theorem (Casorati-Weierstrass):

If $f$ is holomorphic on $\Omega\setminus\left\{{z_0}\right\}$ where $z_0$ is an essential singularity, then for every $V\subset \Omega\setminus\left\{{z_0}\right\}$ , $f(V)$ is dense in ${\mathbf{C}}$ .

Equivalently, suppose $z_{0}$ is an essential isolated singularity of $f(z)$ . Then for every complex number $w_{0}$ , there is a sequence $z_{n} \rightarrow z_{0}$ such that $f\left(z_{n}\right) \rightarrow w_{0}$ .

figures/2022-01-05_05-28-04.png

slogan:

The image of a punctured disc at an essential singularity is dense in ${\mathbf{C}}$ .

proof (of Casorati-Weierstrass):

Let $f$ have an essential singularity. Suppose toward a contradiction that there exists a $z_0\in {\mathbf{C}}$ , a punctured neighborhood $\Omega$ of $z_0$ , and ${\varepsilon}> 0$ with $f(\Omega) \cap{\mathbb{D}}_{\varepsilon}(z_0)$ empty. So ${\left\lvert {f(z) - z_0} \right\rvert}>{\varepsilon}$ for every $z\in \Omega$ . Write $\tilde \Omega \coloneqq\Omega\cup\left\{{z_0}\right\}$ .

Define $g(z) \coloneqq{1\over f(z) - z_0}$ , noting that $z_0$ is the only singularity of $g$ in $\tilde\Omega$ , making $g$ holomorphic on $\Omega$ . We have ${\left\lvert {g(z)} \right\rvert} < {\varepsilon}^{-1}< \infty$ for all $z\in \Omega$ , so $g$ is bounded in $\Omega$ and $z_0$ must be a removable singularity. By Riemann’s removable singularity theorem, $g$ extends to a holomorphic function on all of $\tilde\Omega$ .

We can now write $f(z) = {1\over g(z)} + z_0$ . If $g(z_0) = 0$ , then it is a zero of some finite order for $g$ , and thus a pole of finite order for $f$ , contradicting that $z_0$ is essential for $f$ . Otherwise, if $g(z_0) = w_0\neq 0$ , then $\begin{align*} {\left\lvert { f(z_0)} \right\rvert} = {\left\lvert { {1\over w_0} + z_0} \right\rvert} \leq {\varepsilon}+ {\left\lvert {z_0} \right\rvert} < \infty ,\end{align*}$ making $z_0$ removable and again yielding a contradiction.

proof (of Casorati-Weierstrass, Gamelin):

Pick $w\in {\mathbf{C}}$ and suppose toward a contradiction that $D_R(w) \cap f(V)$ is empty. Consider $\begin{align*} g(z) \coloneqq{1\over f(z) - w} ,\end{align*}$ and use that it’s bounded to conclude that $z_0$ is either removable or a pole for $f$ .

In detail, from Gamelin:

figures/2021-12-10_18-47-34.png

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Exercises

exercise (Entire functions missing a disc):

Show that if $f$ is entire and there exists a disc ${\mathbb{D}}_r(a)$ not intersecting $f({\mathbf{C}})$ , then $f$ must be constant.

#complex/exercise/completed

solution:

Write $a\not\in f({\mathbf{C}})$ , and replace $f$ with $f(z) - a$ so that $f(a) = 0$ and there is some ${\mathbb{D}}_R(0)$ not intersecting $f({\mathbf{C}})$ . Then $f(z) \in {\mathbb{D}}_R^c$ for all $z\in {\mathbf{C}}$ , so ${\left\lvert {f(z)} \right\rvert} > R$ for all $z$ . Defining $G(z) \coloneqq{1\over f(z)}$ yields ${\left\lvert {G(z)} \right\rvert} < R^{-1}$ for all $z$ , making $G$ bounded. The singularity of $G$ at $z=0$ is thus removable, so $G$ extends to an entire function by Riemann’s removable singularity theorem. By Liouville, $G$ is constant, so $f$ is constant.

exercise (Entire functions satisfying a bound):

Find all entire functions $f$ that satisfy $\begin{align*} {\left\lvert {f(z)} \right\rvert} \geq e^{{\left\lvert {z} \right\rvert}} && \forall z\in {\mathbf{C}} .\end{align*}$

#complex/exercise/completed

solution:

Claim: there are no such functions. Consider $g(z) \coloneqq f(z)/e^z$ , which is entire since $e^z$ is nonvanishing. Now $g$ is entire and ${\left\lvert {g} \right\rvert} \geq 1$ everywhere, so $\operatorname{im}(g) \cap{\mathbb{D}}$ is empty. This contradicts Casorati-Weierstrass, which requires that $\operatorname{im}g$ be dense in ${\mathbf{C}}$ .

Alternatively, note $f\neq 0$ by the inequality, so $1/f$ is bounded and entire and thus constant. However, $f(z) = c$ contradicts the inequality, since $e^{{\left\lvert {z} \right\rvert}}\to \infty$ as ${\left\lvert {z} \right\rvert}\to \infty$ .

Alternatively, note that $e^{{\left\lvert {z} \right\rvert}} \geq 1$ for all $z$ , so $\operatorname{im}(f) \cap{\mathbb{D}}$ is empty, again contradicting Casorati-Weierstrass.

exercise (Showing a function is constant):

Let $f$ be entire and define a function $\begin{align*} m: [0, \infty) &\to {\mathbf{R}}\\ r &\mapsto \min_{{\left\lvert {z} \right\rvert} = r} {\left\lvert {f(z)} \right\rvert} .\end{align*}$ Suppose $\tilde m \coloneqq\lim_{r\to \infty}m(r)$ exists and is a finite positive real number, and show that $f$ is constant.

#complex/exercise/work #stuck

Hint: consider $g(z) \coloneqq f(1/z)$ .