\(\)Named after Pierre Alphonse Laurent, a French mathematician and Military Officer, published in the series 1843.

The Laurent series is a representation of a complex function f(z) as a series. Unlike the Taylor series which expresses \(f(z)\) as a series of terms with non-negative powers of \(z\), a Laurent series includes terms with negative powers. Therefore, a Laurent series may be used in cases where a Taylor expansion is not possible.

$$

f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-c)^{n}

$$ where the \(a_n\) and \(c\) are constants defined by

$$

a_{n}={\frac {1}{2\pi i}}\oint _C{\frac {f(z)\,\mathrm {d} z}{(z-c)^{n+1}}}

$$

The contour \(C\) is counterclockwise around a closed, enclosing \(c\) and lying in an annulus \(A\) in which \(f(z)\) analytic.

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see https://math.stackexchange.com/questions/1126321/proof-of-laurent-series-co-efficients-in-complex-residue

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To calculate, use the standard and modified geometric series

$$

\frac{1}{1-z}=

\left\{

\begin{align}

\sum_{n=0}^{\infty}&\ z^n,&&|z|\lt1\nonumber\\

-\sum_{n=1}^{\infty}&\ z^{-n},&&|z|\gt1\nonumber

\end{align}\nonumber

\right.\nonumber

$$

Here \(f(z)=\frac{1}{1-z}\) is analytic everywhere apart from the singularity at \(z=1\). Above are the expansions for \(f(z)\) in the regions inside and outside the unit circle, centered on \(z=0\), where \(|z|\lt1\) is the region inside the circle and \(|z|\gt1\) is the region outside the circle.