Classification of isolated singularities

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Recall

Laurent’s theorem. Let A = {$z \in \mathbb{C}: R \lt |z -a| \lt S$} be an annular region ($0 \le R \lt S \le \infty$) and let f be analytic on A. Then, $f(z) = \sum_{n=-\infty}^\infty c_n(z -a)^n, \forall z \in A$ where $c_n = \frac{1}{2\pi i}\int_{\gamma} \frac{f(w)}{(w-a)^{n+1}}dw$ and $\gamma$ is a positively oriented circle of radius r centered at a and R < r < S ($\gamma$ is inside the annular region).

Classification of isolated singularities

Corollary (aka Rosetta Stone of Complex Analysis). Let f be analytic in $\mathbb{B'}(a; r)$. (i) f has a removable singularity at a if and only if in the Laurent's series expansion of f around a ($\sum_{n = -\infty}^\infty (z-a)^n$) all negative coefficients are zero $c_n = 0, \forall n \lt 0$. (ii) f has a pole of order m ($m \ge 1$) at a if and only if the series stop at the negative power -m, $c_{-m} \ne 0, c_n = 0, \forall n \lt -m$; (iii) f has an isolated essential singularity at a if and only if the series has infinitely many negative terms (it never stops), i.e., there is no m such that $c_n = 0, \forall n \lt -m$

Proof.

Recall that the Laurent Series for a function $f(z)$ around a point $a$ is: $f(z) = \underbrace{\sum_{n=-\infty}^{-1} c_n (z-a)^n}_{\text{Principal Part}} + \underbrace{\sum_{n=0}^{\infty} c_n (z-a)^n}_{\text{Analytic Part}}$

The classification depends entirely on the Principal Part (the negative powers).

(i): Removable Singularity. f has a removable singularity at a if and only if in the Laurent’s series expansion of f around a ($\sum_{n = -\infty}^\infty (z-a)^n$) all negative coefficients are zero $c_n = 0, \forall n \lt 0$.

Assume $c_n = 0, \forall n \lt 0$, then the Laurent series simplifies to contain only non-negative powers: $f(z) = c_0 + c_1(z-a) + c_2(z-a)^2 + \dots$. This is just a standard Taylor Series.

Since power series are continuous inside their radius of convergence, we can simply plug in $z=a: \lim_{z \to a} f(z) = c_0$. The limit exists and is finite.

By the definition of a removable singularity (the limit exists and is finite), f has a removable singularity at $a$. We can “remove” the singularity by defining $f(a) = c_0$, e.g., $f(z) = \frac{\sin z}{z}$. Write the Maclaurin series of $\sin(z): \sin z=z-\frac{z^3}{3!}+\frac{z^5}{5!}-\frac{z^7}{7!}+\cdots$. Then, $\frac{\sin z}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} -\frac{z^6}{7!} + \cdots$. This is a power series in nonnegative powers of z, convergent in all of $\mathbb{C}$, so it’s actually the Taylor (Laurent) expansion about 0 with no negative powers. That means the singularity at z = 0 is removable: we can define $f(0):=\lim_{z\rightarrow 0}\frac{\sin z}{z}=1$, and the resulting function is entire.

Removable Singularities: The “Harmless” Isolated Singularities

A removable singularity is a point $z_0$ where a complex function f(z) is not defined (or not analytic), but can be “filled in” or redefined to make the function perfectly smooth (analytic) at that point. In other words, the apparent breakdown of analyticity at $z_0$ is superficial —something that can be repaired by defining a single value at that point. After this “patch”, the function becomes holomorphic in a full neighborhood of $z_0$.

A point $z_0$ is a removable singularity if it satisfies any of these equivalent conditions:

• Laurent‑Series Criterion. The Laurent expansion series of f(z) centered at $z_0$ contains no negative powers: $f(z) = a_0 + a_1(z-z_0) + a_2(z-z_0)^2 + \dots$. This makes the Laurent series identical to a standard Taylor series.
• Limit Behavior Criterion. The limit of the function as 𝑧 approaches $z_0$ exists and is a finite complex number 𝐿, $\lim_{z \to z_0}f(z) = L~ (|L| \lt \infty)$. This is the most intuitive condition, if the function settles to a finite value as we approach $z_0$, we can simply define $f(z_0)$ to be that value, L.
• Boundedness Criterion (Riemann's Theorem).There exists a punctured neighborhood around $z_0$ where f(z) remains bounded |f(z)| ≤ M (for some constant 𝑀). Riemann’s theorem tells us that boundedness near an isolated singularity forces the singularity to be removable. This is a powerful tool because it allows you to identify a removable singularity even if you cannot easily calculate the limit.
• Analytic Extension Criterion. There exists a function g(z) that is analytic at $z_0$ and exactly matches f(z) everywhere else near $z_0$.

(ii): Pole of Order m. f has a pole of order m ($m \ge 1$) at a if and only if the series stop at the negative power -m, $c_{-m} \ne 0, c_n = 0, \forall n \lt -m$;

Assume the series stops at -m, then the Laurent series simplifies to the following expression: $f(z) = \frac{c_{-m}}{(z-a)^m} + \frac{c_{-m+1}}{(z-a)^{m-1}} + \dots + c_0 + \dots$

Multiply by $(z-a)^m$: $(z-a)^m f(z) = c_{-m} + c_{-m+1}(z-a) + c_{-m+2}(z-a)^2 + \dots$

The reader should remember that to identify a pole, we evaluate the limit of $(z-a)^m f(z)$. Observe that by multiplying by $(z-a)^m$, all the denominators have been eliminated. The result is a regular Power Series.

Take the Limit: As $z \to a$, all terms with $(z-a)$ vanish, so $\lim_{z \to a} (z-a)^m f(z) = c_{-m}$. Since we assumed $c_{-m} \neq 0$, this limit is a finite, non-zero number. This is the precise definition of a Pole of Order m, e.g., $f(z) = \frac{1}{z^2} + \frac{3}{z}$ the Laurent expansion about z=0 is already given, and the lowest power of z is $z^{-2}$. The principal part is $\frac{1}{z^2}+\frac{3}{z}$, which is a finite sum with highest order term $z^{-2}$. That means z = 0 is a pole of order 2.

The Four Equivalent Characterizations of a Pole of Order m

In complex analysis, a pole of order 𝑚 is a type of isolated singularity where a function behaves like $\frac{1}{(z-z_0)^m}$ near a point $z_0$. It is the mildest possible non‑removable isolated singularity. It’s completely controlled, predictable, and algebraic in nature.

A point $z_0$ is defined as a pole of order 𝑚 (where 𝑚 is a positive integer) if it satisfies any of the following equivalent conditions:

• Laurent series expansion. The Laurent series of 𝑓(𝑧) centered at $z_0$ has a finite principal part, i.e. a finite number of terms with negative powers, and the largest such power is 𝑚: $f(z) = \frac{a_{-m}}{(z-z_0)^m} + \frac{a_{-m+1}}{(z-z_0)^{m-1}} + \cdots + \frac{a_{-1}}{(z-z_0)} + \sum_{n=0}^\infty a_n(z-z_0)^n$.
• Analytic multiplier. A pole of order m means you can factor out the singularity: $f(z) = \frac{\phi(z)}{(z-z_0)^m}$ where $\phi(z)$ is analytic (holomorphic) at $z_0$ and $\phi(z_0) \ne 0$. It basically states that the function behaves, acts like, or has the form $(z-z_0)^{-m}$, and everything else is innocuous. It also mirrors the definition of a zero of order m, but inverted.
• Limit characterization (asymptotic behavior). A pole of order m is the smallest positive integer such that the limit exists and is non-zero: $\lim_{z \to z_0}(z-z_0)^mf(z) = L, 0 \lt |L| \lt \infty$. This is the most analytic definition: it captures the precise blow‑up rate. It’s also the easiest way to compute the order of a pole in practice.
• Reciprocal viewpoint. A pole of order m at $z_0$ is equivalent to saying $\frac{1}{f(z)}$ has a zero of order m at $z_0$. This is the most conceptual definition: poles are just zeros of the reciprocal.

(iii): Essential Singularity.

f has an isolated essential singularity at a if and only if the series has infinitely many negative terms (it never stops), i.e., there is no m such that $c_n = 0, \forall n \lt -m$

We know there are only three types of isolated singularities: Removable, Poles, and Essential. We are going to prove it by elimination or exclusion:

• Could it be Removable? No, because Part (i) says removable singularities must have zero negative terms. We have infinitely many.
• Could it be a Pole? No, because Part (ii) says a pole must have a finite number of negative terms (it must stop at some integer $m$). We have infinitely many.
• Since it is not removable and it is not a pole, it must be an essential singularity.

Example: For $f(z) = e^{1/z}$, the Laurent expansion about z = 0 is $e^{1/z} = 1 + \frac{1}{z} + \frac{1}{2! z^2} + \frac{1}{3! z^3} + \dots$ which has infinitely many negative powers of z. That’s exactly the hallmark of an essential singularity at z = 0.

Essential Singularities: The “Chaotic” Isolated Singularities

Among isolated singularities, essential singularities are the most dramatic and extreme. While removable singularities behave tamely and poles blow up in a predictable algebraic way, an essential singularity produces behavior so erratic, wild, and unpredictable that the function becomes nearly uncontrollable in any punctured neighborhood.

An isolated singularity $z_0$ of a complex function f is essential if it meets any of the following equivalent conditions:

• Laurent Series Criterion. The Laurent series of 𝑓(𝑧) centered at $z_0$ contains infinitely many negative‑power terms (the “principal part”): $f(z) = \sum_{n=0}^\infty a_n(z-z_0)^n + \sum_{n=1}^\infty \frac{b_n}{(z-z_0)^n}$ where $b_n \ne 0$ for infinitely many n. This is the algebraic fingerprint of an essential singularity.
• Limit Behavior Criterion (No Limit, No Infinity). The limit $\lim_{z \to z_0}$ does not exist as a finite value, nor does the function tend toward infinity. Near $z_0$, the function’s value depends entirely on the direction of approach. Approaching $z_0$ along different paths yields different value.
• Classification by Exclusion. $z_0$ is an isolated singularity that is neither a removable singularity (finite limit exists) nor a pole (limit is ∞). This is often the quickest way to classify a singularity in practice.

The “wild” behavior of these points is governed by two major theorems:

• Casorati-Weierstrass Theorem: In every punctured neighborhood of an essential singularity, the image f(z) is dense in the complex plane. This means the function gets arbitrarily close to every complex number.
• Great Picard’s Theorem: In every punctured neighborhood of an essential singularity, the function takes every complex value infinitely often, with at most one exception (e.g., $e^{1/z}$ never equals 0, but hits every other complex number infinitely many times.).

Definition. Suppose that f is analytic in $\mathbb{B'}(a; r)$ and that f has a pole at a. The residue of f at a is the unique coefficient $c_1$ of the term $(z-a)^{-1}$ in the Laurent's series expansion of f around a and is denoted by Res{f; a}.

When we expand a function into its Laurent Series around a singularity a, we get a Principal Part (negative powers) and an Analytic Part (positive powers): $f(z) = \dots + \frac{c_{-2}}{(z-a)^2} + \mathbf{\frac{c_{-1}}{z-a}} + c_0 + c_1(z-a) + \dots$

The Residue is simply the coefficient of the $\frac{1}{z-a}$ term, $\text{Res}(f, a) = c_{-1}$. Why this specific term? As we have proved before, when we integrate these terms around a closed loop, the powers like $(z-a)^{-2}$ and $(z-a)^{4}$ all integrate to zero. The term $(z-a)^{-1}$ is the sole survivor, it integrates to $2\pi i$.

Cauchy’s Residue Theorem. If f is analytic inside and on a positively oriented closed contour $\gamma$ except possible for a finite number of singularities $a_1, a_2, \cdots, a_n$ inside $\gamma$, then the integral of f around $\gamma$ is simply $2\pi i$ times the sum of the residues at those points, $\int_{\gamma}f(z)dz = 2\pi i \sum_{k=1}^n Res\{ f; a_k \}$

Consider the contour $\gamma$ as a net. There are multiple “sources” of integration value (the singularities) within the net. According to the theorem, the total value captured by the net is simply the sum of the values produced by each separate source.

Proof.

We want to isolate the “Naughty Behavior” (The Principal Parts).

For each singularity $a_k$, the function $f(z)$ has a specific Principal Part (the negative powers in its Laurent series). Let’s call this singular part $f_k(z)$. There are two possibilities:

• If $a_k$ is a pole of order m: $f_k(z) = \frac{c_{-1}}{z-a_k} + \frac{c_{-2}}{(z-a_k)^2} + \dots + \frac{c_{-m}}{(z-a_k)^m}$
• If $a_k$ is an essential singularity (infinite terms): $f_k(z) = \sum_{j=1}^{\infty} \frac{c_{-j}}{(z-a_k)^j}$

Importantly, each function $f_k(z)$ is analytic everywhere except at its specific point $a_k$, e.g., $\frac{1}{z-2}$ explodes at 2, but is perfectly smooth at 3 or 7.

We define a new, “good” function by taking our original function f(z) and subtracting all the singular parts of all the points, $g(z):= f(z) - \sum_{k=1}^n f_k(z), \forall k = 1, 2, \cdots, n$.

We are identifying the “bad” parts of the function, subtract them out to create a “perfect” function.

Verify $g(z)$ is indeed a “good” (analytic) function.

1. Away from the singularities: f is analytic, and all $f_k$ are analytic (since we are not at any $a_k$), so $g$ is analytic.
2. At a specific singularity $a_k$: Near $a_k$, $f(z)$ looks like “$f_k(z) + \text{regular stuff}$”. When we subtract $f_k(z)$ from $f(z)$, the “bad part” cancels out exactly. The remaining limit exists and is finite, $\lim_{z \to a_k}g(z) = \lim_{z \to a_k}\text{regular stuff} = L_k, |L_k| \lt \infty$. Therefore, $g(z)$ has a Removable Singularity at every $a_k$.

If we fill in these removable holes, $g(z)$ is analytic everywhere inside $\gamma$.

Now, we can apply Cauchy’s Theorem to $g(z)$. Since $g(z)$ is analytic everywhere inside the closed loop $\gamma$, its integral is zero, $\int_{\gamma} g(z) dz = 0$.

The definition of g is then replaced in the previous expression: $\int_{\gamma} \left( f(z) - \sum_{k=1}^{n} f_k(z) \right) dz = 0$

Move the sum to the other side and consider we can swap the integral and sum because it is a finite sum: $\int_{\gamma} f(z) dz = \sum_{k=1}^{n} \int_{\gamma} f_k(z) dz$

We have already proved this lemma. 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.

Now we just need to calculate the integral of each Principal Part $f_k(z)$. $f_k(z) = \frac{c_{-1}}{z-a_k} + \frac{c_{-2}}{(z-a_k)^2} + \dots$ When we integrate this sum around $\gamma$:

• Terms like $\frac{1}{(z-a_k)^2}, \frac{1}{(z-a_k)^3}$ have valid antiderivatives on $\gamma$. Their integrals are 0.
• The term $\frac{c_{-1}}{z-a_k}$ does not have a global antiderivative. Its integral is $c_{-1} \cdot 2\pi i$.
• So, for each $k$ :$\int_{\gamma} f_k(z) dz = 2\pi i \cdot \text{Res}(f, a_k)$

Summing them all up gives the final result:

$\int_{\gamma} f(z)dz = \sum_{k=1}^n \int_{\gamma}f_k(z) dz = 2\pi i\sum_{k=1}^n Res\{ f; a_k \}$