Residues and poles: Worked examples — Maths deep dive

A companion to Residues and poles. Here we don't learn new rules — we drill the ones you already met until no case can surprise you. Every trap the topic can spring, we spring it on purpose and disarm it.

Before we start: two reminders, then the four tools.

Recall What a Laurent series is (cover and recall)

Near an isolated singularity $z_{0}$ every function can be written as a two-sided power series $f (z) = \sum_{n = - \infty}^{\infty} a_{n} (z - z_{0})^{n} = principal (negative) part \dots + \frac{a _{- 2}}{( z - z _{0} ) ^{2}} + \frac{a _{- 1}}{z - z _{0}} + ordinary Taylor part a_{0} + a_{1} (z - z_{0}) + \dots .$ The negative-power terms describe how it blows up; the residue is just one coefficient of this series, namely $a_{- 1}$ . Full construction in Laurent series.

Recall The four residue tools (cover and recall)

Simple pole: $Res (f, z_{0}) = lim_{z \to z_{0}} (z - z_{0}) f (z)$ — the $(z - z_{0})$ factor "cancels the blow-up", then you plug in.
Quotient shortcut (simple pole of $p / q$ ): $\frac{p ( z _{0} )}{q ^{'} ( z _{0} )}$ when $q (z_{0}) = 0, q^{'} (z_{0}) \neq = 0, p (z_{0}) \neq = 0$ .
Order- $m$ pole: $\frac{1}{( m - 1 )!} lim_{z \to z_{0}} \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - z_{0})^{m} f (z)]$ .
Residue theorem: for a positively-oriented (counter-clockwise) closed contour $C$ , $\oint_{C} f d z = 2 π i \sum Res (f, z_{j})$ , poles inside $C$ only. (Reverse the loop → multiply by $- 1$ .)

Two words we lean on. A ==pole of order $m$ == means the worst term in the Laurent series is $a_{- m} (z - z_{0})^{- m}$ with $a_{- m} \neq = 0$ and no worse; order $1$ is a simple pole. A contour $C$ is just a closed loop you integrate around (see Contour integration); "inside" means the region the loop encircles, and unless stated we always travel it counter-clockwise — that sign convention is baked into the $+ 2 π i$ .

The scenario matrix

Every residue problem you will ever meet lands in one of these cells. The examples below are labelled by cell so you can see the whole board is covered.

Cell	What makes it different	Which example
A. Simple pole, cancel-and-plug	one $(z - z_{0})$ factor, plug in	Ex 1
B. Simple pole, quotient shortcut	$p / q$ where factoring is ugly	Ex 2
C. Higher-order pole (derivative)	$m \geq 2$ , must differentiate $m - 1$ times	Ex 3
D. Pole hidden by a zero of numerator	order looks high, cancels down	Ex 4
E. Degenerate: NOT a pole (removable / essential)	formulas silently fail	Ex 5
F. Sign/location: which poles are inside $C$ ?	some poles enclosed, some not	Ex 6
G. Real integral, upper half plane	close the contour, discard arc	Ex 7
H. Word problem (real-world model)	translate then compute	Ex 8
I. Exam twist: an infinite family of trig poles	$sin z = 0$ at every $nπ$	Ex 9

The tricky columns are D, E, F — that is where students lose marks. We hit each head-on.

Cell A — Simple pole, cancel and plug

Worked example Ex 1 (Cell A)

Find $Res (\frac{z + 4}{( z - 2 ) ( z + 3 )}, 2)$ .

Forecast: guess a single number before reading on. (Numerator at $z = 2$ is $6$ ; the other factor at $z = 2$ is $5$ …)

Confirm the pole. At $z = 2$ the factor $(z - 2)$ vanishes; the rest is finite and non-zero → simple pole. Why this step? You must know the order before choosing a formula. One vanishing linear factor = order 1.
Multiply by $(z - 2)$ : $(z - 2) f = \frac{z + 4}{z + 3}$ . Why this step? The $(z - 2)$ factor cancels the blow-up, leaving something finite at $z = 2$ .
Let $z \to 2$ : $\frac{2 + 4}{2 + 3} = \frac{6}{5}$ . Why this step? With the pole cancelled, "the limit" is just substitution.

$Res = \frac{6}{5}$

Verify: independently compute the residue at the other pole $z = - 3$ : $\frac{- 3 + 4}{- 3 - 2} = - \frac{1}{5}$ . As a cross-check we do partial fractions: $\frac{z + 4}{( z - 2 ) ( z + 3 )} = \frac{6/5}{z - 2} + \frac{- 1/5}{z + 3}$ . The coefficients of $\frac{1}{z - 2}$ and $\frac{1}{z + 3}$ read off directly are exactly the two residues $\frac{6}{5}$ and $- \frac{1}{5}$ , confirming our answer. ✔

Cell B — Quotient shortcut

Worked example Ex 2 (Cell B)

Find $Res (\frac{cos z}{z ^{2} + 9}, 3 i)$ .

Forecast: the denominator factors as $(z - 3 i) (z + 3 i)$ , so this is simple. Will the answer be real or imaginary?

Identify $p = cos z$ , $q = z^{2} + 9$ . Check $q (3 i) = (3 i)^{2} + 9 = - 9 + 9 = 0$ . Good, it is a pole. Why this step? The shortcut $p / q^{'}$ needs $q (z_{0}) = 0$ — verify it, never assume.
$q^{'} (z) = 2 z$ , so $q^{'} (3 i) = 6 i \neq = 0$ ; and $p (3 i) = cos (3 i) \neq = 0$ . Why this step? Both conditions ( $q^{'} \neq = 0$ , $p \neq = 0$ ) guarantee a simple pole, licensing the shortcut.
Apply: $Res = \frac{p ( 3 i )}{q ^{'} ( 3 i )} = \frac{cos ( 3 i )}{6 i}$ . Why this step? Faster than factoring $cos z$ ; that's the whole point of the shortcut.
Simplify with $cos (3 i) = cosh 3$ : $Res = \frac{cosh 3}{6 i} = - \frac{i cosh 3}{6}$ . Why this step? $cos (i x) = cosh x$ turns an "imaginary-input cosine" into a real number.

Verify: numerically $cosh 3 \approx 10.0677$ , so $Res \approx - 1.678 i$ — purely imaginary, as expected for a real-even integrand's complex pole. ✔

Cell C — Higher-order pole (the derivative formula)

Worked example Ex 3 (Cell C)

Find $Res (\frac{e ^{z}}{( z - 1 ) ^{3}}, 1)$ .

Forecast: order $m = 3$ , so we differentiate $m - 1 = 2$ times. Guess: something built from $e^{1}$ .

Multiply by $(z - 1)^{3}$ : $(z - 1)^{3} f = e^{z}$ . Why this step? Multiplying by $(z - z_{0})^{m}$ kills every negative power, leaving a clean Taylor series.
Differentiate $m - 1 = 2$ times: $\frac{d ^{2}}{d z ^{2}} e^{z} = e^{z}$ . Why this step? In $a_{- 3} (z - 1)^{- 3} + a_{- 2} (z - 1)^{- 2} + a_{- 1} (z - 1)^{- 1} + \dots$ , multiplying by $(z - 1)^{3}$ pushes $a_{- 1}$ to the $(z - 1)^{2}$ slot; two derivatives + $2!$ extract it.
Evaluate at $z = 1$ and divide by $(m - 1)! = 2!$ : $Res = \frac{1}{2 !} e^{1} = \frac{e}{2}$ . Why this step? The factorial undoes the $\frac{1}{2 !}$ that Taylor coefficients carry.

$Res = \frac{e}{2}$

Verify: Laurent-expand directly: $e^{z} = e \cdot e^{z - 1} = e \sum \frac{( z - 1 ) ^{n}}{n !}$ , so $\frac{e ^{z}}{( z - 1 ) ^{3}} = e \sum \frac{( z - 1 ) ^{n - 3}}{n !}$ . The $(z - 1)^{- 1}$ term is $n = 2$ : coefficient $e /2! = e /2$ . ✔

Cell D — A zero of the numerator lowers the order

This is the sneaky one. The denominator looks order 2, but the numerator vanishes there and cancels a factor.

Cell E — Degenerate inputs: when it's NOT a pole

The formulas do not warn you; you must recognise removable and essential singularities yourself.

Worked example Ex 5 (Cell E — removable AND essential)

(a) $Res (\frac{sin z}{z}, 0)$ . (b) $Res (e^{1/ z}, 0)$ .

Forecast: one of these is $0$ , one is $1$ . Which is which?

Part (a) — removable:

Expand: $sin z = z - \frac{z ^{3}}{6} + \dots$ , so $\frac{sin z}{z} = 1 - \frac{z ^{2}}{6} + \dots$ . Why this step? The Laurent series is the ground truth. No negative powers appear.
Read $a_{- 1} = 0$ . There is no $1/ z$ term. Why this step? Residue is defined as $a_{- 1}$ ; here it's plainly $0$ .

$Res = 0$

Part (b) — essential:

Expand: $e^{1/ z} = \sum_{n = 0}^{\infty} \frac{1}{n !} z^{- n} = 1 + \frac{1}{z} + \frac{1}{2 ! z ^{2}} + \dots$ . Why this step? Infinitely many negative powers → essential; no finite $m$ , so pole formulas cannot be used ( $z^{m} e^{1/ z}$ never becomes analytic).
The $z^{- 1}$ coefficient is $n = 1$ : $a_{- 1} = \frac{1}{1 !} = 1$ . Why this step? For essential singularities the only method is reading $a_{- 1}$ off the series.

$Res = 1$

Verify: (a) $lim_{z \to 0} z \cdot \frac{s i n z}{z} = lim sin z = 0$ — consistent with a removable point (residue $0$ ). (b) apply the residue theorem to the same series: $\oint_{∣ z ∣ = 1} e^{1/ z} d z$ equals $2 π i$ times the coefficient $a_{- 1} = 1$ , because every other power $z^{- n}$ (for $n \neq = 1$ ) integrates to $0$ around the circle. So the loop integral is $2 π i \cdot 1 = 2 π i$ , exactly matching $a_{- 1} = 1$ . ✔

Cell F — Which poles are inside the contour?

The residue theorem sums enclosed poles only. Location (sign of imaginary part, distance from centre) decides membership.

Worked example Ex 6 (Cell F)

Compute $\oint_{∣ z ∣ = 2} \frac{z}{( z - 1 ) ( z - 3 )} d z$ (counter-clockwise).

Forecast: poles at $z = 1$ and $z = 3$ . Circle has radius $2$ . Which lie inside?

The figure below shows the contour and the two poles, so you can see membership rather than just assert it. Read it left-to-right:

the cyan circle is the loop $∣ z ∣ = 2$ ; the small cyan arrow on it points anticlockwise — that anticlockwise direction is exactly what makes the theorem's sign $+ 2 π i$ ;
the filled amber dot at $z = 1$ sits inside the circle → its residue is counted;
the hollow white dot at $z = 3$ sits outside the circle → it is ignored.

Locate poles: $z = 1$ (distance $1$ from origin) and $z = 3$ (distance $3$ ). Why this step? "Inside $∣ z ∣ = 2$ " means $∣ z_{0} ∣ < 2$ . So $z = 1$ is in (the amber dot in the figure), $z = 3$ is out (the hollow white dot outside the cyan circle).
Residue at the enclosed pole $z = 1$ : $lim_{z \to 1} (z - 1) f = \frac{1}{1 - 3} = - \frac{1}{2}$ . Why this step? Only enclosed poles contribute; ignore $z = 3$ entirely.
Apply the theorem (contour is counter-clockwise, so use $+ 2 π i$ ): $\oint = 2 π i (- \frac{1}{2}) = - π i$ . Why this step? $\oint_{C} f = 2 π i \sum_{inside} Res$ for a positively-oriented loop.

$\oint_{∣ z ∣ = 2} f d z = - π i$

Verify: widen the contour to $∣ z ∣ = 4$ (both poles inside) and add $Res (f, 3) = \frac{3}{3 - 1} = \frac{3}{2}$ : total $2 π i (- \frac{1}{2} + \frac{3}{2}) = 2 π i$ . That larger loop should give a different value than the radius-2 loop, and it does — confirming that at radius $2$ only $z = 1$ counts, giving $- π i$ . ✔

Cell G — A real integral by closing the contour

Worked example Ex 7 (Cell G)

Compute $\int_{- \infty}^{\infty} \frac{d x}{x ^{4} + 1}$ .

Forecast: all four poles sit on the unit circle at angles $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ . Only the top two count. Guess the answer is a clean multiple of $π$ .

The figure shows why we may throw the curved part away. Read it as a story:

the flat cyan segment along the real axis is the integral we actually want, traversed left-to-right;
the cyan semicircular arc of radius $R$ closes the loop through the upper half plane (anticlockwise, matching $+ 2 π i$ );
the two filled amber dots ( $z = e^{iπ /4}, e^{i 3 π /4}$ ) are the enclosed poles; the two hollow white dots below the axis are skipped;
the dashed white unit circle is only a guide showing all four poles have modulus $1$ .

Find the poles: $z^{4} = - 1 = e^{iπ}$ , so $z_{k} = e^{i (π + 2 π k) /4}$ , i.e. angles $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ . Why this step? Solving $z^{4} + 1 = 0$ places every pole; the two with positive imaginary part (the amber dots, upper half) are enclosed by the arc.
Show the arc contribution vanishes as $R \to \infty$ . On the arc $∣ z ∣ = R$ , so by the triangle inequality $∣ z^{4} + 1∣ \geq ∣ z^{4} ∣ - 1 = R^{4} - 1$ , hence the integrand obeys $\frac{1}{z ^{4} + 1} \leq \frac{1}{R ^{4} - 1}$ . The arc is a half-circle of radius $R$ , whose length is $\frac{1}{2} (2 π R) = π R$ . Why this step? We must prove the curved piece disappears, or "close the contour" is unjustified. $R^{4} - 1$ comes from the reverse triangle inequality $∣ a + b ∣ \geq ∣ a ∣ - ∣ b ∣$ ; the length $π R$ is just half a circle's circumference $2 π R$ .
Bound the arc integral by (max of integrand) $\times$ (length): $\int_{arc} \leq \frac{1}{R ^{4} - 1} \cdot π R = \frac{π R}{R ^{4} - 1} \to 0$ as $R \to \infty$ (numerator grows like $R$ , denominator like $R^{4}$ ). Why this step? This is the ML-inequality: an integral is at most its integrand's peak size times the path length. Since it tends to $0$ , the closed loop equals the real-line integral alone.
Use the quotient shortcut $Res = \frac{p}{q ^{'}} = \frac{1}{4 z ^{3}}$ at a pole. Since $z^{4} = - 1$ there, $z^{3} = z^{4} / z = - 1/ z$ , so at each pole $Res = \frac{1}{4 z ^{3}} = \frac{1}{4 ( - 1/ z )} = - \frac{z}{4}$ . Why this step? $p = 1, q = z^{4} + 1, q^{'} = 4 z^{3}$ ; rewriting $z^{3} = - 1/ z$ turns the cube into a clean linear expression $- z /4$ evaluated at each enclosed pole.
Evaluate at the two enclosed poles $z_{1} = e^{iπ /4}, z_{2} = e^{i 3 π /4}$ : $- \frac{1}{4} (z_{1} + z_{2}) = - \frac{1}{4} (\frac{2}{2} (1 + i) + \frac{2}{2} (- 1 + i)) = - \frac{1}{4} (2 i) = - \frac{2}{4} i$ . Why this step? The residue theorem adds the residues of enclosed poles; we plug each into $- z /4$ and sum.
Integral $= 2 π i \cdot (- \frac{2}{4} i) = \frac{π 2}{2} = \frac{π}{2}$ . Why this step? $2 π i \times (sum of enclosed residues)$ ; the two $i$ 's multiply to $- 1$ , flipping the sign to positive.

$\int_{- \infty}^{\infty} \frac{d x}{x ^{4} + 1} = \frac{π}{2} \approx 2.2214$

Verify: the integrand is positive and even; a numeric estimate gives $\approx 2.221$ , matching $π / 2$ . ✔

Cell H — Word problem (a real-world model)

Worked example Ex 8 (Cell H)

A signal's frequency response is modelled by $H (x) = \frac{1}{x ^{2} + a ^{2}}$ ( $a > 0$ , a bandwidth). Its total "energy weight" is $E = \int_{- \infty}^{\infty} \frac{d x}{x ^{2} + a ^{2}}$ . Find $E$ in terms of $a$ , then evaluate for $a = 2$ .

Forecast: wider bandwidth $a$ should give a smaller peak but the area might stay simple. Guess $E \propto 1/ a$ .

Poles of $\frac{1}{x ^{2} + a ^{2}}$ : solve $z^{2} + a^{2} = 0 \Rightarrow z = \pm ia$ . Closing in the upper half plane encloses $z = ia$ only. Why this step? Same closing argument as Ex 7. On the arc $∣ z^{2} + a^{2} ∣ \geq R^{2} - a^{2}$ , so the integrand $\leq \frac{1}{R ^{2} - a ^{2}}$ ; times arc length $π R$ gives $\frac{π R}{R ^{2} - a ^{2}} \to 0$ . So the real-line integral equals $2 π i \times$ (upper-half residues).
Residue at $z = ia$ (quotient shortcut, $p = 1$ , $q = z^{2} + a^{2}$ , $q^{'} = 2 z$ ): $\frac{1}{2 ( ia )} = \frac{1}{2 ia}$ . Why this step? Simple pole ( $q^{'} (ia) = 2 ia \neq = 0$ ), so $p / q^{'}$ is instant.
$E = 2 π i \cdot \frac{1}{2 ia} = \frac{π}{a}$ . Why this step? The $i$ 's cancel, leaving a real, positive area — as a physical energy must be.
For $a = 2$ : $E = \frac{π}{2} \approx 1.5708$ . Why this step? Substitute the given bandwidth into the general formula.

$E = \frac{π}{a}, E (a = 2) = \frac{π}{2}$

Verify: units/sanity: larger $a$ → smaller $E$ (a broader, flatter response concentrates less weight), matching intuition; $E > 0$ as an energy must be; and symbolic integration confirms $\int_{- \infty}^{\infty} \frac{d x}{x ^{2} + a ^{2}} = \frac{π}{a}$ . ✔

Cell I — Exam twist: an infinite family of poles (trig)

Worked example Ex 9 (Cell I)

Find $Res (\frac{1}{sin z}, z_{0} = nπ)$ for any integer $n$ .

Forecast: $sin z = 0$ at every $z = nπ$ — infinitely many simple poles. The answer should depend on $n$ only through a sign.

Write $f = \frac{p}{q}$ with $p = 1, q = sin z$ . Check $q (nπ) = sin (nπ) = 0$ → a pole at every multiple of $π$ . Why this step? Confirm the singularities before classifying; here there are countably many along the real axis.
$q^{'} (z) = cos z$ , and $q^{'} (nπ) = cos (nπ) = (- 1)^{n} \neq = 0$ → each is a simple pole. Why this step? $q^{'} \neq = 0$ certifies order 1, so the quotient shortcut applies at every pole at once.
Shortcut: $Res = \frac{p ( nπ )}{q ^{'} ( nπ )} = \frac{1}{( - 1 ) ^{n}} = (- 1)^{n}$ . Why this step? One formula covers the entire infinite family — the elegance of $p / q^{'}$ (note $1/ (- 1)^{n} = (- 1)^{n}$ since $(- 1)^{n} = \pm 1$ ).

$Res (\frac{1}{sin z}, nπ) = (- 1)^{n}$

Verify: at $n = 0$ : residue $= 1$ ; indeed $\frac{1}{s i n z} = \frac{1}{z} \cdot \frac{1}{1 - z ^{2} /6 + \dots} = \frac{1}{z} (1 + \frac{z ^{2}}{6} + \dots)$ , so $a_{- 1} = 1 = (- 1)^{0}$ . At $n = 1$ ( $z = π$ ): $(- 1)^{1} = - 1$ . ✔

The one decision tree for all cells

Active recall

Recall Cover the answers

When a numerator vanishes at the pole, what do you do before choosing a formula? ::: Cancel the common factor first — the reduced form gives the true order. How does using too-large an $m$ in the order formula behave? ::: Safely — extra derivatives of the now-analytic factor add $0$ ; but too-small $m$ gives $\infty$ . Which poles of $\frac{z}{( z - 1 ) ( z - 3 )}$ lie inside $∣ z ∣ = 2$ ? ::: Only $z = 1$ ( $∣1∣ < 2$ ); $z = 3$ is outside. What orientation must $C$ have for $\oint_{C} f d z = + 2 π i \sum Res$ ? ::: Positive, i.e. counter-clockwise; reversing it flips the sign. Why does the big arc vanish in $\int d x / (x^{4} + 1)$ ? ::: ML-inequality: integrand $\leq 1/ (R^{4} - 1)$ times arc length $π R$ gives $π R / (R^{4} - 1) \to 0$ . Residue of $1/ sin z$ at $z = nπ$ ? ::: $(- 1)^{n}$ . Why can't you use pole formulas on $e^{1/ z}$ at $0$ ? ::: It's essential — $z^{m} e^{1/ z}$ never becomes analytic; read $a_{- 1} = 1$ off the series.

Connections

Residues and poles — the parent; these are its cases exhausted.
Laurent series — the only tool for Cell E (removable/essential).
Cauchy's Integral Formula — Ex 3's derivative formula is its higher-order sibling.
Cauchy's Integral Theorem — why the arc-free part of the loop keeps its value under deformation.
Contour integration — Cells G, H live here.
Argument principle / Rouché's theorem — Ex 9's $cot$ -type sums power zero-counting.
Singularities — the classification decision tree above.