Complex integration — contour integrals: Question bank

Three pictures to fix the vocabulary first

Several traps below hinge on words that sound obvious but are easy to fumble: orientation, winding number, and contour deformation. Let's pin them to pictures before the quiz.

Figure — Complex integration — contour integrals

True or false — justify

A closed loop integral of an analytic function is always

0

True only if

f

is analytic on and inside a simply-connected enclosed region; a singularity inside (like

1/ z

0

), or a hole in the domain, breaks it, giving

2 π i

instead of

0

\oint_{∣ z ∣ = 1} \overset{z}{ˉ} d z = 0

by Cauchy's theorem.

False —

\overset{z}{ˉ}

is not analytic (it fails Cauchy–Riemann everywhere), so Cauchy's theorem does not apply; direct parametrisation gives

2 π i

, not

0

\oint_{γ} f d z = 0

for one closed contour, then

f

is analytic inside

γ

False — a zero answer can happen by cancellation. For example

\oint_{∣ z ∣ = 1} 1/ z^{2} d z = 0

even though

z = 0

is a genuine pole; only the

c_{- 1}

term matters, and here it is absent.

Reversing the orientation of

γ

(say from CCW to CW) changes only the sign of the integral.

True — if

γ

is parametrised by

z (t)

t \in [a, b]

, the reversed path is

z (a + b - t)

; its derivative is

- z^{'} (a + b - t)

, so

\int f d z \to - \int f d z

while the magnitude is untouched.

The residue theorem requires the contour to be a circle.

False — it holds for any simple closed positively-oriented (CCW) contour; the circle is just the convenient shape we deform to. What matters is which poles are enclosed.

\int_{γ} f d z

depends only on the endpoints for every function

f

False — path-independence needs

f

analytic on a simply-connected region; for non-analytic

f

(e.g.

\overset{z}{ˉ}

∣ z ∣^{2}

Re z

) the path genuinely matters.

A function analytic everywhere in

C

(entire) has

\oint_{γ} f d z = 0

for every closed

γ

True — an entire function has no singularities anywhere, so every enclosed region is singularity-free and Cauchy's theorem gives

0

If a pole lies on the contour

γ

, the residue theorem gives the answer directly.

False — the integral is not even well-defined in the ordinary sense; the theorem needs poles strictly inside. A pole on the path requires a principal-value / indentation argument instead.

Two contours enclosing the same set of poles give the same integral.

True — the region between them (a ring, drawn in figure s03) is singularity-free, so the difference of the two loop integrals is

0

(deformation invariance); this is exactly the trick behind Cauchy's formula.

Spot the error

\oint_{∣ z ∣ = 3} \frac{1}{z - 5} d z = 2 π i

because there is a pole at

z = 5

Error: the pole

z = 5

lies outside

∣ z ∣ = 3

(since

5 > 3

). No enclosed singularity means the answer is

0

, not

2 π i

\oint_{∣ z ∣ = 1} z^{- 1} d z = \int_{0}^{2 π} \frac{1}{e ^{i t}} d t = \int_{0}^{2 π} e^{- i t} d t = 0

Error: the

d z = z^{'} (t) d t = i e^{i t} d t

factor was dropped. Restoring it gives

\int_{0}^{2 π} e^{- i t} i e^{i t} d t = \int_{0}^{2 π} i d t = 2 π i

f

has a pole at

z_{0}

, so

Res_{z_{0}} f

equals the value

f (z_{0})

Error:

f (z_{0})

is infinite at a pole, so it cannot be the residue. The residue is the Laurent coefficient

c_{- 1}

— the strength of the

1/ (z - z_{0})

term, not a function value.

"For

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

the residue at

z = i

\frac{z}{z ^{2} + 1}_{z = i}

Error: you can't just plug into

p / q

— the denominator vanishes. Use the simple-pole rule

p (z_{0}) / q^{'} (z_{0}) = \frac{i}{2 i} = \frac{1}{2}

"The arc of the big semicircle always vanishes as

R \to \infty

, so I may always close the contour up top."

Error: the arc vanishes only when

∣ f ∣

decays fast enough (here like

1/ R^{2}

against arc length

\sim π R

). For slow decay (Jordan's lemma cases) or the wrong half-plane, the arc need not vanish.

\oint_{∣ z ∣ = 2} \frac{z}{z ^{2} + 1} d z = 2 π i \cdot Res_{z = i} = 2 π i \cdot \frac{1}{2} = π i

Error: both poles

\pm i

lie inside

∣ z ∣ = 2

. Summing both residues (

\frac{1}{2} + \frac{1}{2} = 1

) gives

2 π i

, not

π i

e^{z}

has a pole at infinity, so its loop integral is nonzero."

Error:

e^{z}

is entire (analytic everywhere in the finite plane), so any finite closed loop gives

0

. Behaviour "at infinity" is not enclosed by a finite contour.

Why questions

Why does only the

z^{- 1}

term survive integrating around a closed loop?

Every

z^{n}

with integer

n \neq = - 1

has a single-valued antiderivative

z^{n + 1} / (n + 1)

, so its loop integral returns to start (

0

);

z^{- 1}

's antiderivative is

lo g z

, which is multivalued and jumps by

2 π i

per loop.

Why do the Cauchy–Riemann equations make Green's-theorem integrands vanish?

The two integrands are

- (v_{x} + u_{y})

and

(u_{x} - v_{y})

; the CR equations

u_{x} = v_{y}

and

u_{y} = - v_{x}

set each parenthesis to exactly zero.

Why is

d z

(not

∣ d z ∣

) the right differential for a contour integral?

d z = z^{'} (t) d t

carries direction in the plane; using

∣ d z ∣

(arc length) would discard the phase/orientation information that makes complex integration path- and orientation-sensitive.

Why can we deform

γ

to a tiny circle around a pole without changing the integral?

Between the two curves the integrand is analytic (see the ring in figure s03), so the closed loop bounding that ring integrates to

0

; hence the outer and inner loop integrals are equal.

Why does Cauchy's theorem insist on a simply-connected domain?

The proof deforms

γ

down to a point; if the domain has a hole, a loop around that hole can't shrink to a point without crossing the missing region, so the "

= 0

" conclusion fails — the hole can hide a singularity worth

2 π i

Why does a hard real integral collapse to counting poles?

Closing the real line with a vanishing arc turns it into a closed contour; the residue theorem then replaces the whole integral with

2 π i

times a finite sum of residues — bookkeeping instead of antidifferentiation.

Why must the residue be defined via the Laurent series rather than a Taylor series?

A Taylor series has no negative powers and cannot describe a function that blows up at

z_{0}

; the Laurent expansion supplies the

c_{- 1} (z - z_{0})^{- 1}

term whose coefficient is the residue.

Why does orientation matter at all, when real integrals only flip sign for swapped limits?

It is the same phenomenon: swapping limits flips

d x

's sign; a complex contour reversal flips

d z

's sign everywhere along the path, multiplying the whole integral by

- 1

Edge cases

\oint_{∣ z ∣ = 1} \frac{1}{z ^{2}} d z = ?

0

— the Laurent series is just

z^{- 2}

with no

z^{- 1}

term (

c_{- 1} = 0

), so despite the pole the residue is zero.

\oint_{∣ z ∣ = 1} z^{- 3} d z

and any negative power

z^{- k}

with

k \geq 2

All give

0

; only

k = 1

(the

z^{- 1}

term) survives, so higher-order poles at the origin contribute nothing on their own.

A contour that passes exactly through a pole, e.g.

\oint_{∣ z ∣ = 1} \frac{d z}{z - 1}

Undefined as an ordinary contour integral — the pole

z = 1

sits on

∣ z ∣ = 1

; one must indent the contour or take a principal value; the residue theorem does not apply as stated.

A double pole: residue of

\frac{1}{( z - a ) ^{2}}

z = a

The residue is

0

(coefficient of

(z - a)^{- 1}

is absent). For a general double pole

g (z) / (z - a)^{2}

(with

g

analytic,

g (a) \neq = 0

) the residue is

g^{'} (a)

: expand

g (z) = g (a) + g^{'} (a) (z - a) + \dots

, divide by

(z - a)^{2}

, and the

(z - a)^{- 1}

coefficient is exactly

g^{'} (a)

— hence a derivative, not a plug-in.

Integrating around a loop that encloses no singularities, even for a non-analytic-looking expression like

\frac{e ^{z}}{z - 10}

∣ z ∣ = 1

0

— the only pole

z = 10

is outside; the integrand is analytic inside

∣ z ∣ = 1

, so Cauchy's theorem gives

0

A contour enclosing a pole twice (winding number

n = 2

, figure s02), e.g.

z (t) = e^{2 i t}

t \in [0, 2 π]

, of

1/ z

4 π i

— the integral is

2 π i \cdot n

, and going around twice (

n = 2

) counts the pole twice.

\oint_{∣ z ∣ = 1} \frac{d z}{z}

\oint_{∣ z ∣ = 1} \frac{d z}{z ˉ}

The first is

2 π i

(analytic away from

0

); the second uses

\overset{z}{ˉ} = e^{- i t}

on the circle, giving

\int_{0}^{2 π} e^{i t} i e^{i t} d t = i \int_{0}^{2 π} e^{2 i t} d t = 0

— a reminder that

1/ \overset{z}{ˉ}

is a different, non-analytic beast.

Sum of residues at

z = i

and

z = - i

for

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

— sign check.

Both equal

+ \frac{1}{2}

(not opposite signs), because

Res = z / (2 z) = \frac{1}{2}

regardless of the pole; their sum

1

gives

2 π i

Connections

Analytic functions — the exact class where paths don't matter and loops vanish.
Cauchy–Riemann equations — the hypothesis quietly powering every "it's zero" claim.
Laurent series — where the residue actually lives ( $c_{- 1}$ ).
Green's theorem — the real-analysis engine behind Cauchy's theorem.
Residue theorem applications — turning these traps into working technique.