Prove that if a function \(f\) has two Laurent series expansions in an annulus \(A = \{z \in \mathbb{C} \mid r < |z-a| < R\}\): \[f(z) = \sum_{n=-\infty}^\infty c_n(z-a)^n \quad\text{and}\quad f(z) = \sum_{n=-\infty}^\infty c_n'(z-a)^n\] then \(c_n = c_n'\) for all \(n \in \mathbb{Z}\).
Solution
Fix \(\rho\) with \[ r<\rho<R, \] and let \[ \gamma(t)=a+\rho e^{it},\qquad0\le t\le2\pi. \] For every integer \(k\), define \[ I_k=\frac1{2\pi i}\int_\gamma\frac{f(z)}{(z-a)^{k+1}}\,dz. \]
1. A Laurent series converges uniformly on every circle compactly contained in its annulus of convergence. Hence the first expansion may be integrated term by term on \(\gamma\):\[ I_k =\sum_{n\in\mathbb Z}c_n\frac1{2\pi i}\int_\gamma(z-a)^{n-k-1}\,dz. \]
2. For every integer \(m\),\[ \frac1{2\pi i}\int_\gamma(z-a)^m\,dz =\begin{cases} 1,&m=-1,\\ 0,&m\ne-1. \end{cases} \]Therefore only the term \(n=k\) survives, and\[ I_k=c_k. \]
3. Applying the same computation to the second Laurent expansion gives\[ I_k=c_k'. \]Thus\[ c_k=c_k' \]for every \(k\in\mathbb Z\).
Hence Laurent coefficients on a fixed annulus are unique, with \[ c_k=\frac1{2\pi i}\int_{|z-a|=\rho}\frac{f(z)}{(z-a)^{k+1}}\,dz. \]