Laurent series

The principal part consists of all negative-power terms of a Laurent series about its center. On a punctured neighborhood of an isolated singularity it is zero for a removable singularity, a nonzero finite sum for a pole, and an infinite series for an essential singularity. A pole's order is the largest $N$ with $a_{-N}\ne 0$. At infinity use the local variable $w=1/z$: negative powers of $w$ correspond to positive powers of $z$. A Laurent series on an annulus with strictly positive inner radius may have a principal part without the function being defined close enough to the center to classify an isolated singularity there.

A Laurent coefficient is a coefficient $a_n$ in the Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n(z-z_0{)}^n$ of a holomorphic function on an annulus. Integration around a circle in the annulus gives the displayed formula, including negative indices. The coefficient $a_{-1}$ is the residue.

Suppose a meromorphic function is holomorphic on two concentric annuli about $a$, with only finitely many poles between chosen inner and outer coefficient circles. Let $a_n,b_n$ denote the corresponding Laurent coefficients. Subtract their contour integral formulas and apply the residue theorem to the intervening annular region. It givesThe sum is over the poles crossed, including their multiplicities through the residue. This identity applies to every integer index. Thus changing the annulus can change even the coefficients of nonnegative powers; it is not merely a change in the principal part.

For example, $f(z)=1/\sin z$ has simple poles at $\pm \pi$, each with residue $-1$. Moving a coefficient circle centered at zero from radius below $\pi$ to radius between $\pi$ and $2\pi$ changes its $n$th coefficient by $-{\pi }^{-n-1}-(-\pi {)}^{-n-1}$. Even indices remain unchanged, while odd indices acquire a nonzero correction.

The meromorphic function $(e^z-1{)}^{-1}$ has simple poles of residue one at $2\pi i\mathbb{Z}$. Subtracting $1/z+{\sum }_{r=1}^{n}2z/(z^2+4{\pi }^2{r}^2)$ removes all poles inside radius $2(n+1)\pi$. Comparing its inner Taylor coefficients with the outer Laurent expansion gives annular coefficient $a_1=1/12-{\sum }_{r=1}^{n}1/(2{\pi }^2{r}^2)$. On circles of radius $(2n+1)\pi$, the function is uniformly bounded: outside the strip $|\mathrm{Re}z|\le 1$ use the exponential modulus, and inside it the imaginary part stays near an odd multiple of $\pi$. The Laurent coefficient integral bounds $|a_1|\le C/[(2n+1)\pi ]\to 0$, proving the sum.

A Puiseux series permits fractional exponents with a fixed common denominator: ${\sum }_{n\ge n_0}{a}_n{z}^{n/q}$. These are natural local expansions of branches of algebraic functions. A fractional-power root expansion uses such powers to resolve roots that coalesce as a parameter tends to zero.

Suppose a perturbed polynomial has the local form $A(x-x_0{)}^2+ϵB+O((x-x_0{)}^3,ϵ(x-x_0),{ϵ}^2)$ with $AB\ne 0$. Dominant balance gives $x-x_0={ϵ}^{1/2}z$ and $A{z}^2+B=0$. The two resulting Puiseux series split a double root; their leading displacements can be real or a complex conjugate pair. Ordinary Taylor series in $ϵ$ cannot describe this splitting.

An algebraic root can depend on a parameter through fractional powers even when the polynomial coefficients have ordinary power series. For example, the roots of $t^3={ϵ}^2$ are on the scale ${ϵ}^{2/3}$. Dominant balance for algebraic roots identifies the exponent, and local expansion of the scaled root supplies successive coefficients.

An analytic function on an annulus $r<|z-a|<R$ has a unique locally uniformly convergent expansionwhere $C$ is any positively oriented circle around $a$ in the annulus.