Essential singularities. Casorati-Weierstrass theorem.

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

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.

Introduction to Laurent Series

$\int_\gamma \frac{1}{(z-a)^2} dz = 0$ for any simple closed curve $\gamma$ around a because it has a valid, single-valued antiderivative (primitive) in the punctured domain $\mathbb{C} \setminus \{a\}$.

Using the power rule for integration ($\int x^n dx = \frac{x^{n+1}}{n+1}$), we get: $F(z) = \frac{(z-a)^{-1}}{-1} = -\frac{1}{z-a}$, and this is a valid function in the punctured domain $\mathbb{C} \setminus \{a\}$. It gives a unique complex number and it does’t jump if you walk around “a”.

Now, we could apply the Fundamental Theorem of Calculus. If a function f has an antiderivative F everywhere on the curve $\gamma$, then: $\int_\gamma f(z) dz = F(end_point) - F(start_point)$. Since $\gamma$ is a closed curve, the start and end points are the same ($z_{start_point} = z_{end_point}$), hence: $\int_\gamma \frac{1}{(z-a)^2} dz = F(z_{start}) - F(z_{start}) = 0$

Next, consider two different paths, $\alpha$ and $\beta$, that both start at point P and end at point Q. $\gamma = \beta - \alpha$ represents the closed loop formed by going forward along $\beta$ (from P to Q) and then backward along $\alpha$ (from Q back to P).

Since the integral over the closed loop $\gamma$ is zero, then: $\int_\gamma f(z) dz = 0 \implies \int_\beta f(z) dz - \int_\alpha f(z) dz = 0 \implies \int_\beta f(z) dz = \int_\alpha f(z) dz$. Conclusion: Path independence: Since the integral around the closed loop is zero, the integral from P to Q doesn't depend on which path you take.

However, $g(z) = \frac{1}{z-a}$. Its antiderivative G(z) = ln(z -a). The complex logarithm is multi-valued, $\int_\gamma \frac{1}{z-a} dz = G(\text{end}) - G(\text{start}) = 2\pi i \neq 0$. Because the integral is not zero, no single-valued global antiderivative exists on the punctured disk.

Laurent Series

Any function analytic in an annulus (ring) around a can be written as: $f(z) = \dots + \frac{a_{-2}}{(z-a)^2} + \frac{\mathbf{a_{-1}}}{(z-a)} + a_0 + a_1(z-a) + \dots$

When we integrate this series around a closed curve surrounding a, we integrate term by term.

• The Power Rule Terms ($n \neq -1$): For any integer $n \neq -1$, the antiderivative is $\frac{(z-a)^{n+1}}{n+1}$. This is single-valued, hence $\int_{\gamma} (z-a)^n dz = 0 \quad (\text{for } n \neq -1)$ (this includes our previous example).
• The Logarithmic Term ($n = -1$): This is the only term that survives the integration. $\int_{\gamma} \frac{a_{-1}}{z-a} dz = a_{-1} (2\pi i)$

Conclusion: The integral of the entire function depends only on that one specific coefficient, $a_{-1}$. $\int_{\gamma} f(z) dz = 2\pi i \cdot a_{-1}$ This coefficient $a_{-1}$ is called the Residue because the 1/z term creates a “residue” (non-zero integral), while terms like $1/x^2$ vanish. Furthermore, a function $f(z)$ has an antiderivative in the punctured neighborhood if and only if its residue $a_{-1}$ is zero.

Laurent’s expansion

Let f be analytic in a punctured neighborhood of a, $\mathbb{B'}(a; r)$. Suppose f has a pole of order m at a. This means f(a) is infinite, preventing us from directly using Taylor series to expand f, as this approach requires the function to be nice (“smooth”) and differentiable at a.

We create a helper or auxiliary function h(z) by multiplying f(z) by exactly enough powers of $(z-a)$ to cancel out the infinity, $h(z) = (z-a)^m f(z)$.

Recall that if f has a pole of order m at a, then by definition there exists a holomorphic function g near a such that $f(z)=\frac{g(z)}{(z-a)^m}, g(a) \ne 0$, so h(z) = g(z). Since g is holomorphic at a, the limit exists and equals the value of g at a.

In other words, if $f$ has a pole of order m, it behaves roughly like $\frac{1}{(z-a)^m}$. Multiplying by $(z-a)^m$ neutralizes or cancels the pole. $\lim_{z \to a} h(z) = g(a) = \text{Finite Non-Zero Number}$

Because this limit exists, $h(z) = (z-a)^m f(z)$ has a removable singularity at $a$. We “fill the hole” by defining $h(a)$ as that limit. Now, $h(z)$ is analytic everywhere in the disk $\mathbb{B}(a; r)$.

We define $h(z) = \begin{cases} (z-a)^mf(z), &z \in \mathbb{B'}(a; r)\\\\ \lim_{z \to a}(z-a)^{m}f(z), &z = a \end{cases}$

Since h(z) is analytic at a, we can use Taylor’s Theorem to expand it into a power series with non-negative powers: $h(z) = c_0 + c_1(z -a) + c_2(z -a)^2 + \cdots + c_{m-1}(z -a)^{m-1}+ c_m(z -a)^m + \cdots, \forall z \in \mathbb{B}(a; r)$ where $c_0 = h(a) \ne 0$ (This confirms the pole was exactly order $m$) and the coefficients $c_n$ are just standard Taylor coefficients for h: $c_n = \frac{h^{(n)}(a)}{n!}$.

In particular for $z \ne a, (z-a)^mf(z) = c_0 + c_1(z -a) + c_2(z -a)^2 + \cdots + c_{m-1}(z -a)^{m-1}+ c_m(z -a)^m + \cdots, \forall z \in \mathbb{B'}(a; r)$. Obviously, we want $f(z)$, not $h(z)$, so we divide the entire Taylor series by $(z-a)^m$.

We have split f(z) into two distinct parts: $f(z) = \underbrace{\left[ \frac{c_0}{(z-a)^m} + \dots + \frac{c_{m-1}}{z-a} \right]}_{\text{Principal (Singular) Part}} + \underbrace{\left[ c_m + c_{m+1}(z-a) + \dots \right]}_{\text{Analytic (Regular) Part } f_1(z)}$

This is the Laurent Expansion. The first bracket contains all the negative powers (the “bad” and “naughty” part that causes the problem aka singularity), and the second bracket is a normal power series (the “good” and “nice” part).

Power series (like our $f_1(z)$) and Laurent series converge uniformly on any closed subset of their domain (like the contour $\gamma$). In analysis, uniform convergence is the “safety pass” that allows us to swap the limit (the infinite sum $\Sigma$) and the integral ($\int$), $\int_\gamma \left( \sum c_n (z-a)^n \right) dz = \sum c_n \left( \int_\gamma (z-a)^n dz \right)$

If $\gamma$ is a simple closed contour in $\mathbb{B'}(a; r)$ oriented positively such that $a \in Int(\gamma)$, $\int_{\gamma} f(z) dz =[\text{Sum of integrals of individual terms}] \int_{\gamma} \frac{c_0}{(z-a)^m} + \int_{\gamma}\frac{c_1}{(z-a)^{m-1}} + \cdots + \int_{\gamma} \frac{c_{m-1}}{z-a} + \int_{\gamma} f_1(z) = \int_{\gamma} \frac{c_{m-1}}{z-a} = c_{m-1}\cdot (2\pi i)$

Let’s look at the integral of a general term $\int_\gamma C (z-a)^k dz$. The function $(z-a)^k$ has a valid antiderivative: $F(z) = \frac{(z-a)^{k+1}}{k+1}$. We can apply the Fundamental Theorem of Calculus.

If a function f has an antiderivative F everywhere on the curve $\gamma$, then: $\int_\gamma f(z)dz=F(\text{end point})-F(\text{start point})$. Since the path $\gamma$ is closed, the integral is: $F(\text{end point}) - F(\text{start point}) = 0$. Every term in the series vanishes, except k = 1.

In conclusion, when one integrates the infinite series for f(z), the majority of the terms evaluate to zero, and only one single term survives. $\int_\gamma f(z) dz = 2\pi i \cdot c_{m-1}$. This coefficient $c_{m-1}$ is called the Residue of f at a, often denoted as $\text{Res}(f, a)$.

Lemma. Let f be analytic on and inside a positively oriented simple closed curve $\gamma$ except at a point a inside $\gamma$ where f has a pole of order m. Then, $\int_{\gamma}f(z)dz = c_{m-1}\cdot (2\pi i)$ where $c_{m-1}$ is the coefficient of $\frac{1}{z-a}$ in the singular part of the expansion of f as powers of (z -a) in the neighborhood of a.

• If $c_{m-1} \ne 0$: The integral is non-zero ($2\pi i \cdot c_{m-1}$). The function does not have an antiderivative on the punctured domain.
• If $c_{m-1} = 0$: The integral is zero. The “bad” logarithmic term is missing from the expansion. Therefore, $f(z)$ does have an antiderivative locally.