Consider a function $f(z)$ that is analytic on an annular (i.e. flat ring-shaped) domain, $R_1<|z-z_0|<R_2$ centered on $z_0$. Within that domain $f(z)$ has the following series representation: $f(z)=\sum_{n=-\infty}^{\infty}c_n(z-z_0)^n$ where the coefficient is the following [closed contour integral](Closed%20contour%20integral.md) over curve $C$: $c_n = \frac{1}{2 \pi i} \int_C dz\frac{f(z)}{(z-z_0)^{n+1}}$  Note that in practice we often deal with functions with restricted domains that have truncating Laurent series representations containing few terms. # Correspondence with complex Taylor series Notice that if $f(z)$ is [holomorphic function](Holomorphic%20functions.md) on the whole radius from $z_0$ to $R_2$ then by the [[Cauchy's integral formula]] it becomes $c_n = \frac{1}{2 \pi i} \int_C dz\frac{f(z)}{(z-z_0)^{n+1}} = \frac{f^{(n)}(z_0)}{n!},\;\;\;\; n=0,1,2,...$ Thus reducing Laurent series to a Taylor Series. # [[Residues]] We can quickly extract the residues from the Laurent series since $\mbox{Res}_{z\rightarrow z_0}(f(z))= c_{-1}$ # Connection to [pole](Pole.md)s We can show that if $f(z)$ has an isolated [singularity](Singularity.md) at $z_0$ then $f(z)$ has a Laurent series representation. --- # Proofs and examples ## Proof of Laurent's theorem #MathematicalFoundations/Analysis/ComplexAnalysis