Laurent Series

The Laurent series of a complex function $f(z)$ is a representation of that function as a power series which includes terms of negative degree. It may be used to express complex functions in cases where a Taylor series expansion cannot be applied. It is named after French mathematician Pierre Alphonse Laurent in 1843.

Creating the Laurent Series

Let $0<r_1<|w-z_0|<r_2$ such that

$C_m=\mathcal{C}(t;z_0+r_me^{it},0,2\pi)$

for $m=1,2$ . In order to create the Laurent Series, we need to prove four main theorems.

The First Theorem

Suppose that $f(w)$ is holomorphic for $r_1\le |z-z_0|\le r_2$ . Then

$2\pi f(w)=\oint_{C_2}\frac{f(z)}{z-w}\,dz-\oint_{C_1}\frac{f(z)}{z-w}\,dz$ .

The proof is actually not terrible. Consider some complex valued function $g$ such that

$g(z)=\frac{f(z)-f(w)}{z-w}$

where this exists, and $g(w)=f'(w)$ . Then it follows that $g(z)$ is holomorphic for $r_1\le |z-z_0|\le r_2$ . Hence,

$\oint_{C_1}g(z)\,dz=\oint_{C_2}g(z)\,dz$ .

Now, notice that $\frac{f(z)}{z-w}=g(z)+\frac{f(w)}{z-w}$ for all $z\in C_1,C_2$ . This means we get

$\oint_{C_2}\frac{f(z)}{z-w}\,dz-\oint_{C_1}\frac{f(z)}{z-w}=f(w)\left(\oint_{C_2}\frac{1}{z-w}\,dz-\oint_{C_1}\frac{1}{z-w}\right).$

Clearly we have $w\in\text{Ext}(C_1)$ and $w\in\text{Int}(C_2)$ which means that $\oint_{C_2}\frac{1}{z-w}\,dz=2\pi i$ and $\oint_{C_1}\frac{1}{z-w}\,dz=0$ by a theorem. This implies the result $\square$

The Second Theorem

Suppose that the first theorem holds and let

$a_n=\frac{1}{2\pi i}\oint_{C_m}\frac{f(z)}{(z-z_0)^{n+1}}\,dz$

for all $n\in\mathbb{Z}$ with $m=1,2$ as long as $n\ge 0$ and $n<0$ . We claim that

$f(w)=\sum_{n=-\infty}^{\infty}a_n(w-z_0)^n$ .

Note that from before, we have

$\frac{1}{z-w}=\sum_{n=0}^{\infty}\frac{w-z_0)^n}{(z-z_0)^{n+1}}$

for $z\in|C_2|$ . Now, there must exist an $M$ such that $f(z)\le M$ for all $z\in|C_2|$ by a theorem, so we get the inequality

$\frac{f(z)(w-z_0)^n}{(z-z_0)^{n+1}}\le \frac{M|w-z_0|^n}{r_2^{n+1}}$

for each $z\in|C_2|$ once again. But notice: the right hand side is independent of $z$ ! So, because we already assumed that $r_1<|w-z_0|<r_2$ we see that the sum

$\sum_{n=0}^{\infty}\frac{M|w-z_0|^n}{r_2^{n+1}}$

converges. Hence, the sum $\sum_{n=0}^{\infty}\frac{f(z)(w-z_0)^n}{(z-z_0)^{n+1}}$ converges uniformly on the contour $C_2$ . This means we get $\begin{align*} \oint_{C_2}\frac{f(z)}{z-w}\,dz&=\oint_{C_2}\sum_{n=0}^{\infty}\frac{f(z)(w-z_0)^n}{(z-z_0)^{n+1}}\,dz\\ &=\sum_{n=0}^{\infty}(w-z_0)^n\oint_{C_2}\frac{f(z)}{(z-z_0)^{n+1}}\,dz\\ &=2\pi i\sum_{n=0}^{\infty}(w-z_0)^n. \end{align*}$ But we are not finished just yet! Also notice that $\begin{align*} \frac{1}{w-z}&=\sum_{n=1}^{\infty}\frac{(z-z_0)^{n-1}}{(w-z_0)^n}\\ &=\sum_{n=-\infty}^{-1}\frac{(w-z_0)^n}{(z-z_0)^{n+1}} \end{align*}$ for each $z\in|C_1|$ . This gives

$-\oint_{C_1}\frac{f(z)}{z-w}\,dz=2\pi i\sum_{n=-\infty}^{-1}\frac{(w-z_0)^n}{(z-z_0)^{n+1}}$

and rearrangement gives the desired $\square$

The Third Theorem (The Laurent Series Defined)

We define the term "a ring in a wider sense" to mean the following: the set of points between two concentric circles, a disk without its center, and the exterior of a circle not including $\infty$ . Let $\mathcal{R}$ be a ring in the wider sense, with center $z_0$ and let $f$ be holomorphic on $\mathcal{R}$ . Then there are numbers $a_i$ for each $i\in (-\infty,\infty)$ such that for all $w\in \mathcal{R}$ we have the series

$f(w)=\sum_{n=-\infty}^{\infty}a_n(w-z_0)^n$ .

Choose any $r>0$ such that the circle $C=\mathcal{C}(t;z_0+r_me^{it},0,2\pi)\in\mathcal{R}$ and let

$a_n=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{(z-z_0)^{n+1}}\,dz$

for every $n\in\mathbb{Z}$ . Choose some $w\in\mathcal{R}$ , and let $r_1$ , $r_2$ , $C_1$ and $C_2$ denote the same stuff we used earlier. Then it follows that

$\oint_{C}\frac{f(z)}{(z-z_0)^{n+1}}\,dz=\oint_{C_m}\frac{f(z)}{(z-z_0)^{n+1}}\,dz$

for $m=1,2$ of course. The theorem then follows immediately $\square$

The Fourth Theorem (Uniqueness of the Laurent Series)

Is such a series unique, however? It certainly should be, or else I would not be writing this all down here. Let $\mathcal{R}$ be a ring in the wider sense with center $z_0$ of a circle $C\in\mathcal{R}$ , and let

$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^m$

for every $z\in\mathcal{R}$ . Then we must prove that

$a_n=\frac{1}{2\pi i}\oint_{C}\frac{f(z)}{(z-z_0)^{m+1}}\,dz$

holds for every integer $m$ . We have $\begin{align*} \oint_{C}\frac{f(z)}{(z-z_0)^{m+1}}\,dz&=\oint_{C}\sum_{n=-\infty}^{\infty}a_n(z-z_0)^{n-m-1}\,dz.\\ \end{align*}$ To swap the integrand and the sum, we must note that

$\sum_{n=-\infty}^{\infty}c_n=\sum_{n=0}^{\infty}c_n+\sum_{n=1}^{\infty}c_{-n}.$

Now we can have the following deduction. $\begin{align*} \oint_{C}\frac{f(z)}{(z-z_0)^{m+1}}\,dz&=\oint_{C}\sum_{n=0}^{\infty}a_n(z-z_0)^{n-m-1}\,dz+\oint_{C}\sum_{n=1}^{\infty}a_{-n}(z-z_0)^{n-m-1}\,dz\\ &=\sum_{n=0}^{\infty}a_n\oint_{C}(z-z_0)^{n-m-1}\,dz+\sum_{n=1}^{\infty}a_{-n}\oint_{C}(z-z_0)^{n-m-1}\,dz\\ &=\sum_{n=-\infty}^{\infty}a_n\oint_{C}(z-z_0)^{n-m-1}\,dz.\\ \end{align*}$ Crisis avoided! Sort of. We assumed that $\sum_{n=0}^{\infty}a_n(z-z_0)^{n-m-1}$ and $\sum_{n=1}^{\infty}a_{-n}(z-z_0)^{n-m-1}$ converge uniformly on $C$ . We can fix this! Let the radius of $C$ (remember, $C$ is a circle) be $r$ . Then there exists some $z_1$ such that $|z_1-z_0|>r$ with $z_1\in\mathcal{R}$ . It follows that the series

$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^m$

holds when $z_1=z$ so we get

$f(z)=\sum_{n=0}^{\infty}a_n(z-z_0)^m$

converges. We have to prove uniform convergence though, so back to work. Notice that this result above implies that there is some $k$ such that $|a_n(z-z_0)^n|\le k$ for $n\ge0$ . Hence we get