In deriving Laurent series centered at, say $0$, there are two contours, -- concentric circles with radii $R_{1}, R_{2}$ s.t. $0 < R_{1} < R_{2} < \infty$. The integral on $\{z: |z| = R_{2}\}$ leads to $a_n$, $n\geq0$; and the integral on $\{z:|z|=R_{1}\}$ leads to $a_n$, $n<0$. These two formulas can then be combined into
\begin{equation} a_{n}=\frac{1}{2\pi i}\int_{C}\frac{f\left(w\right)}{w^{n+1}}dw,\quad\forall n \end{equation}
where $C$ is any simple contour in the annulus region.
I must be missing something here... So to get $a_n$, $n\geq 0$, we don't have to begin with the contour $\{z: |z| = R_{2}\}$, just use $C$ instead? Then why is the contour $\{z: |z| = R_{2}\}$ introduced in the first place?? Thanks!
Since $f$ i holomorphic on the annulus, Cauchy's integral theorem implies that $$ \int_{|z|=R_1} \frac{f(w)}{w^{n+1}}\,dw = \int_{|z|=R_2} \frac{f(w)}{w^{n+1}}\,dw = \int_{C} \frac{f(w)}{w^{n+1}}\,dw $$ for all simple contours $C$ within the annulus. The reason we start with $|z|=R_2$ and $|z|=R_1$ for the positive and negative indices, is just a matter of convenience. It's easier to estimate the necessary integrals for circles.