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 “super messy” or “chaotic.” Unlike a pole (which goes to infinity) or a removable singularity (which goes to a specific number), an essential singularity oscillates wildly. This theorem states it oscillates so wild 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.

Proof.

We use the previous classification method:

• Case 1 or 2 (Non-Essential): There exists a real number h (threshold exponent that cleanly separates decay from blow-up) such that limits of $|z-a|^\alpha |f(z)|$ are 0 (for $\alpha > h$) or $\infty$ (for $\alpha < h$).
• Case 3 (Essential): No such threshold exists.

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 bounded away from zero.

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

• Since $\alpha < 0, |z-a| \to 0$, and the term $|z-a|^\alpha$ goes to $\infty$.
• The term $|f(z) - A|$ is at least $\delta$ (a positive number), bounded away from zero.
• Therefore, the product must go to infinity: $\lim_{z \to a} |z-a|^\alpha |f(z) - A| = \infty, \forall \alpha < 0$.

Because this limit is $\infty$ for $\alpha < 0$, the function $g(z) = f(z) - A$ satisfies Condition (ii) of our classification system. 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 $\beta$ such that: $\lim_{z \to a} |z-a|^\beta |f(z) - A| = 0$.

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

Multiply the whole inequality by $|z-a|^\beta, |z-a|^\beta |f(z)| \le |z-a|^\beta |f(z) - A| + |z-a|^\beta |A|$.

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

1. First term on right: We have already established that $\lim_{z \to a} |z-a|^\beta |f(z) - A| = 0$.
2. Second term on right: 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 ($a_n = 0, \forall n \lt 0$). 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, 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., $f(z) = e^\frac{1}{z}$

Example. 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 $k > \frac{1}{\pi r} \implies \frac{1}{k\pi} \lt r$).

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.

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.