Annular expansions
Theorem. If \(f\) is holomorphic on an annulus \(r<|z-a|<R\), then \[ f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n \] converges normally on compact subannuli. The coefficients are \(c_n=(2\pi i)^{-1}\int_\gamma f(\zeta)(\zeta-a)^{-n-1}\,\mathrm{d}\zeta\).
Principal part
Definition. The negative-power part \(\sum_{n<0}c_n(z-a)^n\) is the principal part. It records precisely the obstruction to holomorphic extension at \(a\).
Coefficient independence
Proposition. The coefficient integral is independent of the circle \(\gamma\) inside the annulus. This follows from Cauchy's theorem applied to the annular region between two such circles.
Classification by Laurent coefficients
Corollary. A singularity is removable when all negative coefficients vanish, a pole when finitely many negative coefficients are nonzero, and essential when infinitely many negative coefficients are nonzero.
Residue coefficient
Definition. The coefficient \(c_{-1}\) is the residue of \(f\) at \(a\), denoted \(\operatorname{Res}(f,a)\). It is the only Laurent coefficient detected by contour integration around \(a\).
Essential singularity example
Example. On \(0<|z|<\infty\), \(e^{1/z}=\sum_{n=0}^{\infty}z^{-n}/n!\). Its residue at \(0\) is \(1\), and the infinitely many negative powers show that \(0\) is essential.
Expansion on an annulus
Theorem (Laurent). If \(f\) is holomorphic on \[ A=\{z:r<|z-a|<R\}, \] then uniquely \[ f(z)=\sum_{n=-\infty}^{\infty}c_n(z-a)^n, \qquad c_n=\frac1{2\pi i} \int_{|\zeta-a|=\rho} \frac{f(\zeta)}{(\zeta-a)^{n+1}}\,\mathrm{d}\zeta, \] for any \(r<\rho<R\). The series converges absolutely and locally uniformly on \(A\).
Proof. Choose \(r<\rho_1<|z-a|<\rho_2<R\). Cauchy's formula on the annular boundary gives \[ f(z)=\frac1{2\pi i}\int_{|\zeta-a|=\rho_2} \frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta -\frac1{2\pi i}\int_{|\zeta-a|=\rho_1} \frac{f(\zeta)}{\zeta-z}\,\mathrm{d}\zeta. \] Expand the outer kernel in powers of \((z-a)/(\zeta-a)\) and the inner kernel in powers of \((\zeta-a)/(z-a)\). Both geometric series converge uniformly on the contours, giving positive and negative powers. Cauchy's theorem makes the coefficient integral independent of \(\rho\). Integrating a purported second series against \((z-a)^{-n-1}\) proves uniqueness.
Positive and principal parts
The decomposition \[ f(z)= \underbrace{\sum_{n\ge0}c_n(z-a)^n}_{\text{regular part}} + \underbrace{\sum_{n\ge1}c_{-n}(z-a)^{-n}}_{\text{principal part}} \] separates behavior extendable across \(a\) from the obstruction. The singularity is removable when the principal part vanishes, a pole when it is finite and nonzero, and essential when it has infinitely many terms.
Coefficient estimates on two boundaries
If \(|f|\le M_\rho\) on \(|z-a|=\rho\), then \[ |c_n|\le M_\rho\rho^{-n} \quad(n\in\mathbb Z). \] For positive \(n\), choose large \(\rho<R\); for negative \(n\), choose small \(\rho>r\). These estimates control both tails uniformly on compact subannuli.
Example. The same function can have different Laurent expansions on different annuli: \[ \frac1{z(1-z)} =\sum_{n\ge0}z^{n-1},\qquad 0<|z|<1, \] while for \(|z|>1\), \[ \frac1{z(1-z)} =-\sum_{n\ge0}z^{-n-2}. \] The domain, not just the center, is part of a Laurent expansion.