5.3: Laurent Series

Recall that a function \(f\) of the complex variable \(z\) is analytic at a point \(z_0\) if it has a derivative at each point in some neighbourhood of \(z_0\). An entire function is a function that is analytic at each point in the entire finite plane. If a function \(f\) fails to be analytic at a point \(z_0\) but is analytic at some point in every neighbourhood of \(z_0\), then \(z_0\) is called a singular point, or singularity, of \(f\).

Suppose that \(f(z)\), or any single valued branch of \(f(z)\), if \(f(z)\) is multivalued, is analytic in the region \(0\lt|z-z_0|\lt R\) and not at the point \(z_0\). Then the point \(z_0\) is called an isolated singular point of \(f(z)\).

Now, recall also that any function which is analytic throughout a disk \(|z -z_0|\lt R_0\) must have a Taylor series about \(z_0\). If the function fails to be analytic at a point \(z_0\), it is often possible to find a series representation for \(f(z)\) involving both positive and negative powers of \(z−z_0\). Formally speaking we have the following result:

Theorem \(\PageIndex{1}\)

Suppose that a function \(f\) is analytic throughout an annular domain \(R_1 \lt |z - z_0| \lt R_2\), centred at \(z_0\), and let \(C\) denote any positively oriented simple closed contour around \(z_0\) and lying in that domain. Then, at each point in the domain, \(f(z)\) has the series representation

\(\begin{eqnarray}\label{laurentfunction}
                f(z)=\sum_{n=0}^{\infty} a_n(z-z_0)^n+\sum_{n=1}^{\infty}\frac{b_n}{(z-z_0)^n},
                \end{eqnarray}\)

where

\(\begin{eqnarray}\label{non-principalpart}
                a_n=\frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{n+1}},\quad n=0,1,2,\ldots
                \end{eqnarray}\)

and

\(\begin{eqnarray}\label{principalpart}
                b_n=\frac{1}{2\pi i}\oint_C \frac{f(z)dz}{(z-z_0)^{-n+1}},\quad n=1,2,\ldots
                \end{eqnarray}\)

Figure 1: \(f(z)\) is analytic throughout an annular domain \(R_1 \lt |z - z_0| \lt R_2\).

In practice, the above integral formulae (2) and (3) may not offer the most practical method for computing the coefficients \(a_n\) and \(b_n\) for a given function \(f(z)\); instead, one often pieces together the Laurent series by combining known Taylor expansions. Because the Laurent expansion of a function is unique whenever it exists, any expression of this form that actually equals the given function \(f(z)\) in some annulus must actually be the Laurent expansion of \(f(z)\).

For example, consider the function

\(f(z)=\frac{1}{z(1+z^2)}\)

which has isolated singularities at \(z=0\) and \(z=±i\). In this case, there is a Laurent series representation for the domain \(0\lt |z|\lt 1\) and also one for the domain \(1\lt |z|\lt \infty\), which is exterior to the circle \(|z|=1\). To find each of these Laurent series, we recall the Maclaurin series representation

\(\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n,\quad |z|\lt 1.\)

For the domain \(0\lt |z|\lt 1\), we have

\(\begin{eqnarray*}
            f(z)&=&\frac{1}{z}\frac{1}{1+z^2}=\frac{1}{z}\sum_{n=0}^{\infty}\left(-z^2\right)^n \\
            &=&\sum_{n=0}^{\infty}(-1)^nz^{2n-1}\\
            &=&\frac{1}{z}+\sum_{n=1}^{\infty}(-1)^nz^{2n-1} \\
            &=& \sum_{n=0}^{\infty}(-1)^{n+1}z^{2n+1}+\frac{1}{z}.
            \end{eqnarray*}\)

On the other hand, when \(1\lt |z|\lt \infty\),

\(\begin{eqnarray*}
            f(z)&=&\frac{1}{z^3}\frac{1}{1+\frac{1}{z^2}}=\frac{1}{z^3}\sum_{n=0}^{\infty} \left(-\frac{1}{z^2}
            \right)^n\\
            & =&\sum_{n=0}^{\infty}\frac{(-1)^n}{z^{2n+3}}\\
            &=&\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{z^{2n+1}}
            \end{eqnarray*}\)

In this last part we use the fact that \((-1)^{n-1}=(-1)^{n-1}(-1)^2=(-1)^{n+1}\).

Exercise \(\PageIndex{1}\)

Expand \(f(z)=e^{3/z}\) in a Laurent series valid for \(0\lt |z| \lt \infty\).