There exist some functions f(x) which cannot be expressed as a Power series using Taylor series expansion. We can however use a Laurent series expansion to expand some of these functions into a Power series with negative terms.
Let f(x) be a complex function analytic on the annulus A \colon r_1 < |z - z_0| < r_2. The Laurent series of f(x) around a point z_0 inside the annulus is given by
f(x) = \sum_{n = -\infty}^\infty a_n(z-z_0)^n
where the coefficients a_n are defined by
a_n = \frac{1}{2\pi i} \int_C \frac{f(z)}{(z-z_0)^{n+1}} dz
where C is any circle inside the annulus A.
We first have to prove that the Laurent series expansion exists.
Let f(x) be a complex function analytic on some annulus A. Then the Laurent series of f(x) around z_0 \in A exists.
Note that depending on the annulus, a function will have different Laurent series expansions.
Let f(x) have a Laurent series around some point z_0,
f(x) = \sum_{n = -\infty}^\infty a_n(z-z_0)^n.
Then Residue of f at z_0 is the coefficient a_{-1}, and is denoted
\text{Res}(f,z_0) = a_{-1}.
Let f(z) be analytic on a region D except for a set of isolated singularities S. Then for any simple closed curve C on D
\int_C f(z) dz = 2\pi i \sum_{z_0 \in S} \text{Res}[f,z_0]