Singularities: removable, pole, essential

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2017, 2021, 2023)
• Priority tier: T3
• Marks (count): 10 (1), 15 (2)
• Average solve time: ~9 min
• Difficulty mix: easy 2, medium 1
• Section: A | Dominant type: proof

Why This Chapter Matters

Singularity classification is a gateway concept for the entire residue theory. Questions appear in two flavours: (1) abstract proofs using Liouville’s theorem that exploit the connection between singularity type and growth, and (2) explicit classification of a given function together with its Laurent principal part. The 2017 question is a 4-step Liouville argument (easy once the pattern is seen); the 2021 question is a direct Laurent series argument (very fast); the harder 2023 question requires a careful power-series inversion to find the principal part of $e^z/(z-\sin z)$.

Minimum Theory

Three types of isolated singularity. Let $f$ be analytic on a punctured disc $0 < |z - z_0| < R$ with Laurent series $\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n$. The singularity at $z_0$ is:

• Removable if $a_n = 0$ for all $n < 0$ (no principal part). Equivalently, $f$ is bounded near $z_0$.
• Pole of order $m$ if $a_{-m} \ne 0$ and $a_n = 0$ for all $n < -m$ (principal part has exactly $m$ terms).
• Essential if infinitely many $a_n \ne 0$ for $n < 0$ (principal part has infinitely many terms).

Riemann removable singularity theorem. An isolated singularity $z_0$ of $f$ is removable if and only if $f$ is bounded on some punctured neighbourhood of $z_0$.

Liouville’s theorem. A bounded entire function is constant.

Key pattern: $f$ entire $\leftrightarrow$ $f(1/z)$ at $0$. If $f(z) = \sum_{n \ge 0} a_n z^n$ is entire, then $f(1/z) = \sum_{n \ge 0} a_n z^{-n}$, so the singularity of $f(1/z)$ at $z=0$ is removable if all $a_n = 0$ for $n \ge 1$ (i.e. $f$ is constant), a pole of order $m$ if $f$ is a polynomial of degree $m$, and essential if $f$ is a transcendental entire function (infinitely many nonzero Taylor coefficients).

Question Archetypes

singularity-classification (2 question(s); 2021, 2023)

Recognition Cues — The problem asks to classify an isolated singularity (removable / pole / essential) and/or to find the principal part of the Laurent series. The function is given explicitly. If it involves $f(1/z)$ where $f$ is entire, use the Taylor $\leftrightarrow$ Laurent correspondence directly. If it is a ratio like $e^z/(z - \sin z)$, expand the denominator in a Maclaurin series to find the order of its zero.

Solution Template

1. Find the order of the zero of the denominator (or the number of nonzero negative-power terms) at the singular point.
2. Classify: zero of order $m$ in denominator $\Rightarrow$ pole of order $m$; infinitely many negative powers $\Rightarrow$ essential; none $\Rightarrow$ removable.
3. To find the principal part: expand numerator and denominator as power series, invert using $1/(1-u) = 1 + u + u^2 + \cdots$, multiply, and collect all terms with negative powers.

Worked Example — 2021 (abstract)

2021 Paper 2, 2021-P2-Q3a (15 marks)

Let $f$ be entire with Taylor series at $z=0$ having infinitely many nonzero terms. Show that $z=0$ is an essential singularity of $f(1/z)$.

Setup. $f$ entire means $f(z) = \sum_{n=0}^{\infty} a_n z^n$ converges everywhere, with infinitely many $a_n \ne 0$.

Step 1 — Substitute. Let $g(z) = f(1/z)$. Substituting $z \to 1/z$ in the Taylor series:

$g(z) = f(1/z) = \sum_{n=0}^{\infty} a_n z^{-n} = a_0 + \frac{a_1}{z} + \frac{a_2}{z^2} + \cdots$

This is the Laurent series of $g$ centred at $z = 0$, valid for $z \ne 0$.

Step 2 — Classify. A singularity is essential when the principal part (negative-power terms) has infinitely many nonzero terms. Here the coefficient of $z^{-n}$ is $a_n$. Since infinitely many $a_n \ne 0$, the principal part contains infinitely many nonzero terms.

Therefore $z = 0$ is an essential singularity of $g(z) = f(1/z)$. $\blacksquare$

$\boxed{z=0 \text{ is an essential singularity of } f(1/z).}$

Example: $f(z) = e^z$ has $a_n = 1/n!$ for all $n \ge 0$, so $e^{1/z}$ has essential singularity at $0$ — the textbook illustration of Picard’s theorem.

Worked Example — 2023 (explicit principal part)

2023 Paper 2, 2023-P2-Q4b (15 marks)

Classify the singular point $z=0$ of $f(z) = \dfrac{e^z}{z - \sin z}$ and obtain the principal part of its Laurent series expansion.

Step 1 — Order of the zero of $z - \sin z$ at $0$. Using $\sin z = z - z^3/6 + z^5/120 - \cdots$:

$z - \sin z = \frac{z^3}{6} - \frac{z^5}{120} + \cdots = \frac{z^3}{6}\!\left(1 - \frac{z^2}{20} + \cdots\right).$

The leading term is $z^3/6$, so $z - \sin z$ has a zero of order 3 at $z = 0$.

Step 2 — Classify the singularity. Since $e^0 = 1 \ne 0$, the function $f(z) = e^z/(z - \sin z)$ has a pole of order 3 at $z = 0$.

Step 3 — Expand $1/(z - \sin z)$. Write $u(z) = z^2/20 - z^4/840 + \cdots$ so that

$\frac{1}{z - \sin z} = \frac{6}{z^3} \cdot \frac{1}{1 - u(z)} = \frac{6}{z^3}\!\left(1 + u + u^2 + O(u^3)\right).$

Retaining terms up to $z^4$:

$1 + u + u^2 = 1 + \frac{z^2}{20} + \left(-\frac{1}{840} + \frac{1}{400}\right)z^4 + O(z^6) = 1 + \frac{z^2}{20} + \frac{11z^4}{8400} + O(z^6).$

Therefore:

$\frac{1}{z - \sin z} = \frac{6}{z^3} + \frac{3}{10z} + \frac{11z}{1400} + O(z^3).$

Step 4 — Multiply by $e^z = 1 + z + z^2/2 + z^3/6 + \cdots$ and collect negative powers.

• Coefficient of $z^{-3}$: $\dfrac{6}{z^3} \cdot 1 = \dfrac{6}{z^3}$.
• Coefficient of $z^{-2}$: $\dfrac{6}{z^3} \cdot z = \dfrac{6}{z^2}$.
• Coefficient of $z^{-1}$: $\dfrac{6}{z^3} \cdot \dfrac{z^2}{2} + \dfrac{3}{10z} \cdot 1 = \dfrac{3}{z} + \dfrac{3}{10z} = \dfrac{33}{10z}$.

$\boxed{\text{Principal part} = \frac{6}{z^3} + \frac{6}{z^2} + \frac{33}{10z}.}$

The singularity is a pole of order 3, with residue $33/10$.

Common Traps

• $z - \sin z$ has a zero of order 3, not order 1. The $z$ terms cancel in $z - \sin z$; only the $z^3/6$ term survives as the leading term. Do not check $\sin 0 = 0$ and call it a simple zero.
• To get the $z^{-1}$ coefficient of the full product, you must keep the $z^2/2$ term of $e^z$ when multiplying against $6/z^3$. Missing this gives a wrong residue.
• The expansion $1/(1-u)$ requires the $u^2$ term to correctly evaluate the $z^4$ coefficient; skipping $u^2$ gives a wrong coefficient of $3/(10z)$ (though in this problem the $3/(10z)$ coefficient from $1/(z-\sin z)$ alone is correct).
• Verify: $\lim_{z \to 0} z^3 f(z) = 6 \cdot 1 / 1 = 6 \ne 0$, confirming order exactly 3.

liouville-application (1 question(s); 2017)

Recognition Cues — The problem says $f$ is entire and asks to characterise all such $f$ satisfying some singularity condition on $f(1/z)$. The condition “removable singularity at $0$ of $f(1/z)$” translates to ”$f$ is bounded at infinity”, which combined with entirety and Liouville forces $f$ to be constant.

Solution Template

1. Apply Riemann’s removable singularity theorem: removable $\Leftrightarrow$ $f(1/z)$ bounded near $0$ $\Leftrightarrow$ $f(w)$ bounded for $|w| > R$.
2. On the compact disc $|w| \le R$, $f$ is continuous hence bounded.
3. Combine: $f$ bounded on all of $\mathbb{C}$.
4. Invoke Liouville: bounded entire $\Rightarrow$ constant.
5. Verify the converse: constant functions trivially satisfy the condition.

Worked Example

2017 Paper 2, 2017-P2-Q1d (10 marks)

Determine all entire functions $f(z)$ such that $0$ is a removable singularity of $f\!\left(\dfrac{1}{z}\right)$.

Setup. Let $g(z) = f(1/z)$, which is analytic on $0 < |z| < \infty$ since $f$ is entire and $1/z$ is analytic for $z \ne 0$. We want $z = 0$ to be a removable singularity of $g$.

Step 1 — Removable $\Leftrightarrow$ bounded. By Riemann’s removable singularity theorem, $z=0$ is removable for $g$ if and only if $g$ is bounded on some punctured neighbourhood $0 < |z| < \delta$:

$\exists\, \delta, M > 0 \text{ such that } |f(1/z)| \le M \text{ for } 0 < |z| < \delta.$

Step 2 — Translate to behaviour at $\infty$. Substituting $w = 1/z$, as $z$ ranges over $0 < |z| < \delta$ the variable $w$ ranges over $|w| > 1/\delta =: R$. So the condition becomes

$|f(w)| \le M \quad \text{for all } |w| > R.$

Step 3 — $f$ is bounded on all of $\mathbb{C}$. Since $f$ is entire, it is continuous on the closed disc $\{|w| \le R\}$, which is compact. Continuous functions on compact sets are bounded, so $|f(w)| \le M'$ for $|w| \le R$. Combining:

$|f(w)| \le \max\{M, M'\} \quad \text{for all } w \in \mathbb{C}.$

Step 4 — Liouville’s theorem. A bounded entire function is constant. Therefore $f(z) = c$ for some constant $c \in \mathbb{C}$.

Converse. If $f \equiv c$, then $f(1/z) = c$ is constant on the punctured plane, so $z=0$ is trivially a removable singularity. Every constant satisfies the condition.

$\boxed{f \text{ is entire with } 0 \text{ removable for } f(1/z) \iff f \text{ is constant.}}$

$\blacksquare$

Laurent verification: Write $f(w) = \sum_{n \ge 0} a_n w^n$. Then $f(1/z) = \sum_{n \ge 0} a_n z^{-n}$. Singularity at $0$ is removable iff all $a_n = 0$ for $n \ge 1$, i.e. $f(w) = a_0$ is constant. Agrees exactly with the boxed answer.

Common Traps

• “Removable” is stronger than “isolated”. The correct criterion is boundedness (Riemann), not just that $\lim_{z\to 0} g(z)$ exists in some sense. Do not confuse removable with pole.
• After Step 2 you only know $f$ is bounded outside a disc. You still need the compactness argument on the closed disc $|w| \le R$ before invoking Liouville.
• The Laurent series check is the fastest sanity check: removable $\iff$ no negative-power terms $\iff$ all higher Taylor coefficients of $f$ vanish $\iff$ $f$ is constant.

Marks-Aware Writing

For a 10-mark question (2017 type): state Riemann’s theorem, translate boundedness to $\infty$, invoke compactness, invoke Liouville — four steps clearly labelled, 2–3 marks each. For a 15-mark question (2021, 2023): the 2021 proof needs explicit Laurent series setup and classification by definition; the 2023 question needs the full power-series computation with the denominator expansion shown step by step and the principal-part terms identified term by term. Writing the principal part in any form not broken into $z^{-3}, z^{-2}, z^{-1}$ terms will lose marks.

Practice Set

• 2024-P2-Q3a (15 m) — — Hint: identify each factor’s zero order at $z=0$; note that both $z$ and $\sin \pi z$ vanish there, giving a pole of order 2.
• 2013-P2-Q1d (10 m) — — Hint: standard Liouville or Laurent argument for singularity type.
• 2021-P2-Q8c (20 m) — — Hint: likely involves both classification and principal part; expand denominator carefully.
• 2019-P2-Q4b (10 m) — — Hint: check whether the singularity is removable or a pole by the limit test.
• 2018-P2-Q4b (15 m) — — Hint: combine singularity classification with residue computation.