - a| > r$.
[/step]
[step:Combine the two series and establish the coefficient formula on any circle in $A$]
Adding the two contributions:
\begin{align*}
f(z) = \sum_{n=0}^\infty c_n (z-a)^n + \sum_{n=-\infty}^{-1} c_n (z-a)^n = \sum_{n=-\infty}^\infty c_n (z-a)^n,
\end{align*}
converging for $z \in A$. For any $\rho \in (r, R)$, the coefficient formula $c_n = \frac{1}{2\pi i} \int_{|w-a|=\rho} \frac{f(w)}{(w-a)^{n+1}} \, dw$ holds for all $n \in \mathbb{Z}$: for $n \geq 0$ the formula was derived on $|w-a| = \rho_2$ and for $n < 0$ on $|w-a| = \rho_1$, but the integrand $f(w)/(w-a)^{n+1}$ is holomorphic on $A$, so homotopy invariance shows the integral is independent of the radius $\rho \in (r, R)$.
Uniqueness: if $f(z) = \sum_{n=-\infty}^\infty d_n (z-a)^n$ is another Laurent expansion, multiplying both sides by $(z-a)^{-k-1}$ and integrating over $|z-a| = \rho$ gives $d_k = c_k$ by the orthogonality of the powers $(z-a)^n$ under contour integration.
[/step]
