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)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)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)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)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)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)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 $A(1, 2, 4)$.