Weierstrass-Casorati theorem.html

The Casorati-Weierstrass theorem in complex analysis describes the remarkable behavior of meromorphic functions near essential singularities. It is named for Karl Theodor Wilhelm Weierstrass and Felice Casorati.

Start with an open subset U of the complex plane containing the number z₀, and a holomorphic function f defined on U − {z₀} which has an essential singularity at z₀. The Casorati-Weierstrass theorem then states that

if V is any neighborhood of

z 0

contained in U, then f(V − {z₀}) is dense in C.

This can also be stated as follows:

for any ε > 0 and any complex number w, there exists a complex number z in U with |z - z₀| < ε and |f(z) - w| < ε,

or still, in more descriptive terms,

f comes arbitrarily close to any complex value in every neighbourhood of z₀.

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.

Plot of the function exp(1/z), centered on the essential singularity at z=0. The hue represents the complex argument, the luminance represents the absolute value. This plot shows how approaching the essential singularity from different directions yields different behaviors (as opposed to a pole, which would be uniformly white).

Examples

The function f(z) = exp(1/z) has an essential singularity at z₀ = 0, but the function g(z) = 1/z³ does not (it has a pole at 0).

Consider the function

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

This function has the following Laurent series about the essential singular point at $z = 0$ :

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

Because $f'(z) =\frac{-e^{\frac{1}{z}}}{z^{2}}$ exists for all points $z \neq 0$ we know that $f (z)$ is analytic in the neighborhood of $z = 0$ . Hence it is an isolated singularity like all other essential singularities.

Using a change of variable to polar coordinates $z = r e i θ$ our function, $f(z)=e^{\frac{1}{z}}$ becomes:

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

Taking the absolute value of both sides:

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

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

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

and

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

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:

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

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. Suppose f is meromorphic on some punctured neighborhood V − z₀, and that z₀ is an essential singularity. Suppose also that there is some complex value b and some ε > 0 such that |f(z) − b| ≥ ε for all z in V at which f is defined. Then the function

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

is holomorphic on V - {z₀}, with zeroes at the poles of f, and bounded by 1/ε. It can therefore be holomorphically extended to all of V by Riemann's theorem. So

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

on V - {z₀}. We consider the two possible cases of

$\lim_{z \rarr z_0} g(z).$

If it is 0, then f has a pole at z₀. If it is not 0, then z₀ is a removable singularity of f. Both possibilities contradict the assumption of the theorem. Thus the theorem holds.