Residues and poles: Question bank — Maths deep dive

Before we start, one reminder in plain words so no symbol here is unearned:

A singularity is a point $z_{0}$ where a function stops being nicely differentiable ("analytic").
The Laurent series writes $f$ near $z_{0}$ as $\sum_{n = - \infty}^{\infty} a_{n} (z - z_{0})^{n}$ — an ordinary power series plus a "principal part" of negative powers.
A ==pole of order $m$ == has finitely many negative-power terms, the deepest being $a_{- m} (z - z_{0})^{- m}$ .
The residue is the single coefficient $a_{- 1}$ , the one multiplying $(z - z_{0})^{- 1}$ .
$\oint_{C}$ means "integrate once around a closed loop $C$ ."

Keep the whirlpool picture from the parent note in mind: only the swirl that repeats once per loop survives an integration; everything else cancels. Nearly every trap below is a disguised version of that fact.

See also Cauchy's Integral Theorem, Cauchy's Integral Formula, Contour integration, Argument principle, Rouché's theorem, Singularities.

True or false — justify

The residue of

f

z_{0}

equals

f (z_{0})

False. At a pole

f (z_{0})

is infinite; the residue is the finite coefficient

a_{- 1}

, not a function value. Confusing "value at the pole" with "coefficient of the pole" is the number-one error.

f

is analytic at

z_{0}

(no singularity there), its residue at

z_{0}

0

True. With no negative powers in the expansion,

a_{- 1} = 0

, so nothing survives an encircling loop — exactly what Cauchy's Integral Theorem guarantees.

A larger residue always means a "worse" (higher-order) singularity.

False. The residue is only the

a_{- 1}

coefficient; the order is the depth of the principal part. A simple pole can have residue

1000

while an order-

5

pole can have residue

0

The residue theorem says

\oint_{C} f d z = 2 π i \sum Res

over all singularities of

f

False. Only singularities enclosed by

C

count. Poles outside the contour contribute nothing, since

C

can be deformed away from them without crossing any singularity.

Reversing the orientation of

C

(clockwise instead of counter-clockwise) flips the sign of the integral.

True. The residue theorem assumes positive (counter-clockwise) orientation; a clockwise loop gives

- 2 π i \sum Res

, because

d z = i ρ e^{i θ} d θ

runs

θ

backwards.

Every isolated singularity is either a pole or removable.

False. There is a third type: essential (infinitely many negative powers, like

e^{1/ z}

). These have residues too, but no finite order.

If the residue at

z_{0}

0

, then

\oint_{C} f d z = 0

for a loop around only

z_{0}

True. The loop integral equals

2 π i a_{- 1}

; if

a_{- 1} = 0

the loop yields zero even though

z_{0}

is still a genuine singularity.

A removable singularity has residue

0

True. "Removable" means the principal part is empty (no negative powers at all), so in particular

a_{- 1} = 0

The quotient shortcut

Res = p (z_{0}) / q^{'} (z_{0})

works for any pole of

p / q

False. It only holds when

q (z_{0}) = 0

is a simple zero (

q^{'} (z_{0}) \neq = 0

) and

p (z_{0}) \neq = 0

. A double zero of

q

is an order-

2

pole and needs the derivative formula.

Spot the error

f = \frac{1}{z ^{2}}

0

: simple pole, so

Res = lim_{z \to 0} z \cdot \frac{1}{z ^{2}} = lim \frac{1}{z} = \infty

, so the residue is infinite."

Error: wrong order. This is an order-

2

pole, not simple. The blowing-up limit is the diagnostic warning that you chose

m

too small. The true residue (coefficient of

z^{- 1}

) is

0

, since

1/ z^{2}

has no

z^{- 1}

term.

"For

\frac{e ^{z}}{z ^{2} + 1}

z = i

, use

p / q^{'}

with

p = e^{z}, q = z^{2} + 1

Res = \frac{e ^{i}}{2 i ^{2}} = \frac{e ^{i}}{- 2}

Error: $q^{'}$ evaluated wrong.

q^{'} (z) = 2 z

, so

q^{'} (i) = 2 i

, not

2 i^{2}

. The correct residue is

e^{i} / (2 i)

. Substitute

z = z_{0}

into

q^{'}

, don't square it.

\int_{- \infty}^{\infty} \frac{d x}{x ^{2} + 1}

: poles at

\pm i

, sum both residues, integral

= 2 π i (\frac{1}{2 i} + \frac{- 1}{2 i}) = 0

Error: included a pole outside the contour. Closing in the upper half-plane encloses only

z = i

. The pole

z = - i

lies below and must be excluded, giving

2 π i \cdot \frac{1}{2 i} = π

e^{1/ z}

has an order-

1

pole at

0

, so

Res = lim_{z \to 0} z e^{1/ z}

Error: it's essential, not a pole.

e^{1/ z} = \sum \frac{1}{n !} z^{- n}

has infinitely many negative powers, so no finite order exists and

z^{m} e^{1/ z}

never becomes analytic. Read

a_{- 1} = 1

straight from the series.

\frac{sin z}{z}

0

has a pole because of the

z

in the denominator, residue

= lim z \cdot \frac{s i n z}{z} = 0

, and

0

residue means it's a pole of residue

0

Error: it's removable, not a pole. Since

sin z = z - \frac{z ^{3}}{6} + \dots

, the

z

cancels and there are no negative powers. It's a removable singularity; there is no pole at all.

"The residue theorem gives

\oint_{C} \frac{1}{z} d z

; since

\frac{1}{z}

is odd, positive and negative halves cancel and the integral is

0

Error: real-symmetry reasoning fails for complex loops. On a closed complex contour the answer is

2 π i \cdot Res (\frac{1}{z}, 0) = 2 π i

, not

0

. The "cancellation" intuition applies to real intervals, not to closed loops around a pole.

Why questions

Why does every Laurent term except

n = - 1

integrate to zero around a loop?

On a circle each term becomes

i ρ^{n + 1} \int_{0}^{2 π} e^{i (n + 1) θ} d θ

, and a full-circle average of

e^{ik θ}

0

unless

k = 0

, i.e. unless

n + 1 = 0

. Only

n = - 1

makes the exponent vanish.

Why is

2 π i

the magic factor and not, say,

π

2 π

For

n = - 1

the exponent

k = n + 1 = 0

, so

\int_{0}^{2 π} e^{0} d θ = 2 π

, and the leftover

i

comes from

d z = i ρ e^{i θ} d θ

. Together:

i \cdot 2 π = 2 π i

Why must we multiply by

(z - z_{0})^{m}

before taking a limit?

Multiplying kills the negative powers, turning

f

into an ordinary (Taylor) series that is finite at

z_{0}

. Only then can a limit or derivative extract a coefficient without hitting infinity.

Why the

(m - 1)

-th derivative for an order-

m

pole, and not just a limit?

After multiplying by

(z - z_{0})^{m}

, the residue coefficient

a_{- 1}

has been shifted to sit as the

(m - 1)

-th Taylor coefficient. Differentiating

m - 1

times and dividing by

(m - 1)!

is exactly the Taylor formula that plucks out that coefficient.

Why does the big semicircle contribute nothing in the real-integral example

\int \frac{d x}{x ^{2} + 1}

Because

∣1/ (z^{2} + 1) ∣

decays like

1/ R^{2}

while the arc length grows like

R

, so the arc integral is bounded by

\sim R \cdot R^{- 2} = R^{- 1} \to 0

as the radius

R \to \infty

. Only the real axis survives.

Why can we split a big loop into a sum over individual poles?

By Cauchy's Integral Theorem the region between the big loop and tiny circles around each pole is singularity-free, so the big-loop integral equals the sum of the small-circle integrals — one

2 π i a_{- 1}

per pole.

Why does Cauchy's Integral Formula count as a special case of residues?

Writing

\frac{f ( z )}{z - z _{0}}

makes a simple pole at

z_{0}

with residue

f (z_{0})

; the residue theorem then reads

\oint \frac{f ( z )}{z - z _{0}} d z = 2 π i f (z_{0})

, which is Cauchy's formula.

Why do residues appear in the Argument principle and Rouché's theorem?

Integrating

f^{'} / f

produces simple poles whose residues are

+ 1

at each zero and

- 1

at each pole of

f

; summing them counts zeros minus poles inside the contour.

Edge cases

What is the residue of a function that is analytic (no singularity) at

z_{0}

Zero — the Laurent series is a plain Taylor series with no

a_{- 1}

term, so an encircling loop returns nothing.

Two simple poles collide onto the same point

z_{0}

; what happens to the pole order?

The orders add: two coincident simple poles become an order-

2

pole, and the simple-pole formula fails — you must use the derivative formula with

m = 2

Does an essential singularity have a residue at all?

Yes —

a_{- 1}

still exists in its (infinite) Laurent series, e.g.

Res (e^{1/ z}, 0) = 1

. What fails is the pole formulas, not the concept of residue.

Can the residue at a genuine pole be

0

Yes. For example

1/ z^{2}

has an order-

2

pole at

0

but no

z^{- 1}

term, so its residue is

0

; the singularity is real even though the loop integral vanishes.

What if a suspected pole is actually cancelled by a zero of the numerator?

Then it isn't a pole — the singularity is removable (or of lower order). Always simplify

p / q

by common factors before classifying; a factor of

(z - z_{0})

in the numerator can downgrade or delete the pole.

What is

\oint_{C} f d z

C

encloses no singularity of

f

Exactly

0

, directly from Cauchy's Integral Theorem — with nothing to survive inside, the whole loop cancels.

What happens at a pole sitting exactly on the contour

C

The residue theorem does not apply — the integral is not even well-defined as an ordinary contour integral. One must indent the contour (a small detour around the pole) and take a principal value, which typically contributes

π i \cdot Res

(half the loop) instead of

2 π i

f

has infinitely many poles inside

C

, does the residue theorem still work?

No — the theorem requires finitely many isolated singularities inside

C

. Infinitely many accumulating poles break the "shrink onto each one" argument.

Recall

Recall One-line diagnostics

Residue vs value: residue is the coefficient $a_{- 1}$ , never $f (z_{0})$ .
Wrong- $m$ signal: if $(z - z_{0})^{m} f$ still blows up at $z_{0}$ , your $m$ is too small.
Contour filter: only enclosed poles count.
Essential: no order $m$ — read $a_{- 1}$ off the series directly.

Connections

Residues and poles — parent; this bank stress-tests its definitions.
Laurent series — the source of every $a_{n}$ , including the residue.
Cauchy's Integral Theorem — why non-enclosed / non-singular loops vanish.
Cauchy's Integral Formula — the residue theorem's flagship special case.
Contour integration — where "which pole is inside" decides everything.
Argument principle / Rouché's theorem — residues of $f^{'} / f$ counting zeros and poles.
Singularities — the pole vs removable vs essential trichotomy.