Residues: computation at poles of various orders

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2019, 2024, 2025)
• Priority tier: T3
• Marks (count): 15 (3)
• Average solve time: ~13 min
• Difficulty mix: medium 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Residue computation is the workhorse of complex integration. All three questions are 15-mark problems in Section A, and each tests a different facet: 2019 asks for the theoretical proof that pole order $m$ is equivalent to the $\phi/(z-z_0)^m$ form and derives the residue formula; 2024 requires locating poles (including a subtle order-2 pole at $z=0$ where two denominator factors vanish simultaneously), determining orders, and computing residues; 2025 requires evaluating a contour integral by finding residues at both a second-order and a third-order pole using the derivative formula. Mastering the $\frac{1}{(m-1)!} \frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$ formula for arbitrary $m$ covers all three.

Minimum Theory

Residue at an isolated singularity. The residue of $f$ at an isolated singularity $z_0$ is the coefficient $a_{-1}$ in the Laurent expansion $f(z) = \sum a_n (z-z_0)^n$.

Residue formula for a pole of order $m$. If $z_0$ is a pole of order $m \ge 1$:

$\operatorname{Res}_{z=z_0} f = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}}\!\left[(z-z_0)^m f(z)\right].$

For a simple pole ($m=1$): $\operatorname{Res} = \lim_{z \to z_0}(z-z_0)f(z)$. For a simple zero of the denominator $q$ when numerator $p(z_0) \ne 0$: $\operatorname{Res} = p(z_0)/q'(z_0)$.

Residue theorem. If $f$ is analytic inside and on a simple closed positively-oriented contour $C$ except at finitely many poles $z_1, \ldots, z_k$ inside $C$:

$\oint_C f(z)\,dz = 2\pi i \sum_{j=1}^k \operatorname{Res}_{z=z_j} f.$

Pole-order characterisation. An isolated singularity $z_0$ is a pole of order $m$ if and only if $f(z) = \phi(z)/(z-z_0)^m$ where $\phi$ is analytic at $z_0$ and $\phi(z_0) \ne 0$. The residue is then $\phi^{(m-1)}(z_0)/(m-1)!$.

Question Archetypes

residue-theory-proof (1 question(s); 2019)

Recognition Cues — The problem says “show that” or “prove” an equivalence between a pole of order $m$ and the $\phi/(z-z_0)^m$ representation. Expect both directions ($\Rightarrow$ and $\Leftarrow$) and a separate derivation of the residue formula. This is a pure proof question, not a computation.

Solution Template

1. State the definition: $z_0$ is a pole of order $m$ iff Laurent series starts at $(z-z_0)^{-m}$ with nonzero coefficient.
2. ($\Rightarrow$) Multiply the Laurent series by $(z-z_0)^m$ to obtain $\phi$; show $\phi$ extends analytically to $z_0$ and $\phi(z_0) = a_{-m} \ne 0$.
3. ($\Leftarrow$) Taylor-expand $\phi$ at $z_0$, divide by $(z-z_0)^m$; the lowest power is $-m$ with coefficient $\phi(z_0) \ne 0$, confirming a pole of order exactly $m$.
4. Read off $a_{-1}$ (the residue) as the Taylor coefficient at $k = m-1$: $a_{-1} = \phi^{(m-1)}(z_0)/(m-1)!$.

Worked Example

2019 Paper 2, 2019-P2-Q2d (15 marks)

Show that an isolated singular point $z_0$ of $f(z)$ is a pole of order $m$ if and only if $f(z) = \dfrac{\phi(z)}{(z-z_0)^m}$ where $\phi(z)$ is analytic and non-zero at $z_0$. Moreover $\operatorname{Res}_{z=z_0} f(z) = \dfrac{\phi^{(m-1)}(z_0)}{(m-1)!}$ if $m \ge 1$.

Let $z_0$ be an isolated singularity of $f$, with Laurent expansion on $0 < |z-z_0| < R$:

$f(z) = \sum_{n=-\infty}^{\infty} a_n (z-z_0)^n.$

By definition, $z_0$ is a pole of order $m$ ($m \ge 1$) iff $a_{-m} \ne 0$ and $a_n = 0$ for all $n < -m$.

Step 1 — ($\Rightarrow$) Pole of order $m$ implies the form $\phi/(z-z_0)^m$.

If $z_0$ is a pole of order $m$, the above Laurent series starts at $n = -m$. Multiply by $(z-z_0)^m$:

$\phi(z) := (z-z_0)^m f(z) = \sum_{k \ge 0} a_{k-m}(z-z_0)^k.$

This is a convergent power series on $0 < |z-z_0| < R$, hence defines an analytic function on the full disc $|z-z_0| < R$ (the singularity is removed), with $\phi(z_0) = a_{-m} \ne 0$. Thus $f(z) = \phi(z)/(z-z_0)^m$ with $\phi$ analytic at $z_0$ and $\phi(z_0) \ne 0$.

Step 2 — ($\Leftarrow$) The form $\phi/(z-z_0)^m$ implies a pole of order $m$.

Suppose $f(z) = \phi(z)/(z-z_0)^m$ with $\phi$ analytic at $z_0$ and $\phi(z_0) \ne 0$. Taylor-expand $\phi$:

$\phi(z) = \sum_{k \ge 0} \frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^k.$

Then:

$f(z) = \sum_{k \ge 0} \frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^{k-m}.$

The lowest power is $(z-z_0)^{-m}$ with coefficient $\phi(z_0) \ne 0$, and there are no powers below $-m$. So the principal part has exactly $m$ nonzero terms: $z_0$ is a pole of order exactly $m$. $\blacksquare$

Step 3 — Residue formula.

The residue is $a_{-1}$, i.e. the coefficient of $(z-z_0)^{-1}$ in the last display. This occurs at $k - m = -1$, i.e. $k = m-1$:

$\operatorname{Res}_{z=z_0} f = a_{-1} = \frac{\phi^{(m-1)}(z_0)}{(m-1)!}.$

Since $\phi = (z-z_0)^m f$, this is equivalently $\dfrac{1}{(m-1)!}\lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}}\!\left[(z-z_0)^m f(z)\right]$.

$\boxed{z_0 \text{ pole of order } m \iff f = \frac{\phi(z)}{(z-z_0)^m},\ \phi \text{ analytic},\ \phi(z_0) \ne 0;\quad \operatorname{Res}_{z=z_0} f = \frac{\phi^{(m-1)}(z_0)}{(m-1)!}.}$

Common Traps

• Both directions require $\phi(z_0) \ne 0$. If $\phi(z_0) = 0$, the actual pole order is less than $m$ and the characterisation fails.
• In Step 1, the removed-singularity argument — that the power series for $\phi$ converges on the full disc, making $\phi$ analytic at $z_0$ — must be stated explicitly.
• The residue is at $k = m-1$, giving $(m-1)!$ in the denominator. Off-by-one (writing $m!$ or $(m-2)!$) is the classic slip. For $m=1$: $\phi^{(0)}(z_0)/0! = \phi(z_0)$, i.e. $\lim_{z \to z_0}(z-z_0)f(z)$. Check this agrees with the simple-pole formula.

poles-and-residues (1 question(s); 2024)

Recognition Cues — “Locate the poles and their order; find the residue.” The function has several factors in the denominator, each potentially vanishing at different points. The key challenge is counting multiplicity correctly when two factors vanish at the same point.

Solution Template

1. Identify all zeros of the denominator by setting each factor to zero.
2. At each singularity, determine whether it is a zero of the denominator of multiplicity $\ge 2$ (from coinciding zeros of different factors).
3. Compute the residue using the formula appropriate to the pole order: limit formula for simple poles; derivative formula for order 2.
4. For simple poles where the denominator has a simple zero: use $\operatorname{Res} = \text{numerator}/(\text{denominator}')$ evaluated at the pole.

Worked Example

2024 Paper 2, 2024-P2-Q3a (15 marks)

Locate the poles and their order for the function $f(z) = \dfrac{1}{z(\sin\pi z)\bigl(z+\tfrac{1}{2}\bigr)}$ and find the residue of $f(z)$ at these poles.

Step 1 — Identify singularities. The denominator vanishes when:

• $z = 0$ (from the $z$ factor);
• $z = -1/2$ (from the $z + 1/2$ factor);
• $\sin \pi z = 0$, i.e. $z = n$ for any integer $n$ (from the $\sin \pi z$ factor).

At $z = 0$: both $z$ and $\sin \pi z$ vanish (each with a simple zero), so the pole is order 2. At $z = -1/2$: only $z + 1/2$ vanishes (order 1), and $\sin\pi(-1/2) = -1 \ne 0$ and $-1/2 \ne 0$. Simple pole. At $z = n$, $n \ne 0$ integer: only $\sin\pi z$ vanishes (order 1), and $n \ne 0$, $n + 1/2 \ne 0$. Simple pole.

Step 2 — Residue at $z = 0$ (order 2). Using $\operatorname{Res}_{z=0} f = \lim_{z\to 0} \frac{d}{dz}[z^2 f(z)]$:

$z^2 f(z) = \frac{z}{(\sin\pi z)(z+1/2)}.$

Near $z = 0$: $\sin\pi z = \pi z(1 - \pi^2 z^2/6 + \cdots)$, so $z/\sin\pi z = (1/\pi)(1 + \pi^2 z^2/6 + \cdots)$. Thus

$z^2 f(z) = \frac{1}{\pi(z+1/2)}\!\left(1 + \frac{\pi^2 z^2}{6} + \cdots\right).$

Differentiating at $z = 0$ (the $O(z^2)$ correction contributes an $O(z)$ term that vanishes at 0):

$\operatorname{Res}_{z=0} f = \frac{d}{dz}\!\left[\frac{1}{\pi(z+1/2)}\right]_{z=0} = \left.\frac{-1}{\pi(z+1/2)^2}\right|_{z=0} = -\frac{1}{\pi \cdot 1/4} = -\frac{4}{\pi}.$

Step 3 — Residue at $z = -1/2$ (simple pole).

$\operatorname{Res}_{z=-1/2} f = \lim_{z \to -1/2} (z+\tfrac12) f(z) = \frac{1}{(-1/2)\sin(-\pi/2)} = \frac{1}{(-1/2)(-1)} = 2.$

Step 4 — Residue at $z = n$, $n \in \mathbb{Z} \setminus \{0\}$ (simple pole). Writing the denominator as $(\sin\pi z) \cdot [z(z+1/2)]$ and applying the simple-zero formula:

$\operatorname{Res}_{z=n} f = \frac{1/[n(n+1/2)]}{(\sin\pi z)'\big|_{z=n}} = \frac{1}{n(n+1/2) \cdot \pi(-1)^n} = \frac{(-1)^n \cdot 2}{\pi n(2n+1)}.$

$\boxed{\operatorname{Res}_{z=0} = -\frac{4}{\pi}, \quad \operatorname{Res}_{z=-1/2} = 2, \quad \operatorname{Res}_{z=n} = \frac{(-1)^n \cdot 2}{\pi n(2n+1)}, \; n \in \mathbb{Z}\setminus\{0\}.}$

Common Traps

• At $z=0$ both $z$ and $\sin\pi z$ vanish; the pole is order 2, not 1. Checking only that $f \to \infty$ as $z \to 0$ without counting multiplicity gives the wrong order.
• $\sin(-\pi/2) = -1$, not $+1$. The sign error in the residue at $z = -1/2$ is very common.
• The simple-pole formula $\operatorname{Res} = p(z_0)/q'(z_0)$ applies when $q(z_0)=0$ and $q'(z_0) \ne 0$; for the $\sin\pi z$ zeros, $q'(n) = \pi\cos(\pi n) = \pi(-1)^n$.

residue-evaluation (1 question(s); 2025)

Recognition Cues — “Evaluate $\oint_C f(z)\,dz$” where $f$ has explicit higher-order poles and the contour is a circle of radius large enough to enclose both. The method is: identify poles and their orders, verify which are inside $C$, compute residues using the derivative formula, sum and multiply by $2\pi i$.

Solution Template

1. Factorise the denominator and list the poles with their orders.
2. Check which poles lie strictly inside the contour $C$.
3. For each enclosed pole of order $m$: compute $\frac{1}{(m-1)!}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$ evaluated at $z = z_0$.
4. Sum residues and multiply by $2\pi i$.

Worked Example

2025 Paper 2, 2025-P2-Q3a (15 marks)

Evaluate the integral $\displaystyle\oint_C \dfrac{e^z}{z^2(z+1)^3}\,dz$, $C : |z| = 2$.

Step 1 — Poles. The denominator $z^2(z+1)^3$ gives:

• $z = 0$, order 2;
• $z = -1$, order 3.

Both $|0| = 0 < 2$ and $|-1| = 1 < 2$, so both lie inside $C$.

By the residue theorem:

$\oint_C \frac{e^z}{z^2(z+1)^3}\,dz = 2\pi i\!\left[\operatorname{Res}_{z=0} + \operatorname{Res}_{z=-1}\right].$

Step 2 — Residue at $z = 0$ (order 2).

$\operatorname{Res}_{z=0} = \frac{1}{1!}\lim_{z \to 0} \frac{d}{dz}\!\left[\frac{e^z}{(z+1)^3}\right].$

Using the quotient rule:

$\frac{d}{dz}\frac{e^z}{(z+1)^3} = \frac{e^z(z+1)^3 - 3e^z(z+1)^2}{(z+1)^6} = \frac{e^z(z-2)}{(z+1)^4}.$

At $z = 0$: $e^0(0-2)/1^4 = -2$.

$\operatorname{Res}_{z=0} = -2.$

Step 3 — Residue at $z = -1$ (order 3).

$\operatorname{Res}_{z=-1} = \frac{1}{2!}\lim_{z \to -1} \frac{d^2}{dz^2}\!\left[\frac{e^z}{z^2}\right].$

Let $h(z) = e^z z^{-2}$. Differentiate twice:

$h'(z) = e^z z^{-2} - 2e^z z^{-3} = e^z\!\left(\frac{1}{z^2} - \frac{2}{z^3}\right).$

$h''(z) = e^z\!\left(\frac{1}{z^2} - \frac{2}{z^3}\right) + e^z\!\left(-\frac{2}{z^3} + \frac{6}{z^4}\right) = e^z\!\left(\frac{1}{z^2} - \frac{4}{z^3} + \frac{6}{z^4}\right).$

At $z = -1$: $e^{-1}(1 - 4/(-1) + 6/1) = e^{-1}(1 + 4 + 6) = 11e^{-1}$.

$\operatorname{Res}_{z=-1} = \frac{1}{2} \cdot 11e^{-1} = \frac{11}{2e}.$

Step 4 — Combine.

$\oint_C \frac{e^z}{z^2(z+1)^3}\,dz = 2\pi i\!\left(-2 + \frac{11}{2e}\right) = 2\pi i \cdot \frac{11 - 4e}{2e} = \frac{\pi i(11 - 4e)}{e}.$

$\boxed{\oint_{|z|=2} \frac{e^z}{z^2(z+1)^3}\,dz = \frac{\pi i(11 - 4e)}{e} = \pi i\!\left(\frac{11}{e} - 4\right) \approx 0.147\,i.}$

Common Traps

• The order at $z = 0$ is 2 (from $z^2$) and at $z = -1$ is 3 (from $(z+1)^3$). Using $m=1$ for either gives the wrong derivative formula.
• For the order-3 residue, the second derivative formula requires differentiating $h(z) = e^z/z^2$ twice; forgetting the $6/z^4$ term from the second differentiation gives a wrong result.
• Both poles satisfy $|z_0| < 2$ and are inside $C$; missing one (e.g. computing only the residue at $z=0$) loses half the marks.
• The factorial in the denominator: order 3 residue uses $(3-1)! = 2! = 2$, so the formula has a factor of $1/2$. Writing $1/3!$ or $1/1!$ is wrong.

Marks-Aware Writing

All three questions are 15 marks. The examiner expects: (a) explicit identification of each pole and its order with justification, (b) the residue formula written out before substitution, (c) all differentiation steps shown (quotient rule or product rule), and (d) arithmetic shown cleanly. For the proof question (2019), both directions of the iff must be clearly separated and labelled; omitting either direction loses about 5 marks. For the evaluation questions (2024, 2025), verify that poles are inside the contour before applying the residue theorem.

Practice Set

• 2023-P2-Q2c (20 m) — — Hint: likely a contour integral needing several residues; check all poles against the contour radius.
• 2014-P2-Q3c (20 m) — — Hint: use residues; watch for coinciding zeros in the denominator.
• 2015-P2-Q3a (15 m) — — Hint: identify pole orders and apply the derivative formula.
• 2022-P2-Q3a (15 m) — — Hint: standard residue evaluation; verify which poles lie inside the given contour.
• 2021-P2-Q1d (10 m) — — Hint: compulsory Q1 residue; likely a simple or order-2 pole.
• 2016-P2-Q3c (15 m) — — Hint: residue computation with trigonometric denominator.
• 2025-P2-Q2c (20 m) — — Hint: extended residue calculation; keep track of all poles and their orders.