Laurent Series

(Lecture 21 of Mathematical Methods II.)

Laurent Series generalizes Taylor Series to include negative integers in the powers of the series:

\begin{equation} \tag{1} \sum_{n=-\infty}^\infty c_n(z-z_0)^n \end{equation}

If we spell this out:

\begin{equation} \tag{2} c_0+c_1(z-z_0)+c_2(z-z_0)^2+c_3(z-z_0)^3+\cdots+\frac{c_{-1}}{z-z_0}+\frac{c_{-2}}{(z-z_0)^2}+\frac{c_{-3}}{(z-z_0)^3}+\cdots \end{equation}

A first important type of Laurent Series is to describe functions that are not analytic at a given point, about which one can not compute the Taylor Series. Let us remember the case of the function

\begin{equation} \tag{3} f(z)=\exp(-1/z^2)\,. \end{equation}

It is analytic everywhere but at the origin. For any nonzero~$z$, we can formally represent the Series representation of the exponential $\exp(z)=\sum_{k=0}^\infty z^k/k!$ and substitute $z$ for $-1/z^2$. This gives us the Laurent series of the:

\begin{equation} \tag{4} f(z)=\sum_{k=0}^\infty\frac{(-1)^k}{k!}\frac{1}{z^{2k}}= 1-\frac{1}{z^2}+\frac{1}{2z^4}-\frac{1}{6z^6}+\cdots \end{equation}

which is a Laurent Series, convergent for all~$|z|>0$. We will see later the consequence of having an infinite number of negative coefficients. Not all Series are like this, of course, for instance:

\begin{equation} \tag{5} \frac{e^z}{z^3}=\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{2z}+\frac{1}{3!}+\frac{z}{4!}+\frac{z^2}{5!}+\cdots \end{equation}

This time, the number of terms with negative powers is finite. We will see in next lecture how such Laurent expansions allow to classify singularities.

There can be various domains of convergence for a Laurent Series. They are annuli of the type:

\begin{equation} \tag{6} \mathcal{A}(z_0,r,R)=\{z\,:\,r<|z|<R\} \end{equation}

and the expression of the Laurent Series on various annuli can be different, although the Laurent expansion of an analytic function on a given annulus is unique.

Let us first illustrate the above with an example. Consider the function

\begin{equation} \tag{7} f(z)=\frac{3}{2+z-z^2}\,. \end{equation}

The denominator factors out as $(1+z)(2-z)$ so the partial fraction decomposition yields:

\begin{equation} \tag{8} f(z)=\frac{1}{z+1}+\frac{1}{2}\frac{1}{1-z/2}\,. \end{equation}

This gives us four possible series decompositions: in and out the circle of convergences:

\begin{align} \frac{1}{1+z}&=\sum_{k=0}^\infty(-z)^k\,,\quad\text{for }|z|<1\tag{9}\\ \frac{1}{1-z/2}&=\sum_{k=0}^\infty(z/2)^k\,,\quad\text{for }|z|<2\tag{10}\\ \frac{1}{1+z}&=\frac{1}{z}\frac{1}{1+\frac{1}{z}}=\sum_{k=0}^\infty\frac{(-1)^k}{z^{k+1}}\,,\quad\text{for }|z|>1\tag{11}\\ \frac{1}{1-z/2}&=\frac{2}{z}\frac{-1}{1-2/z}=-\sum_{k=0}^\infty\frac{2^{k+1}}{z^{k+1}}\,,\quad\text{for }|z|>2\tag{12} \end{align}

yielding three types of Laurent Series. In the unit disk, we sum Eqs.~((9)) and~((10)):

\begin{equation} \tag{13} f(z)=\sum_{k=0}^\infty((-1)^k+(1/2)^{k+1})z^k=\frac{3}{2}-\frac{3}{4}z+\frac{9}{8}z^2-\frac{15}{16}z^3+\cdots\quad\text{for }|z|<1\,, \end{equation}

which is the Maclaurin Series. In the annulus $\mathcal{A}(0,1,2)$, we sum Eqs.~((10)) and~((11)):

\begin{equation} \tag{14} f(z)=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{z^{k}}+\sum_{k=0}^\infty\frac{z^k}{2^{k+1}}=\cdots+\frac{1}{z^3}-\frac{1}{z^2}+\frac{1}{z}+\frac{1}{2}+\frac{z}{4}+\frac{z^2}{8}+\cdots\,, \end{equation}

for $1<|z|<2$ and for~$|z|>2$, i.e., in the annulus $\mathcal{A}(0,2,\infty)$, summing Eqs.~((11)) and ((12)):

\begin{equation} \tag{15} f(z)=\sum_{k=1}^\infty\frac{(-1)^{k-1}-2^{k-1}}{z^k}=-\frac{3}{z^2}-\frac{3}{z^3}-\frac{9}{z^4}-\frac{15}{z^5}-\cdots,\quad\text{for }|z|>2\,. \end{equation}

There is a Laurent theorema analogous to the Taylor theorem that provides the Taylor expansion of an analytic function. It reads: An analytic function~$f$ in a domain that contains two concentric circles~$\mathcal{C}_1$ and~$\mathcal{C}_2$ with center~$z_0$ can be represented in the annulus that separates them as:

\begin{equation} \tag{16} f(z)=\sum_{n=-\infty}^\infty\left[\frac{1}{2i\pi}\oint_\mathcal{C}\frac{f(w)}{(w-z_0)^{n+1}}\,dw\right](z-z_0)^n\,, \end{equation}

with~$\mathcal{C}$ any single closed path within the annulus. The proof is an extension of the proof of Taylor's theorem and is left as an exercise.

When the inner circle can be shrinked to a single point to exclude only one point, the series of the negative terms is called the principal part of~$f$ at~$z_0$.

We conclude with a powerful result on the division of power series: if $f$ and $g$ are two analytic functions at~$z_0$ and $g(z_0)\neq0$, with Taylor series representations $f(z)=\sum_{k=0}^\infty a_k(z-z_0)^k$ and $g(z)=\sum_{k=0}b_k(z-z_0)^k$, then the quotient $f/g$ has the Taylor series:

\begin{equation} \tag{17} \frac{f(z)}{g(z)}=\sum_{k=0}^\infty c_k(z-z_0)^k\,, \end{equation}

where the coefficients $c_k$ satisfy the long division algorithm:

\begin{equation} \tag{18} a_n=b_0c_n+\cdots+b_nc_0 \end{equation}

i.e., $a_0=b_0c_0$, $a_1=b_0c_1+b_1c_0$, $a_2=b_0c_2+b_1c_1+b_2c_0$, etc., so that $c_0=a_0/b_0$, $c_1=a_1/b_0-a_0b_1/b_0^2$, etc., although it is usually easier to compute recursively.

For example, the Taylor Series of $1/\cos(z)$ is:

\begin{equation} \tag{19} \frac{1}{\cos(z)}=\frac{1}{\displaystyle 1-\frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+\cdots} \end{equation}

so that $a_0=1$ and all other $a_n=0$, and $b_0=1$, $b_1=0$, $b_2=-1/2$, $b_3=0$, $b_4=1/4!$, etc. This gives:

\begin{equation} \tag{20} \frac{1}{\cos(z)}=1+\frac{1}{2}z^2+\frac{5}{24}z^4+\frac{61}{720}z^5\cdots \end{equation}

Since the $k$th coefficient is also given by the formal Taylor series as $f^{(k)}(0)/k!$, we see that, for instance:

\begin{equation} \tag{21} (1/\cos(z))'''''|_{z=0}=61\,. \end{equation}