Laurent series — principal part, annulus of convergence

4.10.4 · Maths › Advanced Topics (Elite Level)

WHAT is a Laurent series?

Yeh formula kyun? (Derivation scratch se.) Hum $c_{n}$ chahte hain expansion $f (z) = \sum_{m} c_{m} (z - a)^{m}$ ke liye. $(z - a)^{- n - 1}$ se multiply karo aur annulus mein loop $γ$ ke around integrate karo: $\oint_{γ} \frac{f ( z )}{( z - a ) ^{n + 1}} d z = \sum_{m} c_{m} \oint_{γ} (z - a)^{m - n - 1} d z .$ Yeh step kyun? Hum contour integration ke under powers ki orthogonality use kar rahe hain. Ab ek key fact yaad karo (ise derive karo!): $\oint_{γ} (z - a)^{k} d z = {2 π i 0 k = - 1 k \neq = - 1$ Key fact ka proof: parametrise karo $z = a + ρ e^{i θ}$ , $d z = i ρ e^{i θ} d θ$ , toh integral ban jaata hai $\int_{0}^{2 π} ρ^{k} e^{ik θ} i ρ e^{i θ} d θ = i ρ^{k + 1} \int_{0}^{2 π} e^{i (k + 1) θ} d θ$ . $k \neq = - 1$ ke liye $θ$ -integral $0$ hai; $k = - 1$ ke liye yeh $2 π$ hai. ✔

Toh sirf woh term bachti hai jismein $m - n - 1 = - 1$ , yaani $m = n$ : $\oint_{γ} \frac{f ( z )}{( z - a ) ^{n + 1}} d z = c_{n} \cdot 2 π i \Rightarrow c_{n} = \frac{1}{2 π i} \oint_{γ} \frac{f ( z )}{( z - a ) ^{n + 1}} d z .$ Yeh uniqueness bhi prove karta hai: coefficients forced hote hain.

WHY ek annulus, disk nahi?

Analytic part $\sum_{n \geq 0} c_{n} (z - a)^{n}$ ek disk $∣ z - a ∣ < R$ ke andar converge karta hai (normal power series ki tarah). $R =$ bahar ki nearest singularity ki doori.
Principal part $\sum_{n \geq 1} c_{- n} (z - a)^{- n}$ ek power series hai $w = \frac{1}{z - a}$ mein, toh yeh $∣ w ∣ < \frac{1}{r}$ , yaani $∣ z - a ∣ > r$ ke liye converge karta hai. $r =$ $a$ par ya andar ki nearest singularity ki doori.

Laurent series wahan converge karti hai jahan dono converge karein: annulus $r < ∣ z - a ∣ < R$ .

Figure — Laurent series — principal part, annulus of convergence

Singularity ko principal part se classify karna

Coefficient $c_{- 1}$ $a$ par $f$ ka residue hai — woh single number jo integration mein survive karta hai (kyunki sirf $k = - 1$ hi $2 π i$ deta hai). Isliye residues poori contour integration theory chalate hain.

Worked examples

Worked example 2.

$f (z) = \frac{1}{( z - 1 ) ( z - 2 )}$ — annulus par depend karta hai Pehle partial fractions: $\frac{1}{( z - 1 ) ( z - 2 )} = \frac{- 1}{z - 1} + \frac{1}{z - 2}$ . Kyun? Har simple piece ek geometric series hai jise hum sahi direction mein expand kar sakte hain.

Region A: $∣ z ∣ < 1$ (dono ko $∣ z ∣$ small maanke expand karo): $\frac{- 1}{z - 1} = \frac{1}{1 - z} = \sum_{n \geq 0} z^{n}$ , aur $\frac{1}{z - 2} = \frac{- 1}{2} \cdot \frac{1}{1 - z /2} = - \sum_{n \geq 0} \frac{z ^{n}}{2 ^{n + 1}}$ . Sum: $\sum_{n \geq 0} (1 - \frac{1}{2 ^{n + 1}}) z^{n}$ . Koi principal part nahi — pure Taylor (andar koi singularity nahi).

Region B: $1 < ∣ z ∣ < 2$ — Yahan split kyun? $1$ par singularity ab andar hai, toh us piece ke liye negative powers chahiye. $\frac{- 1}{z - 1} = \frac{- 1}{z} \cdot \frac{1}{1 - 1/ z} = - \sum_{n \geq 0} z^{- n - 1}$ (valid $∣ z ∣ > 1$ ), $\frac{1}{z - 2} = - \sum_{n \geq 0} \frac{z ^{n}}{2 ^{n + 1}}$ rakhte hain (valid $∣ z ∣ < 2$ ). Ab ek principal part hai: $- \sum_{n \geq 1} z^{- n}$ .

Region C: $∣ z ∣ > 2$ — dono singularities andar; dono ko $1/ z$ mein expand karo: $\frac{- 1}{z - 1} = - \sum_{n \geq 0} z^{- n - 1}$ , $\frac{1}{z - 2} = \frac{1}{z} \cdot \frac{1}{1 - 2/ z} = \sum_{n \geq 0} 2^{n} z^{- n - 1}$ . Sum $= \sum_{n \geq 0} (2^{n} - 1) z^{- n - 1}$ . Saari negative powers.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho ek gaana hai jo speaker ki taraf chalne par tez aur tez hota jaata hai — speaker ke bilkul paas jaake unbearably loud ho jaata hai. Ek normal (Taylor) recipe sirf quiet, decent sounds describe kar sakti hai, toh woh speaker ke paas fail ho jaati hai. Laurent recipe special "scream terms" ( $\frac{1}{z}$ , $\frac{1}{z ^{2}}$ , …) add karti hai jo speaker ke paas bahut bade ho jaate hain — toh yeh gaane ko uske bahut paas bhi describe kar sakti hai, lekin sirf ring-shaped region mein jo "bahut paas" aur "ek aur speaker aur dur" ke beech ho. Scream terms ki count batati hai ki speaker kitna nasty hai: koi nahi = fake speaker, thode = normal loud speaker (pole), infinitely many = ek bilkul crazy wala (essential).

Active recall

Laurent series mein Taylor series se zyada kya extra terms hoti hain?

Negative-power terms

(z - a)^{- n}

, jo principal part banate hain.

Laurent series ke principal part ko define karo.

Saari negative-power terms ka sum

\sum_{n \geq 1} c_{- n} (z - a)^{- n}

Laurent coefficient

c_{n}

ka formula kya hai?

c_{n} = \frac{1}{2 π i} \oint_{γ} \frac{f ( z )}{( z - a ) ^{n + 1}} d z

Kaunsa contour integral fact coefficient formula ko kaam karata hai?

\oint_{γ} (z - a)^{k} d z = 2 π i

agar

k = - 1

ho, warna

0

Convergence ka region annulus kyun hota hai, disk kyun nahi?

Analytic part

∣ z - a ∣ < R

ke andar converge karta hai; principal part

∣ z - a ∣ > r

ke bahar converge karta hai; dono sirf ring

r < ∣ z - a ∣ < R

par hold karte hain.

Principal part se singularity ka type kaise padhte hain?

Zero terms ⇒ removable; finitely many (

\leq m

) ⇒ pole of order

m

; infinitely many ⇒ essential.

Laurent coefficients ke terms mein residue kya hota hai?

Coefficient

c_{- 1}

Kya kisi function ki Laurent series unique hoti hai?

Unique per annulus hoti hai; ek hi function ki alag-alag annuli mein alag-alag Laurent series ho sakti hain.

∣ z ∣ > 1

region mein

1/ (z - 1)

ko kaise expand karte hain?

z

factor out karo:

\frac{1}{z} \cdot \frac{1}{1 - 1/ z} = \sum_{n \geq 0} z^{- n - 1}

e^{1/ z}

0

ke around principal part kya hai, aur iska kya matlab hai?

\sum_{n \geq 1} \frac{1}{n ! z ^{n}}

(infinitely many terms) ⇒ essential singularity.

Connections

Taylor series — empty principal part wala special case.
Residue theorem — $c_{- 1}$ use karta hai; Laurent iska foundation hai.
Poles and singularities — principal part se classify hote hain.
Geometric series — Laurent expansions banane ka workhorse.
Cauchy integral formula — $c_{n}$ formula dene ke liye generalise kiya gaya.
Annulus of convergence — radius of convergence ka two-radius analogue.

WHAT is a Laurent series?

WHY ek annulus, disk nahi?

Singularity ko principal part se classify karna

Worked examples

Active recall

Connections

Concept Map