Residues and poles: Question bank — Hinglish deep dive

4.10.5 · D5 · Maths › Advanced Topics (Elite Level) › Residues and poles

Shuru karne se pehle, plain words mein ek reminder taaki koi bhi symbol yahan be-wajah na ho:

Singularity woh point $z_{0}$ hai jahan ek function achhi tarah se differentiable ("analytic") hona band kar deta hai.
Laurent series $f$ ko $z_{0}$ ke paas $\sum_{n = - \infty}^{\infty} a_{n} (z - z_{0})^{n}$ ke roop mein likhti hai — ek ordinary power series plus negative powers ka ek "principal part".
==Pole of order $m$ == mein finitely many negative-power terms hote hain, jinka sabse gehra $a_{- m} (z - z_{0})^{- m}$ hota hai.
Residue woh single coefficient $a_{- 1}$ hai, jo $(z - z_{0})^{- 1}$ ko multiply karta hai.
$\oint_{C}$ ka matlab hai "ek closed loop $C$ ke around ek baar integrate karo."

Parent note ka whirlpool picture dimaag mein rakho: sirf woh swirl jo ek baar per loop repeat hota hai integration mein bachta hai; baaki sab cancel ho jaata hai. Neeche har trap usi fact ka ek chhupa hua version hai.

Yeh bhi dekho 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. Ek pole par

f (z_{0})

infinite hota hai; residue woh finite coefficient

a_{- 1}

hai, function value nahi. "Pole par value" ko "pole ka coefficient" samajhna sabse badi number-one error hai.

Agar

f

z_{0}

analytic hai (wahan koi singularity nahi), to

z_{0}

par uska residue

0

hai.

True. Expansion mein koi negative powers nahi honge, isliye

a_{- 1} = 0

, aur ek encircling loop mein kuch bhi survive nahi karta — bilkul wahi jo Cauchy's Integral Theorem guarantee karta hai.

Ek bada residue hamesha ek "worse" (higher-order) singularity ka matlab hota hai.

False. Residue sirf

a_{- 1}

coefficient hai; order principal part ki depth hai. Ek simple pole ka residue

1000

ho sakta hai jabki ek order-

5

pole ka residue

0

ho sakta hai.

Residue theorem kehta hai ki

\oint_{C} f d z = 2 π i \sum Res

sabhi singularities of

f

par.

False. Sirf woh singularities count hoti hain jo

C

ke andar enclosed hain. Contour ke bahar ke poles kuch contribute nahi karte, kyunki

C

ko unse door kiya ja sakta hai bina kisi singularity cross kiye.

C

ki orientation reverse karna (counter-clockwise ki jagah clockwise) integral ka sign flip kar deta hai.

True. Residue theorem positive (counter-clockwise) orientation assume karta hai; ek clockwise loop

- 2 π i \sum Res

deta hai, kyunki

d z = i ρ e^{i θ} d θ

mein

θ

ulta run karta hai.

Har isolated singularity ya to ek pole hai ya removable.

False. Ek teesra type bhi hai: essential (infinitely many negative powers, jaise

e^{1/ z}

). Inke residues bhi hote hain, lekin koi finite order nahi hota.

Agar

z_{0}

par residue

0

hai, to

z_{0}

ke around sirf ek loop ke liye

\oint_{C} f d z = 0

hai.

True. Loop integral

2 π i a_{- 1}

ke barabar hota hai; agar

a_{- 1} = 0

to loop zero yield karta hai chahe

z_{0}

abhi bhi ek genuine singularity ho.

Ek removable singularity ka residue

0

hota hai.

True. "Removable" ka matlab hai principal part empty hai (bilkul bhi koi negative powers nahi), isliye khaas taur par

a_{- 1} = 0

Quotient shortcut

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

kisi bhi pole of

p / q

ke liye kaam karta hai.

False. Yeh sirf tab kaam karta hai jab

q (z_{0}) = 0

ek simple zero ho (

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

) aur

p (z_{0}) \neq = 0

q

ka double zero ek order-

2

pole hai aur derivative formula ki zaroorat padti hai.

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: galat order. Yeh ek order-

2

pole hai, simple nahi. Blowing-up limit woh diagnostic warning hai ki aapne

m

bahut chhota choose kiya. True residue (coefficient of

z^{- 1}

)

0

hai, kyunki

1/ z^{2}

mein koi

z^{- 1}

term nahi hai.

"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: $q^{'}$ galat evaluate kiya.

q^{'} (z) = 2 z

, isliye

q^{'} (i) = 2 i

hai, na ki

2 i^{2}

. Correct residue

e^{i} / (2 i)

hai.

z = z_{0}

q^{'}

mein substitute karo, use square mat karo.

\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: contour ke bahar ka pole include kar liya. Upper half-plane mein close karna sirf

z = i

ko enclose karta hai. Pole

z = - i

neeche hai aur exclude kiya jaana chahiye, jo

2 π i \cdot \frac{1}{2 i} = π

deta hai.

e^{1/ z}

has an order-

1

pole at

0

, so

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

Error: yeh essential hai, pole nahi.

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

mein infinitely many negative powers hain, isliye koi finite order exist nahi karta aur

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

kabhi analytic nahi banta. Series se seedha

a_{- 1} = 1

padho.

\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: yeh removable hai, pole nahi. Kyunki

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

, isliye

z

cancel ho jaata hai aur koi bhi negative powers nahi hote. Yeh ek removable singularity hai; wahan koi pole hi nahi hai.

"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 complex loops ke liye fail karta hai. Ek closed complex contour par answer

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

hai,

0

nahi. "Cancellation" ka intuition real intervals par apply hota hai, pole ke around closed loops par nahi.

Why questions

Kyun loop ke around sirf

n = - 1

wala Laurent term integrate hoke survive karta hai, baaki sab zero ho jaate hain?

Ek circle par har term

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

ban jaata hai, aur

e^{ik θ}

ka full-circle average

0

hota hai jab tak

k = 0

na ho, yaani jab tak

n + 1 = 0

na ho. Sirf

n = - 1

exponent ko vanish karta hai.

2 π i

magic factor kyun hai,

π

2 π

kyun nahi?

n = - 1

ke liye exponent

k = n + 1 = 0

hai, isliye

\int_{0}^{2 π} e^{0} d θ = 2 π

, aur bacha hua

i

d z = i ρ e^{i θ} d θ

se aata hai. Milake:

i \cdot 2 π = 2 π i

Limit lene se pehle hum

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

se multiply kyun karte hain?

Multiply karne se negative powers khatam ho jaate hain,

f

ko ek ordinary (Taylor) series mein badal deta hai jo

z_{0}

par finite hoti hai. Tabhi ek limit ya derivative ek coefficient extract kar sakta hai bina infinity se takraye.

Order-

m

pole ke liye

(m - 1)

-th derivative kyun, sirf limit kyun nahi?

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

se multiply karne ke baad, residue coefficient

a_{- 1}

shift ho jaata hai aur

(m - 1)

-th Taylor coefficient ban jaata hai.

m - 1

baar differentiate karna aur

(m - 1)!

se divide karna exactly wahi Taylor formula hai jo us coefficient ko nikaalta hai.

Real-integral example

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

mein bada semicircle kuch contribute kyun nahi karta?

Kyunki

∣1/ (z^{2} + 1) ∣

1/ R^{2}

ki tarah decay karta hai jabki arc length

R

ki tarah badhti hai, isliye arc integral

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

se bound hota hai jab radius

R \to \infty

. Sirf real axis survive karta hai.

Hum ek bade loop ko individual poles par sum mein kyun split kar sakte hain?

Cauchy's Integral Theorem se, bade loop aur har pole ke aas-paas tiny circles ke beech ka region singularity-free hai, isliye bada-loop integral chhote-circle integrals ka sum hota hai — har pole ke liye ek

2 π i a_{- 1}

Cauchy's Integral Formula residues ka ek special case kyun count hota hai?

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

likhne se

z_{0}

par residue

f (z_{0})

ke saath ek simple pole banta hai; residue theorem phir

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

padhta hai, jo Cauchy's formula hi hai.

Residues Argument principle aur Rouché's theorem mein kyun appear karte hain?

f^{'} / f

integrate karne se simple poles bante hain jinke residues

f

ke har zero par

+ 1

aur har pole par

- 1

hote hain; unhe sum karna contour ke andar zeros minus poles count karta hai.

Edge cases

Kisi function ka residue kya hai jo

z_{0}

par analytic hai (koi singularity nahi)?

Zero — Laurent series ek plain Taylor series hai jisme koi

a_{- 1}

term nahi hai, isliye ek encircling loop kuch return nahi karta.

Do simple poles ek hi point

z_{0}

par collide karte hain; pole order ka kya hoga?

Orders add ho jaate hain: do coincident simple poles ek order-

2

pole ban jaate hain, aur simple-pole formula fail ho jaata hai — aapko

m = 2

ke saath derivative formula use karna hoga.

Kya essential singularity ka residue hota hai?

Haan —

a_{- 1}

uski (infinite) Laurent series mein abhi bhi exist karta hai, e.g.

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

. Jo fail hota hai woh pole formulas hain, residue ka concept nahi.

Kya ek genuine pole par residue

0

ho sakta hai?

Haan. For example

1/ z^{2}

0

par order-

2

pole hai lekin koi

z^{- 1}

term nahi hai, isliye uska residue

0

hai; singularity real hai chahe loop integral vanish ho jaaye.

Kya hoga agar suspected pole actually numerator ke ek zero se cancel ho jaaye?

Tab woh pole nahi hai — singularity removable hai (ya lower order ki). Classify karne se pehle

p / q

ko common factors se simplify karo; numerator mein

(z - z_{0})

ka ek factor pole ko downgrade ya delete kar sakta hai.

Agar

C

mein

f

ki koi singularity enclosed na ho to

\oint_{C} f d z

kya hoga?

Exactly

0

, seedha Cauchy's Integral Theorem se — andar survive karne ke liye kuch nahi hai, to poora loop cancel ho jaata hai.

Kya hota hai agar contour

C

par exactly ek pole ho?

Residue theorem apply nahi hota — integral ordinary contour integral ke roop mein well-defined bhi nahi hota. Contour ko indent karna padta hai (pole ke around ek chhota detour) aur principal value lena padta hai, jo typically

π i \cdot Res

contribute karta hai (

2 π i

ki jagah half loop).

Agar

f

ke andar

C

mein infinitely many poles hain, kya residue theorem abhi bhi kaam karta hai?

Nahi — theorem ko

C

ke andar finitely many isolated singularities chahiye. Infinitely many accumulating poles "shrink onto each one" argument ko tod dete hain.

Recall

Recall One-line diagnostics

Residue vs value: residue coefficient $a_{- 1}$ hai, kabhi $f (z_{0})$ nahi.
Wrong- $m$ signal: agar $(z - z_{0})^{m} f$ abhi bhi $z_{0}$ par blow up karta hai, aapka $m$ bahut chhota hai.
Contour filter: sirf enclosed poles count hote hain.
Essential: koi order $m$ nahi — seedha series se $a_{- 1}$ padho.

Connections

Residues and poles — parent; yeh bank uski definitions ko stress-test karta hai.
Laurent series — har $a_{n}$ ka source, including residue.
Cauchy's Integral Theorem — kyun non-enclosed / non-singular loops vanish ho jaate hain.
Cauchy's Integral Formula — residue theorem ka flagship special case.
Contour integration — jahan "kaun sa pole andar hai" sab decide karta hai.
Argument principle / Rouché's theorem — $f^{'} / f$ ke residues zeros aur poles count karte hain.
Singularities — pole vs removable vs essential trichotomy.