Weierstrass–Casorati theorem

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

The Casorati-Weierstrass theorem states that

if f has an essential singularity at

z 0

, and V is any neighborhood of

z 0

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

Or, spelled out: if ε > 0 and w is any complex number, then there exists a complex number z in U with |z - z₀| < ε and |f(z) - w| < ε.

Start with an open subset U of the complex plane containing the number z₀, and a holomorphic function f defined on U − {z₀}. The complex number $z 0$ is called an essential singularity if there is no natural number n such that the limit

$\lim_{z \to z_0} f(z) \cdot (z - z_0)^n$

exists.

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:

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

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

Taking the absolute value of both sides:

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

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

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

For values of $θ$ such that $cos \theta >0, f(z)\rightarrow$ infinity as $r \rightarrow 0$ , and for $cos \theta <0, f(z) \rightarrow 0$ as $r \rightarrow 0$ .

Consider what happens, for example when r 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,

$f(z) = e^{R} \left[ cos \left( Rtan \theta \right) - isin \left( Rtan \theta \right) \right]$

and

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

Thus, $\left| f(z) \right|$ may take any positive value otherthan 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:

$\left[ cos \left( Rtan \theta \right) - isin \left( Rtan \theta \right) \right]$

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

Theorem:

F (z)

has an essential singularity at

z = z 0

then for any complex number

w

f (z)

becomes arbitrarily close to

w

in a neighborhood of

z 0

. That is, given

w

, and

ε > 0

δ > 0

, there is a

z

such that

| f (z) - w | < ε

whenever

0 < | z - z 0 | < δ

Proof:

The proof is by contradiction. So, first we suppose the opposite: that $| f (z) - w | > ε$ whenever $| z - z 0 | δ$ , where $δ$ is small enough such that $f (z)$ is analytic in the punctured neighborhood of radius $δ$ around $z 0$ . In this region, $0 < | z - z 0 | < δ$ we can then assume that the function:

$h(z)=\frac{1}{f(z)-w}$

must be analytic, because $f(z)-w \neq 0$ . Because it is analytic it is bounded in the region, so $|h(z)|<\frac{1}{\epsilon}$ . The function $f (z)$ is not identically constant, because it if was both bounded and constant it would be analytic and it would not posses any singular point. Because $h (z)$ is analytic and bounded, it is representable by a power series with the form:

$h(z)=\displaystyle\sum_{n=o}^{\infty}C_{n}(z-z_{0})^{n}$

thus, it must have a removable singularity, because it is bounded and has a power series. By choosing $C 0 = h (z)$ , it follows that $h (z)$ is analytic for $| z - z 0 | < δ$ . So we have:

$f(z)= w + \frac{1}{h(z)}$

and $f (z)$ is either analytic with $h(z) \neq 0$ or else $f (z)$ has a pole of order $N$ strength $C N$ , where $C N$ is the first non-zero coefficient of the term $(z - z 0) n$ in the Taylor series representation of $h (z)$ . In either case, this contradicts the hypothesis that $f (z)$ has an essential singular point in the neighborhood of $z = z 0$ Both possibilities contradict the assumption of the theorem. Thus the theorem holds.

Retrieved from "http://en.wikipedia.org/wiki/Weierstrass%E2%80%93Casorati_theorem"

Categories: Complex analysis | Mathematical theorems

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