Complex integration — contour integrals: Question bank

4.10.2 · D5 · Maths › Advanced Topics (Elite Level) › Complex integration — contour integrals

Pehle vocabulary fix karne ke liye teen pictures

Neeche kai traps un words par hinge karte hain jo obvious lagte hain par fumble karna aasaan hai: orientation, winding number, aur contour deformation. Quiz se pehle inhe pictures se pin karte hain.

Figure — Complex integration — contour integrals

True or false — justify karo

Ek analytic function ka closed loop integral hamesha

0

hota hai.

True sirf tabhi jab

f

analytic ho on and inside ek simply-connected enclosed region par; andar koi singularity (jaise

1/ z

0

), ya domain mein hole, isse tod deta hai aur

0

ki jagah

2 π i

milta hai.

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

Cauchy's theorem se.

False —

\overset{z}{ˉ}

analytic nahi hai (yeh Cauchy–Riemann equations har jagah fail karta hai), isliye Cauchy's theorem apply nahi hota; direct parametrisation se

2 π i

milta hai,

0

nahi.

Agar ek closed contour ke liye

\oint_{γ} f d z = 0

ho, toh

f

γ

ke andar analytic hai.

False — zero answer cancellation se bhi ho sakta hai. Jaise

\oint_{∣ z ∣ = 1} 1/ z^{2} d z = 0

hota hai jabki

z = 0

ek genuine pole hai; sirf

c_{- 1}

term matter karta hai, aur yahan woh absent hai.

γ

ki orientation reverse karna (say CCW se CW) sirf integral ka sign change karta hai.

True — agar

γ

z (t)

se parametrise kiya gaya hai

t \in [a, b]

par, toh reversed path

z (a + b - t)

hai; uska derivative

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

hai, isliye

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

ho jaata hai jabki magnitude untouched rehti hai.

Residue theorem ke liye contour ka circle hona zaroori hai.

False — yeh kisi bhi simple closed positively-oriented (CCW) contour ke liye valid hai; circle sirf wo convenient shape hai jis par hum deform karte hain. Kya matter karta hai yeh hai ki kaun se poles enclosed hain.

\int_{γ} f d z

har function

f

ke liye sirf endpoints par depend karta hai.

False — path-independence ke liye

f

ka simply-connected region par analytic hona zaroori hai; non-analytic

f

ke liye (jaise

\overset{z}{ˉ}

∣ z ∣^{2}

Re z

) path genuinely matter karta hai.

Ek function jo poore

C

mein analytic ho (entire) ka har closed

γ

ke liye

\oint_{γ} f d z = 0

hoga.

True — ek entire function mein kahin koi singularity nahi hoti, isliye har enclosed region singularity-free hai aur Cauchy's theorem

0

deta hai.

Agar ek pole contour

γ

par ho, toh residue theorem directly answer deta hai.

False — integral ordinary sense mein define bhi nahi hota; theorem ko poles strictly inside chahiye. Path par pole ke liye principal-value / indentation argument ki zaroorat hoti hai.

Dono contours jo same set of poles enclose karte hain, unka integral same hota hai.

True — unke beech ka region (ek ring, figure s03 mein drawn) singularity-free hai, isliye dono loop integrals ka difference

0

hai (deformation invariance); yahi trick Cauchy's formula ke peeche kaam karti hai.

Spot the error

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

kyunki

z = 5

par ek pole hai."

Error: pole

z = 5

∣ z ∣ = 3

ke bahar hai (kyunki

5 > 3

). Koi enclosed singularity nahi hai, isliye answer

0

hai,

2 π i

nahi.

\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:

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

factor drop ho gaya. Ise restore karne par milta hai

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

f

z_{0}

par pole hai, isliye

Res_{z_{0}} f

equals value

f (z_{0})

hai."

Error:

f (z_{0})

pole par infinite hai, isliye woh residue nahi ho sakta. Residue Laurent coefficient

c_{- 1}

hai —

1/ (z - z_{0})

term ki strength, koi function value nahi.

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

ka residue

z = i

par

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

hai."

Error: tum directly

p / q

mein plug in nahi kar sakte — denominator zero ho jaata hai. Simple-pole rule use karo:

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

"Bade semicircle ka arc hamesha

R \to \infty

par vanish ho jaata hai, isliye main hamesha contour upar close kar sakta hoon."

Error: arc tabhi vanish hota hai jab

∣ f ∣

kaafi fast decay kare (yahan

1/ R^{2}

jaisi, arc length

\sim π R

ke against). Slow decay (Jordan's lemma cases) ya wrong half-plane ke liye, arc zaroor nahi vanish hota.

\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: dono poles

\pm i

∣ z ∣ = 2

ke andar hain. Dono residues ka sum (

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

) karne par

2 π i

milta hai,

π i

nahi.

e^{z}

ka infinity par ek pole hai, isliye uska loop integral nonzero hai."

Error:

e^{z}

entire hai (finite plane mein har jagah analytic), isliye koi bhi finite closed loop

0

deta hai. "Infinity par" ka behaviour kisi finite contour se enclosed nahi hota.

Why questions

Kisi closed loop ke around integrate karne par sirf

z^{- 1}

term kyun bachta hai?

Integer

n \neq = - 1

wale har

z^{n}

ka ek single-valued antiderivative

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

hota hai, isliye uska loop integral start par wapis aata hai (

0

);

z^{- 1}

ka antiderivative

lo g z

hai, jo multivalued hai aur har loop par

2 π i

jump karta hai.

Cauchy–Riemann equations Green's-theorem integrands ko zero kyun banate hain?

Dono integrands hain

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

aur

(u_{x} - v_{y})

; CR equations

u_{x} = v_{y}

aur

u_{y} = - v_{x}

har parenthesis ko exactly zero kar dete hain.

d z

(na ki

∣ d z ∣

) contour integral ke liye sahi differential kyun hai?

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

plane mein direction carry karta hai;

∣ d z ∣

(arc length) use karne se woh phase/orientation information discard ho jaati jo complex integration ko path- aur orientation-sensitive banati hai.

Hum

γ

ko pole ke around ek tiny circle mein deform kar sakte hain bina integral change kiye — kyun?

Dono curves ke beech integrand analytic hai (figure s03 mein ring dekho), isliye us ring ko bound karne wala closed loop

0

integrate karta hai; isliye outer aur inner loop integrals equal hain.

Cauchy's theorem simply-connected domain par kyun insist karta hai?

Proof

γ

ko ek point tak deform karta hai; agar domain mein hole ho, toh us hole ke around ek loop bina missing region cross kiye ek point tak shrink nahi ho sakta, isliye "

= 0

" conclusion fail ho jaata hai — hole

2 π i

worth ki singularity chhupa sakta hai.

Ek mushkil real integral poles count karne mein kyun collapse ho jaata hai?

Real line ko ek vanishing arc se close karne par woh ek closed contour ban jaata hai; residue theorem phir poore integral ko

2 π i

times residues ke finite sum se replace kar deta hai — antidifferentiation ki jagah bookkeeping.

Residue ko Laurent series se define kyun karna padta hai, Taylor series se kyun nahi?

Taylor series mein koi negative powers nahi hoti aur woh ek aise function ka description nahi de sakti jo

z_{0}

par blow up kare; Laurent expansion

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

term provide karta hai jiska coefficient hi residue hai.

Orientation matter kyun karta hai, jab real integrals sirf swapped limits ke liye sign flip karte hain?

Yeh same phenomenon hai: limits swap karna

d x

ka sign flip karta hai; ek complex contour reversal

d z

ka sign path par har jagah flip karta hai, jo poore integral ko

- 1

se multiply kar deta hai.

Edge cases

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

0

— Laurent series sirf

z^{- 2}

hai jisme koi

z^{- 1}

term nahi hai (

c_{- 1} = 0

), isliye pole hone ke bawajood residue zero hai.

\oint_{∣ z ∣ = 1} z^{- 3} d z

aur koi bhi negative power

z^{- k}

jisme

k \geq 2

ho.

Sab

0

dete hain; sirf

k = 1

(yaani

z^{- 1}

term) bachta hai, isliye origin par higher-order poles akele kuch contribute nahi karte.

Ek contour jo exactly ek pole se guzarta ho, jaise

\oint_{∣ z ∣ = 1} \frac{d z}{z - 1}

Ordinary contour integral ke sense mein undefined — pole

z = 1

∣ z ∣ = 1

par baithta hai; contour ko indent karna padega ya principal value leni padegi; residue theorem as stated apply nahi hota.

Double pole:

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

ka residue

z = a

par.

Residue

0

hai (

(z - a)^{- 1}

ka coefficient absent hai). General double pole

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

ke liye (jahan

g

analytic ho,

g (a) \neq = 0

) residue

g^{'} (a)

hai:

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

expand karo,

(z - a)^{2}

se divide karo, aur

(z - a)^{- 1}

coefficient exactly

g^{'} (a)

aata hai — isliye ek derivative hai, plug-in nahi.

Ek aisa loop jo koi singularity enclose na kare, even

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

jaisi expression ke liye

∣ z ∣ = 1

par.

0

— sirf pole

z = 10

bahar hai; integrand

∣ z ∣ = 1

ke andar analytic hai, isliye Cauchy's theorem

0

deta hai.

Ek contour jo ek pole ko do baar enclose kare (winding number

n = 2

, figure s02), jaise

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

t \in [0, 2 π]

1/ z

ke liye.

4 π i

— integral

2 π i \cdot n

hai, aur do baar ghoomne par (

n = 2

) pole do baar count hota hai.

\oint_{∣ z ∣ = 1} \frac{d z}{z}

\oint_{∣ z ∣ = 1} \frac{d z}{z ˉ}

Pehla

2 π i

hai (analytic away from

0

); doosre mein circle par

\overset{z}{ˉ} = e^{- i t}

use hota hai, jo deta hai

\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

— ek reminder ki

1/ \overset{z}{ˉ}

ek alag, non-analytic cheez hai.

z = i

aur

z = - i

par

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

ke residues ka sum — sign check.

Dono

+ \frac{1}{2}

hain (opposite signs nahi), kyunki

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

pole se regardless hai; unka sum

1

2 π i

milta hai.

Connections

Analytic functions — exact class jahan paths matter nahi karte aur loops vanish ho jaate hain.
Cauchy–Riemann equations — woh hypothesis jo quietly har "it's zero" claim ko power karti hai.
Laurent series — jahan residue actually rehta hai ( $c_{- 1}$ ).
Green's theorem — Cauchy's theorem ke peeche ka real-analysis engine.
Residue theorem applications — in traps ko working technique mein convert karna.