Laurent series

When a function $f$ has an isolated singularity at $z_0$ — analytic on some punctured disk $0<|z-z_0|<R$ but not at $z_0$ — there is no Taylor series at $z_0$. There is a generalization, the Laurent series, which permits negative powers:

$f(z)={\sum }_{n=-\infty }^{\infty }{a}_n(z-z_0{)}^n.$

The terms with $n<0$ are called the principal part; the terms with $n\ge 0$ are the regular part. The principal part detects the singularity.

Theorem (Laurent)

If $f$ is analytic on the annulus $r_1<|z-z_0|<r_2$ (with $r_1$ possibly $0$ and $r_2$ possibly $\infty$), then on that annulus

$f(z)={\sum }_{n=-\infty }^{\infty }{a}_n(z-z_0{)}^n,$

with

$a_n=\frac{1}{2\pi i}{\oint }_{\Gamma }\frac{f(z)}{(z-z_0{)}^{n+1}}dz$

for any circle $\Gamma$ in the annulus.

The coefficient formula is what you’d guess by analogy with the generalized CIF, extended to negative $n$.

Sketch of proof

Apply the Deformation principle in a “keyhole” enclosing a target $z$ in the annulus: an outer circle $C_2$ (radius slightly less than $r_2$, counterclockwise) and an inner circle $C_1$ (radius slightly more than $r_1$, clockwise). By Cauchy’s theorem on the simply connected region between,

$f(z)=\frac{1}{2\pi i}\left[{\oint }_{C_2}\frac{f(w)}{w-z}dw-{\oint }_{C_1}\frac{f(w)}{w-z}dw\right].$

On the outer circle $|w-z_0|>|z-z_0|$: expand $1/(w-z)$ as a geometric series in non-negative powers of $(z-z_0)$. On the inner circle $|w-z_0|<|z-z_0|$: expand the same kernel as a geometric series in $(w-z_0)/(z-z_0)$, which extracts negative powers of $(z-z_0)$.

Term by term integration gives the two-sided Laurent series. ∎

In practice: manipulate known series

You rarely compute Laurent coefficients from the integral formula. Instead, manipulate known Taylor series of related functions.

Example. $f(z)=e^z/z^2$ around $z_0=0$. Use $e^z=1+z+z^2/2!+z^3/3!+\cdots$:

$\frac{e^z}{z^2}=\frac{1}{z^2}\left(1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots \right)=\frac{1}{z^2}+\frac{1}{z}+\frac{1}{2!}+\frac{z}{3!}+\cdots .$

Two negative-power terms ($1/z^2$ and $1/z$); then a Taylor-like tail. So $a_{-2}=1$, $a_{-1}=1$, $a_0=1/2$, $a_1=1/6$, etc.

Example. $f(z)=1/(z(z-1))$ around $z_0=0$, in the annulus $0<|z|<1$.

Partial fractions: $1/(z(z-1))=-1/z+1/(z-1)=-1/z-1/(1-z)=-1/z-\sum z^n$ for $|z|<1$.

Laurent series: $-1/z-1-z-z^2-\cdots$ on $0<|z|<1$.

Same function, larger annulus. Same $f$ on $|z|>1$. Now we can’t use $\sum z^n$ (diverges). Rewrite:

$\frac{1}{z-1}=\frac{1}{z}\cdot \frac{1}{1-1/z}=\frac{1}{z}\sum (1/z{)}^n=\sum z^{-n-1}=\frac{1}{z}+\frac{1}{z^2}+\cdots$

for $|z|>1$. So on this annulus, $f=-1/z+(1/z+1/z^2+\cdots )=1/z^2+1/z^3+\cdots$.

Different Laurent series for the same function on different annuli. The series depends on the annulus, not just on the function. Which annulus matters in your problem depends on what set you’re working on.

Singularity classification

The negative-power terms of the Laurent series at $z_0$ classify the singularity:

• Removable singularity. No negative terms; the Laurent series is actually a Taylor series. $f$ has a finite limit at $z_0$ and can be extended analytically by setting $f(z_0)=$ that limit. Example: $\sin z/z$ at $z=0$, with $f(0)=1$.
• Pole of order $m$. Finitely many negative terms, with the most negative being $a_{-m}/(z-z_0{)}^m$ where $a_{-m}\ne 0$ and $a_n=0$ for $n<-m$. Example: $1/z^2+1/z+\cdots$ has a pole of order 2 at 0.
• Essential singularity. Infinitely many negative terms. Example: $e^{1/z}=1+1/z+1/(2!z^2)+\cdots$ has an essential singularity at 0.

Each type has different residue formulas; see Residue (complex analysis) for computing them.

The residue: coefficient of $1/(z-z_0)$

The single most important coefficient is $a_{-1}$ — the coefficient of $1/(z-z_0)$ in the Laurent series. It’s called the residue of $f$ at $z_0$:

$\mathrm{Res}(f,z_0)=a_{-1}.$

Why this coefficient? Because ${\oint }_C(z-z_0{)}^{n}dz=0$ for all $n\ne -1$, and ${\oint }_C(z-z_0{)}^{-1}dz=2\pi i$. Integrating the Laurent series term by term, only the $a_{-1}$ term contributes to a closed-contour integral around $z_0$. So

${\oint }_{C}f(z)dz=2\pi i{a}_{-1}=2\pi i\mathrm{Res}(f,z_0).$

This is the kernel of the Residue theorem.

In context

Laurent series are the working tool for the Residue theorem. Every closed-contour integral over a function with isolated singularities reduces to summing residues, and computing residues either uses direct Laurent expansion or specialized shortcuts that effectively read off $a_{-1}$ without writing the full series.