Laurent's Theorem for Analytic Complex Functions

Laurent's Theorem for Analytic Complex Functions

Recall from the Laurent Series of Analytic Complex Functions page that if $f$ is an analytic function on the annulus $A(z_0, r_1, r_2)$ then the Laurent series of $f$ centered at $z_0$ on $A(z_0, r_1, r_2)$ is defined as the following series:

(1)
\begin{align} \quad \sum_{n=-\infty}^{\infty} a_n(z - z_0)^n \end{align}

Where $\displaystyle{a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z_0)^{n+1}} \: dw}$ for each $n \in \mathbb{Z}$.

We will now look at a remarkable theorem which generalizes Taylor's Theorem for Analytic Complex Functions.

Theorem 1 (Laurent's Theorem): Let $f$ be a function that is analytic on the annulus $A(z_0, r_1, r_2)$. Then:
a) The Laurent series of $f$ on $A(z_0, r_1, r_2)$ converges to $f(z)$ on $A(z_0, r_1, r_2)$, i.e., $\displaystyle{f(z) = \sum_{n=-\infty}^{\infty} a_n (z - z_0)^n}$ ($\displaystyle{a_n = \frac{1}{2\pi i} \int_{\gamma} \frac{f(w)}{(w - z_0))^{n+1}} \: dw}$ for all $n \in \mathbb{N}$ where $\gamma$ is any circle contained in the annulus $A(z_0, r_1, r_2)$).
b) For all $\rho_1, \rho_2 > 0$ such that $r_1 < \rho_1 < rho_2 < r_2$, the Laurent series of $f$ on $A(z_0, r_1, r_2)$ converges uniformly to $f(z)$.
Theorem 2: Let $f$ be a function that is analytic on the annulus $A(z_0, 0, r)$ and let $\displaystyle{\sum_{n=-\infty}^{\infty} a_n(z - z_0)^n}$ be the corresponding Laurent series expansion. If $z_0$ is an isolated singularity of $f$ then:
a) $z_0$ is a removable singularity of $f$ if and only if $a_n = 0$ for all $n < 0$.
b) $z_0$ is a pole singularity of order $k$ of $f$ if and only if $a_n = 0$ for all $n < k$ and $a_k \neq 0$.
c) $z_0$ is an essential singularity of $f$ if and only if there $a_n \neq 0$ for infinitely many $n < 0$.

For example, consider the function $\displaystyle{f(z) = \sin \left ( \frac{1}{z} \right )}$. This function is analytic on $\mathbb{C} \setminus \{ 0 \}$ and the Laurent series expansion of this function is given (and valid of $\mathbb{C} \setminus \{ 0 \}$) by:

(2)
\begin{align} \quad \sin \left ( \frac{1}{z} \right ) = \sum_{n=0}^{\infty} \frac{(-1)^n}{z^{2n+1} (2n+1)!} = \frac{1}{z} - \frac{1}{3! z^3} + \frac{1}{5! z^5} - ... \end{align}

Note that $a_n \neq 0$ for ALL $n < 0$. So $0$ is an essential singularity of $f$ by Theorem 2.

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