Essential Singularities and the Casorati-Weierstrass Theorem

All things are difficult before they are easy, Thomas Fuller.

Casorati-Weierstrass theorem

Casorati-Weierstrass theorem. Let f be analytic in a punctured neighborhood of a, $\mathbb{B'}(a; r)$ and let a be an essential singularity. Then, the image of any neighborhood of a under f is dense in $\mathbb{C}$. In other words, f(z) comes arbitrarily close to any complex value A as z approaches a

An essential singularity is mathematically “chaotic.” Unlike a pole (which cleanly shoots off to infinity) or a removable singularity (which settles on a specific number), an essential singularity oscillates wildly. This theorem states that it oscillates so violently that if you pick any arbitrary target number A in the complex plane, f(z) will get infinitely close to A somewhere near the singularity.

Basically, it does not settle, nor does it blow up in just one direction. It behaves so erratically that its outputs densely fill the complex plane.

Proof.

We use the previous classification method:

• Case 1 or 2 (Non-Essential): There exists a real threshold h such that the limit of $|z-a|^\alpha |f(z)|$ are 0 (for $\alpha > h$) or $\infty$ (for $\alpha < h$).
• Case 3 (Essential): No such threshold exists.

Step 1: The Assumption

Suppose the theorem is false. This means f(z) does not get arbitrarily close to some complex value $A \in \mathbb{C}$. Mathematically, there exists a “safety zone” or more specifically, a “bubble” of radius $\delta > 0$ around A such that f(z) never enters it, $|f(z) - A| \ge \delta, \forall z \in \mathbb{B}'(a; r)$

This condition is not trivial at all, but quite significant. It tell us that the function f(z) - A is strictly bounded away from zero.

Step 2: Testing the Shifted Function

Consider the limit of this shifted function multiplied by a negative exponent $\alpha < 0$: $\lim_{z \to a} |z-a|^\alpha |f(z) - A|$

• Since $\alpha < 0$, we can rewrite the term as a fraction, $|z-a|^\alpha = \frac{1}{|z-a|^{|\alpha|}}$. As $z \to a$, the denominator shrinks to 0, meaning the fraction shoots to $\infty$.
• The term $|f(z) - A|$ is at least $\delta$ (a positive, non-zero number), bounded away from zero.
• Therefore, a number bounded away from zero multiplied by infinity must go to infinity: $\lim_{z \to a} |z-a|^\alpha |f(z) - A| = \infty, \forall \alpha < 0$.

Step 3: Finding the Contradiction

Because this limit is $\infty$ for $\alpha < 0$, the function $g(z) = f(z) - A$ satisfies the criteria for a non-essential singularity (ii). This means g(z) is not an essential singularity and we know from the previous analysis that there exists some “threshold integer” h and a large enough $\beta$ capable of “crushing” the function’s growth: $\lim_{z \to a} |z-a|^\beta |f(z) - A| = 0$.

Recall the growth classification of isolated singularities: If there exists a real threshold h such that $\lim_{z\to a}|z-a|^\alpha |f(z)|$ equals 0 for $\alpha>h$ and $\infty$ for $\alpha < h$, then the singularity is removable or a pole. If no such threshold exists, the singularity is essential.

$|f(z)| = |(f(z) - A) + A| \le[\text{Triangle Inequality}] |f(z) - A| + |A|$

Multiply this entire inequality by our crushing weight $|z-a|^\beta, |z-a|^\beta |f(z)| \le |z-a|^\beta |f(z) - A| + |z-a|^\beta |A|$.

Now, take the limit as $z \to a$:

1. First term on the right side: We have already established that $\lim_{z \to a} |z-a|^\beta |f(z) - A| = 0$.
2. Second term on the right side: Since A is constant and $\beta > 0$, $\lim_{z \to a} |z-a|^\beta |A| = 0$.
3. Therefore, $\lim_{z \to a} |z-a|^\beta |f(z)| \le 0 + 0 = 0$. Since the absolute value is non-negative, by the Sandwich theorem, the limit is exactly 0.

We have shown that $\lim_{z \to a} |z - a|^\beta |f(z)| = 0$. This satisfies Condition (i) of the classification system. If a function satisfies Condition (i) for some $\beta$, it falls into Case 1 or 2 (removable or pole). Therefore, a is NOT an essential singularity.

This contradicts the given information that a is an essential singularity. Thus, our initial assumption (that f avoids A) must be false. f(z) must get arbitrarily close to every complex value A.

Classification of isolated singularities. To classify the isolated singularity of a function f(z) at $z_0$, we examine its Laurent series expansion $\sum _{n=-\infty}^{\infty} a_n(z-z_0)^n$ valid in a deleted neighborhood $0 \lt |z - z_0| \lt R$. The part with negative powers, $\sum_{n=1}^{\infty} a_{-n}(z-z_0)^{-n}$ is called the principal part.

1. A singularity is removable if the principal part is zero ($a_n = 0, \forall n \lt 0$), no negative powers. In this case, the function can be made analytic at $z_0$ by defining $f(z_0) = a_0$. Condition: $\lim_{z \to z_0}f(z)$ exists and is finite, e.g., $f(z) = \frac{sin(z)}{z}, z_0 = 0$.

2. A singularity is a pole of order 𝑚 if the principal part has a finite number of non-zero terms, finitely many negative powers, ending at $(z-z_0)^{-m}$. Form: $\frac{a_{-m}}{(z-z_0)^m} + \cdots + \frac{a_{-1}}{z-z_0} + \sum_{n=0}^{\infty} a_n(z-z_0)^n$. Condition: $\lim_{z \to z_0}|f(z)| = \infty,$ e.g., $f(z)=\frac{1}{z^2}$ has a pole of order 2 at $z_0 = 0.$

3. A singularity is essential if the principal part has infinitely many non-zero terms with negative powers. Near an essential singularity, the function exhibits extreme behavior, taking on every possible complex value (except possibly one) infinitely often in any neighborhood of $z_0$ (Casorati-Weierstrass Theorem), e.g., $f(z) = e^\frac{1}{z}$

Examples

Let’s illustrates the chaotic nature of an essential singularity.

$f(z) = e^\frac{1}{z}$ is analytic in $\mathbb{B'}(0; 1)$.

If we expand $e^{1/z}$ using the standard Taylor series for $e^w$ (where $w = 1/z$), we get: $e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2! z^2} + \frac{1}{3! z^3} + \cdots$. Because this Laurent series has infinitely many terms with negative powers of z (principal part), z = 0 is an essential singularity.

To see why the limit does not exist, let’s approach the origin ($z \to 0$) along different paths.

• Approaching along the Imaginary Axis ($z = -ri, r \in \mathbb{R}$).

$\frac{1}{z} = \frac{1}{-ri} = \frac{1}{-i} \cdot \frac{1}{r} = i \cdot \frac{1}{r}$. Then, $f(-ri) = e^{i(1/r)} \implies |f(-ri)| = |e^{i(1/r)}| = 1$

Recall Euler’s formula: $e^{i\theta} = \cos(\theta) + i\sin(\theta)$. Here $\theta = 1/r$.

As $r \to 0$, the value $1/r$ goes to infinity. The angle $\theta$ spins infinitely fast! The function does not go to infinity or zero; instead, it spins wildly around the unit circle (our function stays bounded), hitting every point on that circle infinitely many times.

• Approaching along the Positive Real Axis (z = r). Let z = r where r > 0 is a real number. As $z \to 0^+$, $r \to 0^+, \frac{1}{r} \to \infty$, then $\lim_{r \to 0^+} e^{1/r} = +\infty$. The function explodes to infinity.
• Approaching along the Negative Real Axis ($z = -r$). Let $z = -r$ where $r > 0$. As $z \to 0$, $1/z = -1/r \to -\infty$, then $f(-r) = e^{-1/r} = \frac{1}{e^{1/r}} \to 0$. In words, the function is crushed to zero.

Therefore, 0 is an essential singularity and by Casorati-Weierstrass theorem we also know that the function comes arbitrarily close to any complex value in every neighborhood of z = 0.

Can we find a $z$ very close to 0 such that $f(z) = A$? Set up the equation $e^{1/z} = A$.

Let’s express A in polar form. Let $A = R e^{i\phi}$, where $R = |A|$ and $\phi$ is the argument. Then, $e^{1/z} = e^{\ln R + i\phi}$.

Solve for the exponent: $\frac{1}{z} = \ln R + i(\phi + 2\pi n), \forall n \in \mathbb{Z}$ and solve for z: $z_n = \frac{1}{\ln R + i(\phi + 2\pi n)}$. Notice that as $n \to \infty, z_n \to 0$ (the series {$z_n$} converge to the origin) and $f(z_n)$ is exactly our target value A.

Non-isolated singularities

sin(z) is zeros whenever z is an integer multiple of $\pi: sin(z) = 0 \iff z = k\pi, k \in \mathbb{Z}$.

The Cosecant function $\csc(w) = \frac{1}{\sin(w)}$ has singularities wherever the sine is zero. These are simple poles (order 1) because the zeros of sine are simple. $\csc(z)$ has simple poles at $k\pi, k \in \mathbb{Z}$.

Now consider our specific function $\csc(\frac{1}{z})$ (composition of cosecant with 1/z). The singularities occur when the input to the sine function is an integer multiple of $\pi: \frac{1}{z} = k\pi, k \neq 0$.

Solving for z, we get a sequence of singularities: $z_k = \frac{1}{k\pi}, k = \pm 1, \pm 2, \pm 3, \dots$. As $k \to \infty, z_k$ get closer and closer to 0, $\lim_{k \to \infty} z_k = 0$. The point z = 0 is a singularity because $\csc(1/0)$ is undefined. However, it is fundamentally different from the others.

• The points $z_k$ are isolated singularities. If we zoom in on any $z_k = \frac{1}{k\pi}$, we can draw a tiny circle around it that contains no other singularities.
• The point 0 is a non-isolated singularity (or a limit point of singularities). If we draw a circle of any radius r around 0 (no matter how small), it will contain infinitely many poles $z_k$ (specifically, all poles where $k > \frac{1}{\pi r} \implies \frac{1}{k\pi} \lt r$).

It has simple poles at $\frac{1}{k\pi}, k \in \mathbb{Z}$, and there are indeed infinitely many $\frac{1}{k\pi}$’s in any $\mathbb{B'}(0; r)$, hence 0 is a limit point of these singularities. Therefore, 0 is definitely not an isolated singularity (as every neighborhood of 0 contains infinitely many singularities $\frac{1}{k\pi}$’s).

Because standard complex analysis tools (like Laurent Series or Residues) require a simplified “annulus” free of other singularities, they break down completely at z = 0.

Definition. A function f is meromorphic on an open set $G \subset \mathbb{C}$ if it is analytic on G except for a set of poles P, provided that P has no limit point in G.

Note: A pole at a means: $f(z)=\frac{g(z)}{(z-a)^m},\qquad g(a)\neq 0$, so f blows up like $|z-a|^{-m}$ near a.
This representation is only meaningful if a is an isolated singularity: you need a punctured disk around a where f is analytic except at a.

The reader should consider the following facts:

1. Poles must be isolated (they must be spaced apart), you cannot have a cluster of poles piling up inside our domain. Every pole must have a little space or breathing room around it.
2. If the domain G is compact (like the Extended Complex Plane/Riemann Sphere), then a set of points with no limit point must be finite. Then, a function meromorphic on the entire Riemann Sphere has only finitely many poles.
3. On the domain $\mathbb{C} \setminus \{0\}$, $\csc(1/z)$ is analytic except at the points $1/k\pi$. These points do not accumulate in this domain (because 0 is excluded). So it is meromorphic on the punctured plane. However, on the domain $\mathbb{C}$ (including 0), $\csc(1/z)$ is not meromorphic because the poles accumulate at 0, which is indeed inside the domain.