This tag is for questions about finding a Laurent series of functions and their convergence. The Laurent series is a generalisation of the power series which allows negative indices and is essential for investigating the behaviour of functions near poles.

Laurent series:Suppose that $~f(z)~$ is analytic on the annulus $~A : r_1 <|z − z_0| < r_2~$. Then $~f(z)~$ can be expressed as a series $$f(z) = \sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}+\sum_{n=0}^{\infty}a_n(z-z_0)^n$$ The coefficients have the formulas $$a_n=\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z_0)^{n+1}}\, dw$$and$$b_n=\frac{1}{2\pi i}\int_\gamma {f(w)}{(w-z_0)^{n-1}}\, dw$$where $~\gamma~$ is any circle $~|w − z_0| = r~$ inside the annulus, i.e. $~r_1 < r < r_2~.~$The entire series is called the Laurent series for $~f~$ around $~z_0~$.

**Notes:** $~(a)~~~$ The series $$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$ is called the **analytic** or **regular part** of the Laurent series.

$(b)~~~$ The series $$\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$ is called the **singular** or **principal part** of the Laurent series.

$(c)~~~$ Since $~f(z)~$ may not be analytic (or even defined) at $~z_0~$ we don’t have any formulas for the coefficients using derivatives.

**Remarks:**

- The series $$\sum_{n=0}^{\infty}a_n(z-z_0)^n$$converges to an analytic function for $~|z − z_0| < r_2~$.
- The series $$\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n}$$converges to an analytic function for $~ |z − z_0| > r_1~$.
- Together, the series both converge on the annulus $~A~$ where $~f~$ is analytic.

The Laurent series is calculated over contour integrals of counterclockwise self-avoiding rectifiable paths of the function. For holomorphic functions the Taylor series and Laurent series are identical.

The Laurent series has a principal part, which consists entirely of negative-degree terms. When the principal part vanishes (there are no negative indices) the function is holomorphic; when it is an infinite sum the function has an essential pole.

**Reference:**