# Ngô Quốc Anh

## May 19, 2009

### Casorati–Weierstrass theorem, the behavior of meromorphic functions near essential singularities

Filed under: Giải tích 7 (MA4247) — Tags: , — Ngô Quốc Anh @ 14:42

In complex analysis, a branch of mathematics, the Casorati–Weierstrass theorem describes the remarkable behavior of meromorphic functions near essential singularities. It is named for Karl Theodor Wilhelm Weierstrass and Felice Casorati.

Formal statement of the theorem.

Start with some open subset $U$ in the complex plane containing the number $z_0$, and a function $f$ that is holomorphic on $U\backslash \left\{ {{z_0}} \right\}$, but has an essential singularity at $z_0$. The Casorati–Weierstrass theorem then states that

if $V$ is any neighborhood of $z_0$ contained in $U$, then $f(V \backslash \{z_0\})$ is dense in $\mathbb C$.

This can also be stated as follows

for any $\varepsilon > 0$ and any complex number $w$, there exists a complex number $z$ in $U$ with $|z-z_0| < \varepsilon$ and $|f(z)-w| < \varepsilon$.

Or in still more descriptive terms

$f$ comes arbitrarily close to any complex value in every neighbourhood of $z_0$.

This form of the theorem also applies if $f$ is only meromorphic. The theorem is considerably strengthened by Picard’s great theorem, which states, in the notation above, that $f$ assumes every complex value, with one possible exception, infinitely often on $V$.

Examples.

The function $f(z) = e^\frac{1}{z}$ has an essential singularity at $z_0 = 0$, but the function $g(z) = \frac{1}{z^3}$ does not (it has a pole at $0$). Consider the function

$\displaystyle f(z)=e^{1/z}$.

This function has the following Laurent series about the essential singular point at $z_0$

$\displaystyle f(z)=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!z^{n}}$.

Because

$\displaystyle f'(z) =\frac{-e^{\frac{1}{z}}}{z^{2}}$

exists for all points $z\ne 0$ we know that $f(z)$ is analytic in the neighborhood of $z = 0$. Hence it is an isolated singularity, as well as being and essential singularity. Using a change of variable to polar coordinates $z = re^{i\theta}$ our function, $f(z) = e^\frac{1}{z}$ becomes

$\displaystyle f(z)=e^{\frac{1}{r}e^{-i\theta}}=e^{\frac{1}{r}\cos(\theta)}e^{-\frac{1}{r}i \sin(\theta)}$.

Taking the absolute value of both sides

$\displaystyle\left| f(z) \right| = \left| e^{\frac{1}{r}cos \theta} \right| \left| e^{-\frac{1}{r}i \sin(\theta)} \right |$.

Thus, for values of $\theta$ such that $\cos \theta > 0$, we have $f(z)\rightarrow\infty$ as $r \rightarrow 0$, and for $\cos \theta < 0$, $f(z) \rightarrow 0$ as $r \rightarrow 0$.

Consider what happens, for example when $z$ takes values on a circle of diameter $\frac{1}{R}$ tangent to the imaginary axis. This circle is given by $r = \frac{1}{R} \cos \theta$. Then,

$\displaystyle f(z) = e^{R} \left[ \cos \left( R\tan \theta \right) - i \sin \left( R\tan \theta \right) \right]$

and

$\displaystyle \left| f(z) \right| = e^R$.

Thus, $\left| f(z) \right|$ may take any positive value other than zero by the appropriate choice of $R$. As $z \rightarrow 0$ on the circle, $\theta \rightarrow \frac{\pi}{2}$ with $R$ fixed. So this part of the equation

$\displaystyle\left[ \cos \left( R \tan \theta \right) - i \sin \left( R \tan \theta \right) \right]$

takes on all values on the unit circle infinitely often. Hence $f(z)$ takes on all the value of every number in the complex plane except for zero infinitely often.

Proof of the theorem.

A short proof of the theorem is as follows: Take as given that function $f$ is meromorphic on some punctured neighborhood $V\backslash \left\{ {{z_0}} \right\}$, and that $z_0$ is an essential singularity. Assume by way of contradiction that some value $b$ exists that the function can never get close to; that is: assume that there is some complex value $b$ and some $\varepsilon > 0$ such that $|f(z)-b| \geq \varepsilon$ for all $z$ in $V$ at which $f$ is defined.

Then the new function

$g(z) = \frac{1}{f(z) - b}$

must be holomorphic on $V\backslash \left\{ {{z_0}} \right\}$, with zeroes at the poles of $f$, and bounded by $\frac{1}{\varepsilon}$. It can therefore be analytically continued (or continuously extended, or holomorphically extended) to all of $V$ by Riemann’s analytic continuation theorem. So the original function can be expressed in terms of $g$

$f(z) = \frac{1}{g(z)} + b$

for all arguments $z$ in $V\backslash \left\{ {{z_0}} \right\}$. Consider the two possible cases for $\lim_{z \to z_0} g(z)$. If the limit is $0$, then $f$ has a pole at $z_0$. If the limit is not $0$, then $z_0$ is a removable singularity of $f$. Both possibilities contradict the assumption that the the point $z_0$ is an essential singularity of the function $f$. Hence the assumption is false and the theorem holds.