Criterion for Classifying Isolated Removable Sing. of Analytic Functs.

# Criterion for Classifying Isolated Removable Singularities of Analytic Functions

Recall from the Singularities of Analytic Complex Functions page that a point $z_0$ is said to be a singularity of $f$ if $f$ is not analytic at $z_0$ and moreover, $z_0$ is said to be an isolated singularity of $f$ if $f$ is analytic on a punctured disk $D(z_0, r) \setminus \{ z_0 \}$.

We also looked at three classifications of isolated singularities. We will focus on the first type on this page - the removable singularities. Recall that $z_0$ is a removable singularity of $f$ if $\displaystyle{\lim_{z \to z_0} f(z) }$ exists. The following theorem gives us a nice classification of these singularities.

Theorem 1: Let $f$ and $g$ be functions that are analytic at $z_0$ and such that $f(z_0) = 0$ and $g(z_0) = 0$ (i.e., $z_0$ is a root of $f$ and of $g$). Let $\displaystyle{h = \frac{f}{g}}$. If the order of $z_0$ for $f$ is greater than or equal to the order of $z_0$ for $g$ then $z_0$ is a removable singularity of $h$. |

- **Proof: Let $f$ and $g$ be analytic at $z_0$. Then there exists open disks centered at $z_0$ for which $f$ and $g$ are analytic. In particular, this means that $z_0$ is an isolated singularity for the function $h$.

- Let $r > 0$ be such that $f$ and $g$ have no other roots in the open disk $D(z_0, r)$. Then $h$ is analytic on the puncture open disk $D(z_0, r) \setminus \{ 0 \}$. Let $k$ be the order of $z_0$ for $f$ and let $s$ be the order of $z_0$ for $g$. Then $f$ and $g$ assume the following forms:

\begin{align} \quad f(z) = (z - z_0)^k f^*(z) \end{align}

(2)
\begin{align} \quad g(z) = (z - z_0)^s g^*(z) \end{align}

- Where $f^*$ and $g^*$ are analytic on $D(z_0, r)$ and $f^*(z_0) \neq 0$, $g^*(z_0) \neq 0$. So we may write:

\begin{align} \quad h(z) = \frac{f(z)}{g(z)} = \frac{(z - z_0)^k f^*(z)}{(z - z_0)^s g^*(z)} = (z - z_0)^{k-s} \frac{f^*(z)}{g^*(z)} \end{align}

- If the order of $z_0$ for $f$ is equal to the order of $z_0$ for $g$ ($k = s$) then:

\begin{align} \quad \lim_{z \to z_0} h(z) = lim_{z \to z_0} \frac{f^*(z)}{g^*(z)} = \frac{f^*(z_0)}{g^*(z_0)} \quad (\mathrm{By \: the \: continuity \: of \:} f^*, g^*) \end{align}

- If the order of $z_0$ for $f$ is greater than the order of $z_0$ for $g$ ($k > s$) then:

\begin{align} \quad \lim_{z \to z_0} h(z) = lim_{z \to z_0} (z - z_0)^{k-s} \frac{f^*(z)}{g^*(z)} = 0 \cdot \frac{f^*(z_0)}{g^*(z_0)} = 0 \end{align}

- In both cases we see that the limit exists and so $z_0$ is a removable singularity of $h$. $\blacksquare$