Laurent Series of Analytic Complex Functions

# Laurent Series of Analytic Complex Functions

So far we have looked at Taylor series of analytic complex functions. We are about to look at a more general type of series expansion for a complex analytic function known as a Laurent series. We will first need to define a special type of domain called an annulus.

 Definition: Let $r_1, r_2 > 0$ with $0 \leq r_1 < r_2 \leq \infty$. The Annulus centered at $z_0$ with inner radius $r_1$ and outer radius $r_2$ is the set $A(z_0, r_1, r_2) = \{ z \in \mathbb{C} : r_1 < \mid z - z_0 \mid < r_2 \}$.

We can now define the Laurent series of $f$ on the annulus $A(z_0, r_1, r_2)$.

 Definition: Let $f$ be an analytic function on the annulus $A(z_0, r_1, r_2)$. Then the Laurent Series of $f$ centered at $z_0$ on this annulus is the series $\displaystyle{\sum_{n=-\infty}^{\infty} a_n(z - z_0)^n}$ where $\displaystyle{a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z_0)^{n+1}} \: dw}$ where $\gamma$ is an circle centered at $z_0$ contained in the $A(z_0, r_1, r_2)$.

For example, consider the following function:

(1)
\begin{align} \quad f(z) = \frac{1}{(z + 1)(z + 3)} \end{align}

This function is analytic on $\mathbb{C} \setminus \{ -1, -3 \}$. Suppose that we want to find the Laurent series of $f$ centered at $z_0 = 1$ on the annulus with inner radius $2$ and outer radius $4$, i.e., on $A(1, 2, 4)$.

We will first use partial fractions to rewrite $f$ in a simpler form. Note that there exists $A, B \in \mathbb{C}$ such that:

(2)
\begin{align} \quad f(z) = \frac{1}{(z + 1)(z + 3)} = \frac{A}{z + 1} + \frac{B}{z + 3} \end{align}

Therefore $A(z + 3) + B(z + 1) = 1$. Set $z = -1$. Then $2A = 1$ so $\displaystyle{A = \frac{1}{2}}$. Now set $z = -3$. Then $-2B = 1$ so $\displaystyle{B = -\frac{1}{2}}$. So the partial fraction decomposition of $f$ is:

(3)
\begin{align} \quad f(z) = \frac{1}{2} \frac{1}{z + 1} - \frac{1}{2} \frac{1}{z + 3} = \frac{1}{2} \left [ \frac{1}{z + 1} - \frac{1}{z + 3} \right ] \end{align}

Now here is the idea. We want a series expansion for $\displaystyle{\frac{1}{z + 3}}$ that is valid for for $\mid z - 1 \mid < 4$, and we want a series expansion for $\displaystyle{\frac{1}{z + 1}}$ that is valid for $\mid z - 1 \mid > 2$. Then the combined expansion will be valid for $2 < \mid z -1 \mid < 4$ which is the set of points in the annulus $A(1, 2, 4)$.

For the series $\displaystyle{\frac{1}{z + 3}}$ we have:

(4)
\begin{align} \quad \frac{1}{z + 3} = \frac{1}{4 + (z - 1)} = \frac{1}{4} \frac{1}{1 + \frac{z - 1}{4}} = \frac{1}{4} \sum_{n=0}^{\infty} \left ( -\frac{z - 1}{4} \right )^n = \sum_{n=0}^{\infty} \frac{(-1)^n (z - 1)^n}{4^{n+1}} \quad (*) \end{align}

Note that this series is valid for $\biggr \lvert \frac{z - 1}{4} \biggr \rvert < 1$, i.e., for $\mid z - 1 \mid < 4$!

For the series $\displaystyle{\frac{1}{z + 1}}$ we have that:

(5)
\begin{align} \quad \frac{1}{z + 1} = \frac{1}{(z - 1) + 2} = \frac{1}{z - 1} \frac{1}{1 + \frac{2}{z - 1}} = \frac{1}{z - 1} \sum_{n=0}^{\infty} \left ( -\frac{2}{z - 1} \right )^n = \sum_{n=0}^{\infty} \frac{(-1)^n 2^n}{(z - 1)^{n+1}} \quad (**) \end{align}

Note that this series is valid for $\biggr \lvert \frac{2}{z - 1} \biggr \rvert < 1$, i.e., for $2 < \mid z - 1 \mid$.

We combine the series in $(*)$ and $(**)$ to get the Laurent series of $f$ on the annulus $A(1, 2, 4)$:

(6)
\begin{align} \quad \sum_{n=0}^{\infty} \frac{(-1)^n (z - 1)^n}{4^{n+1}} + \sum_{n=0}^{\infty} \frac{(-1)^n 2^n}{(z - 1)^{n+1}} \end{align}

Note that the Laurent series is valid for $2 < \mid z -1 \mid < 4$, i.e., on the annulus $A(1, 2, 4)$. We will soon se that this series indeed converges to $f$ on $A(1, 2, 4)$.