Badly-Behaved Points

Previously, we discussed Taylor series and their cousin Laurent series:

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k (z-z_0)^k,\quad z \in \mathrm{Ann}(z_0, r_1, r_2) ,$

where

$\displaystyle c_n = \frac 1{2\pi i} \oint_{\partial B(z_0, r)} \frac{f(w)}{(w-z_0)^{n+1}}\, \mathrm dw ,\quad r_1 < r < r_2.$

The Laurent series exists (and converges uniformly) whenever $f$ is holomorphic on $\overline{\text{Ann}} (z_0, r_1, r_2)$ .

Lemma 1. Let $f : \mathbb C \to \mathbb C$ be holomorphic on $\mathrm{Ann}(z_0, r_1, r_2)$ with Laurent series

$\displaystyle f(z) = \sum_{k=-\infty}^{\infty} c_k (z-z_0)^k,\quad z \in \mathrm{Ann}(z_0, r_1, r_2).$

If $c_k = 0$ for $k < 0$ , then there exists a unique function $F : \mathbb C \to \mathbb C$ holomorphic on $B(z_0, r_2)$ such that $F|_{\mathrm{Ann}(z_0, r_1, r_2) } = f|_{\mathrm{Ann}(z_0, r_1, r_2) }$ .

Definition 1. Given the setup in Lemma 1, we call $z_0$ an isolated singularity if $f$ is holomorphic on $\bar B(z_0, r) \backslash \{z_0\}$ for some $r > 0$ . Furthermore, we call $z_0$ :

a removable singularity if $c_k = 0$ for $k < 0$ ,
a pole of order $m$ if $m$ is the largest integer such that $a_{-m} \neq 0$ ,
an essential singularity if $c_k \neq 0$ for infinitely many $k < 0$ .

A pole of order $1$ (resp. $2$ ) is called a simple (resp. double) pole.

Denote $\hat{\mathbb C} := \mathbb C \cup \{ \infty \}$ for brevity.

Lemma 2. An isolated singularity $z_0$ of $f$ is removable if and only if $f$ approaches a limit in $\mathbb C$ as $z \to z_0$ .

Proof. In the direction $(\Rightarrow)$ , $f(z) \to c_0$ as $z \to z_0$ . In the direction $(\Leftarrow)$ , suppose $z_0 = 0$ without loss of generality. Since $f$ has a finite limit at $0$ , $f$ is bounded on some $B(0, r) \supseteq \bar B(0, r) \backslash \{0\}$ . Assume $|f| \leq 1$ without loss of generality. Hence, we can compute the Laurent coefficients $c_{-k}$ , $k \in \mathbb N^+$ by

$\begin{aligned}2 \pi i \cdot c_{-k} &= \oint_{\partial B(0, r)} \frac{f(w)}{w^{-k+1}}\, \mathrm dw \end{aligned}$

For any $w \in \partial B(0, r)$ , $|f(w)/w^{-k+1}| = |f(w)| / |w|^{-k+1} \leq 1 / r^{-k+1}$ , By the ML-inequality,

$\begin{aligned}2\pi \cdot |c_{-k}| = |2 \pi i \cdot c_{-k}| &\leq \frac 1{r^{-k+1}} \cdot 2 \pi r = 2 \pi r^k \quad \Rightarrow \quad |c_{-k}| \leq r^k.\end{aligned}$

Taking $r \to 0^+$ , $c_{-k} = 0$ .

Lemma 3. An isolated singularity $z_0$ of $f$ is a pole of order $m$ if and only if $(z-z_0)^m f(z)$ has a limit in $\mathbb C \backslash \{0\}$ as $z \to z_0$ .

Proof. In the direction $(\Rightarrow)$ , the Laurent series of $f$ is given by

$\displaystyle f(z) = \sum_{k=-m}^{\infty} c_k (z-z_0)^k,\quad z \in \bar B(z_0, r) \backslash \{z_0\}.$

Multiplying by $(z-z_0)^m$ ,

$\displaystyle (z-z_0)^mf(z) = \sum_{k=-m}^{\infty} c_k (z-z_0)^{m+k} = \sum_{k=0}^{\infty} c_{k-m} (z-z_0)^{k}.$

By Lemma 2, since the right-hand side has a limit $c_{-m}$ as $z \to z_0$ , so also $(z-z_0)^mf(z) \to c_{-m} \neq 0$ .

Lemma 4. An isolated singularity $z_0$ of $f$ is a pole of some order if and only if $|f(z)| \to \infty$ as $z \to z_0$ .

Proof. Suppose $z_0 = 0$ for simplicity. In the direction $(\Rightarrow)$ , $f$ has a Laurent series given by

$\displaystyle f(z) = \frac{a_{-m}}{z^m} + \sum_{k=1}^{m-1} \frac{a_{-k}}{z^m} \cdot z^{m-k} + \sum_{k=0}^{\infty} a_k z^k.$

The Taylor series portion converges to $a_0$ as $z \to 0$ . For the other portions, we use the reverse triangle inequality for sufficiently small $|z|$ to obtain $|f(z)| \to \infty$ .

For the reverse implication, suppose $|f(z)| \to \infty$ as $z \to 0$ . First, restrict $r >0$ such that $|f| |_{B(0, r) \backslash \{0\}} > 1$ . Hence, $g:=1/f$ is bounded on $B(0, r) \backslash \{0\}$ . Using the argument in Lemma 2, $0$ is a removable singularity of $g$ . Since $|f(z)| \to \infty$ , $g(z) \to 0$ as $z \to 0$ . Extend $g$ to a holomorphic function $G$ on $B(0, r) \backslash \{0\}$ , so that

$\displaystyle G(0) = \lim_{z \to 0} G(z) = \lim_{z \to 0} g(z) = 0.$

Find the largest $m$ such that $g(z) = z^m \cdot h(z)$ , where $h$ is a holomorphic function that has no root in $B(0, r)$ . Then

$\displaystyle f(z) = \frac 1{z^m} \cdot \frac 1{h(z)}.$

Since $1/h$ has no root in $B(0, r)$ , we can expand it into its Taylor series, so that

$\displaystyle f(z) = \frac 1{z^m} \sum_{k=0}^\infty d_k z^k,\quad d_0 \neq 0.$

In particular, $f$ has a pole of order $m$ .

Lemma 5. An isolated singularity $z_0$ of $f$ is an essential singularity if and only if $f$ has no limit in $\hat{\mathbb C}$ as $z \to z_0$ .

Proof. Lemmas 2, 3, and 4.

Essential singularities are therefore relatively bad-behaved points.

Theorem 1 (Casorati-Weierstrass Theorem). Let $z_0$ be an essential singularity of $f$ . Then for any $\delta > 0, w \in \mathbb C, \epsilon > 0$ ,

$f(B(z_0, \delta)\backslash \{z_0\}) \cap B(w,\epsilon) \neq \emptyset.$

Proof. Assume $z_0 = 0$ for simplicity. Suppose for a contradiction there exists $\delta, w, \epsilon$ such that the intersection is empty. Then for any $z \in B(0, \delta) \backslash \{0\}$ , $|f(z) - w| \geq \epsilon > 0$ . Define $g$ by $g(z) := 1/(f(z) - w)$ , which is bounded above by $1/\epsilon$ , and so has a removable singularity at $0$ . Find the largest integer $m$ such that $g(z) = z^m \cdot h(z)$ for some holomorphic $h$ that does not have a root in $B(0,\delta)$ . Therefore,