Laurent series — principal part, annulus of convergence: Worked examples

4.10.4 · D3 · Maths › Advanced Topics (Elite Level) › Laurent series — principal part, annulus of convergence

Yeh page ek drill hai. Parent note ne theory build ki; yahan hum ise har tarah ke inputs ke through push karte hain jo yeh topic throw kar sakta hai. Har example se pehle: apna answer khud forecast karo. Drill ka point wohi surprise hai jab tumhara guess galat ho.

The scenario matrix

Har Laurent problem is grid ke exactly ek cell mein land karta hai. Columns hain "singularity kitni buri hai", rows hain "problem ko kya awkward banata hai".

Awkwardness ↓ \ Singularity type →	Removable (no principal part)	Pole of order $m$ (finite principal part)	Essential (infinite principal part)
Simplest / textbook centre	Ex 1	Ex 2	Ex 6
Multiple annuli, one function	—	Ex 3 (three rings)	—
Centre ≠ singularity (shifted $a$ )	—	Ex 4	—
Degenerate / limiting input	Ex 5 (limit $z \to 0$ )	Ex 7 ( $m \to \infty$ boundary)	Ex 6
Real-world / word problem	—	Ex 8 (signal decay)	—
Exam twist (product of two blow-ups)	—	Ex 9	—

"—" se mark kiye gaye cells ek neighbour mein collapse ho jaate hain (jaise ek removable singularity ka sirf ek interesting annulus hota hai, isliye use koi multi-ring row ki zaroorat nahi). Hum saare 9 examples cover karte hain, har reachable cell ko touch karte hue.

Shuru karne se pehle, ek reusable tool jo har example kaam aata hai.

Figure — Laurent series — principal part, annulus of convergence

Upar ka figure dikhata hai kyun wahi $\frac{1}{z - 1}$ do tarah se split hota hai: agar $z$ $1$ ki fence ke andar hai toh hum constant factor out karte hain; agar bahar hai toh $z$ factor out karte hain. Yeh picture apne dimag mein rakho — yeh 80% kaam hai.

Removable singularities

Worked example

Ex 1 — Cell: Removable, textbook centre. $f (z) = \frac{sin z}{z}$ about $a = 0$ . Forecast: guess karo — kya isme koi $\frac{1}{z}$ terms hain?

Steps.

Numerator ki Taylor series likho: $sin z = z - \frac{z ^{3}}{3 !} + \frac{z ^{5}}{5 !} - \dots$ . Yeh step kyun? Numerator analytic hai (ek genuine Taylor citizen, dekho Taylor series); baad mein divide karna sirf powers shift karta hai, jo trivial hai jab ek baar series mil jaaye.
Har term ko $z$ se divide karo: $\frac{s i n z}{z} = 1 - \frac{z ^{2}}{3 !} + \frac{z ^{4}}{5 !} - \dots .$ Yeh step kyun? $z$ se division har power ko ek se kam kar deta hai. Sabse chhota term $z^{1}$ tha, toh sabse chhota ho jaata hai $z^{0}$ — koi negative power nahi aata.
Principal part padho: hai hi nahi.

Result: principal part $= 0$ ⇒ removable singularity. $0$ par "hole" fake hai; $f (0) = 1$ define karne se $f$ analytic ho jaata hai. Valid hai $0 < ∣ z ∣ < \infty$ mein.

Verify: $lim_{z \to 0} \frac{s i n z}{z} = 1$ , jo constant term $c_{0} = 1$ ke barabar hai jo humne nikala. Ek removable singularity ka hamesha finite limit hota hai — sanity check pass.

Worked example

Ex 5 — Cell: Removable, limiting/degenerate input. $f (z) = \frac{1 - cos z}{z ^{2}}$ about $0$ . Forecast: denominator $z^{2}$ hai — surely double pole? Padhne se pehle guess karo.

Steps.

$cos z = 1 - \frac{z ^{2}}{2 !} + \frac{z ^{4}}{4 !} - \dots$ , toh $1 - cos z = \frac{z ^{2}}{2 !} - \frac{z ^{4}}{4 !} + \dots$ . Yeh step kyun? Leading $1$ ka cancellation — yahi poori trick hai — numerator secretly $z^{2}$ se start karta hai, denominator se match karte hue.
$z^{2}$ se divide karo: $\frac{1 - c o s z}{z ^{2}} = \frac{1}{2 !} - \frac{z ^{2}}{4 !} + \frac{z ^{4}}{6 !} - \dots .$ Yeh step kyun? Numerator aur denominator ka $z^{2}$ ek doosre ko cancel kar deta hai. "Double pole" ek illusion tha.
Principal part $= 0$ phir se ⇒ removable.

Verify: $lim_{z \to 0} \frac{1 - c o s z}{z ^{2}} = \frac{1}{2}$ (ek standard limit). Hamara constant term hai $c_{0} = \frac{1}{2 !} = \frac{1}{2}$ . ✔ Yeh degenerate case hai: bada denominator pole guarantee nahi karta — tumhe check karna hoga ki numerator matching order tak vanish hota hai ya nahi.

Poles

Worked example

Ex 2 — Cell: Pole, textbook centre. $f (z) = \frac{cos z}{z ^{3}}$ about $0$ . Forecast: kis order ka pole hai, aur residue $c_{- 1}$ kya hai?

Steps.

$cos z = 1 - \frac{z ^{2}}{2 !} + \frac{z ^{4}}{4 !} - \dots$ . Yeh step kyun? Pehle jaisi wajah — analytic numerator expand karo, phir shift karo.
$z^{3}$ se divide karo: $\frac{c o s z}{z ^{3}} = principal part \frac{1}{z ^{3}} - \frac{1}{2 !} \frac{1}{z} + analytic part \frac{z}{4 !} - \dots .$ Yeh step kyun? Har term 3 powers girta hai. Sabse negative $z^{- 3}$ hai, aur wahan ruk jaata hai (kuch aur negative nahi) — yahi finiteness ise pole banati hai, essential nahi.
Sabse negative power $z^{- 3}$ hai ⇒ pole of order 3. Residue hai $z^{- 1}$ ka coefficient: $c_{- 1} = - \frac{1}{2 !} = - \frac{1}{2}$ .

Verify: Order $m = 3$ ke pole ke liye, $Res = \frac{1}{2 !} lim_{z \to 0} \frac{d ^{2}}{d z ^{2}} (z^{3} f (z)) = \frac{1}{2} cos^{''} (0) = \frac{1}{2} (- cos 0) = - \frac{1}{2}$ . ✔ Match karta hai (yeh Residue theorem shortcut hai).

Worked example

Ex 3 — Cell: Pole, multiple annuli. $f (z) = \frac{1}{z ( z - 3 )}$ about $0$ — teeno rings karo. Forecast: ek function ke kitne distinct Laurent series yahan hain? Number guess karo.

Singularities $z = 0$ aur $z = 3$ par hain. Woh plane ko ( $0$ ke centre par) rings mein cut karte hain: $0 < ∣ z ∣ < 3$ aur $∣ z ∣ > 3$ . Figure dekho.

Steps.

Partial fractions: $\frac{1}{z ( z - 3 )} = \frac{- 1/3}{z} + \frac{1/3}{z - 3}$ . Yeh step kyun? Har simple piece ek single fence hai, toh woh ek clean geometric series ban jaata hai jab hum sahi expansion direction chunein (Ex-1 picture).
Ring A: $0 < ∣ z ∣ < 3$ . $\frac{- 1/3}{z}$ piece already ek single negative power hai (ise rakhho). Doosre ke liye, $∣ z ∣ < 3$ toh $∣ z /3∣ < 1$ : $\frac{1/3}{z - 3} = \frac{1/3}{- 3} \cdot \frac{1}{1 - z /3} = - \frac{1}{9} \sum_{n \geq 0} (\frac{z}{3})^{n} .$ Yeh step kyun? Humne badi quantity $3$ factor out ki toh ratio $z /3$ ka modulus $< 1$ hai — yeh parent note ka "wrong way round" pitfall hai, jo avoid ho gaya. Full series: $f (z) = - \frac{1}{3 z} - \frac{1}{9} n \geq 0 \sum \frac{z ^{n}}{3 ^{n}} = - \frac{1}{3 z} - \frac{1}{9} - \frac{z}{27} - \dots$ . Principal part $= - \frac{1}{3 z}$ ⇒ simple pole at $0$ , residue $c_{- 1} = - \frac{1}{3}$ .
Ring B: $∣ z ∣ > 3$ . Ab $∣3/ z ∣ < 1$ hai, toh $z$ factor out karo: $\frac{1/3}{z - 3} = \frac{1/3}{z} \cdot \frac{1}{1 - 3/ z} = \frac{1}{3 z} \sum_{n \geq 0} (\frac{3}{z})^{n} = \sum_{n \geq 0} \frac{3 ^{n}}{3} z^{- n - 1} .$ Yeh step kyun? Outer ring mein $z$ badi quantity hai, toh $3/ z$ chhota ratio hai. $- \frac{1}{3 z}$ ko add karo: $f (z) = - \frac{1}{3 z} + n \geq 0 \sum \frac{3 ^{n - 1}}{1} z^{- n - 1} = n \geq 1 \sum (\frac{3 ^{n - 1}}{1}) z^{- n - 1} + \dots$ . $n = 0$ term hai $\frac{1}{3 z}$ , exactly $- \frac{1}{3 z}$ cancel karta hai, sirf deeper negative powers bachti hain: $f (z) = \frac{1}{z ^{2}} + \frac{3}{z ^{3}} + \frac{9}{z ^{4}} + \dots$ . Saari negative powers.

Result: ek function ke liye do distinct Laurent series (rings A aur B).

Verify: $0$ par residue (ring A se, woh ring jo $0$ ko encircle karti hai) hai $c_{- 1} = - \frac{1}{3}$ . Cross-check: $Res_{z = 0} = lim_{z \to 0} z \cdot \frac{1}{z ( z - 3 )} = \frac{1}{0 - 3} = - \frac{1}{3}$ . ✔ Ring B mein $z^{- 1}$ ka coefficient $0$ hai — sahi hai, kyunki ring B us residue-theorem sense mein residue ke around wind nahi karta jo us ring ke liye use hota hai.

Worked example

Ex 4 — Cell: Pole, centre ≠ singularity. $f (z) = \frac{1}{z - 2}$ about $a = 1$ , ring $0 < ∣ z - 1∣ < 1$ mein. Forecast: singularity $2$ par hai, lekin hum $1$ ke baare mein expand kar rahe hain. Kya $(z - 1)$ ke negative powers honge?

Steps.

Sab kuch variable $w = z - 1$ mein rewrite karo (toh centre origin ban jaaye): $f = \frac{1}{( w + 1 ) - 2} = \frac{1}{w - 1} .$ Yeh step kyun? Laurent hamesha "centre ke baare mein" hota hai; rename karne se centre $w = 0$ ho jaata hai toh hum apna ek tool directly use kar sakte hain.
Nearest singularity $w = 1$ par hai, yaani $∣ z - 1∣ = 1$ . $0 < ∣ z - 1∣ < 1$ mein $∣ w ∣ < 1$ hai, toh constant factor out karo: $\frac{1}{w - 1} = \frac{- 1}{1 - w} = - \sum_{n \geq 0} w^{n} = - 1 - w - w^{2} - \dots .$ Yeh step kyun? $∣ w ∣ < 1$ se $u = w$ ek valid ratio banta hai; $- 1$ factor out karne se $\frac{1}{1 - u}$ ki shape milti hai.
$w = z - 1$ back-substitute karo: $f (z) = - \sum_{n \geq 0} (z - 1)^{n} .$

Result: Is ring mein koi principal part nahi — centre $1$ $f$ ka ek ordinary analytic point hai, toh us ke baare mein hum pure Taylor series lete hain. Singularity boundary $∣ z - 1∣ = 1$ par rehti hai, andar nahi.

Verify: $z = 1$ par sum hai $- 1$ , aur $f (1) = \frac{1}{1 - 2} = - 1$ . ✔ Sabak: function ka pole sirf us ring mein "principal-part" pole hota hai jo use actually encircle kare.

Essential singularities aur limiting behaviour

Worked example

Ex 6 — Cell: Essential. $f (z) = z^{2} e^{1/ z}$ about $0$ . Forecast: finite ya infinite principal part? Aur residue kya hai?

Steps.

$e^{1/ z} = \sum_{n \geq 0} \frac{1}{n !} z^{- n} = 1 + \frac{1}{z} + \frac{1}{2 ! z ^{2}} + \frac{1}{3 ! z ^{3}} + \dots$ expand karo. Yeh step kyun? $e^{w} = \sum w^{n} / n!$ har jagah; $w = 1/ z$ substitute karna $z \neq = 0$ ke liye valid hai aur turant infinitely many negative powers deta hai.
$z^{2}$ se multiply karo: $z^{2} e^{1/ z} = z^{2} + z + \frac{1}{2 !} + \frac{1}{3 ! z} + \frac{1}{4 ! z ^{2}} + \dots .$ Yeh step kyun? $z^{2}$ se multiply karna har power ko 2 se utha deta hai. Poori tail upar shift hoti hai lekin infinitely long negative tail nahi kati.
Abhi bhi infinitely many negative powers hain ⇒ essential singularity. Residue $c_{- 1} = \frac{1}{3 !} = \frac{1}{6}$ .

Verify: $c_{- 1}$ hai $z^{- 1}$ ka coefficient: woh $e^{1/ z}$ ke $\frac{1}{3 ! z ^{3}}$ term se aaya times $z^{2}$ , jo deta hai $\frac{1}{3 !} z^{- 1}$ . Toh $c_{- 1} = \frac{1}{6}$ . ✔ Note karo ki $z^{2}$ se multiply karne se essential singularity pole mein nahi badi — sirf finite truncation kar sakti thi, aur $e^{1/ z}$ kabhi truncate nahi hota.

Worked example

Ex 7 — Cell: Pole aur essential ke beech boundary. $g (z) = \frac{1}{z ^{5}}$ vs. $h (z) = \frac{1}{sin ( 1/ z )}$ -style thinking about $0$ compare karo. Forecast: kaun sa "sirf bada pole" hai aur kaun sa "genuinely essential"? Guess karo.

Steps.

$g (z) = z^{- 5}$ . Principal part sirf ek term $z^{- 5}$ hai; yeh ruk jaata hai. Yeh step kyun? Order $m$ ka pole define hota hai sabse negative power ke finite hone se. Yahan $m = 5$ , done.
$sin (1/ z) = \sum_{n \geq 0} \frac{( - 1 ) ^{n}}{( 2 n + 1 )!} z^{- (2 n + 1)} = \frac{1}{z} - \frac{1}{3 ! z ^{3}} + \dots$ jaisi function ke liye negative tail kabhi nahi rukti. Yeh step kyun? Yeh limiting case hai $m \to \infty$ : ek pole "ek essential singularity hai jo finite hoti hai." Jab order infinite ho jaaye, hum use essential naam dete hain.
Toh classification ek clean dichotomy hai: largest negative power finite ⇒ us order ka pole; finite nahi ⇒ essential. Removable sirf "order $0$ " hai.

Verify: $g$ ke liye, residue $c_{- 1} = 0$ ( $z^{- 5}$ mein koi $z^{- 1}$ term nahi), pure power ke saath consistent. $sin (1/ z)$ ke liye, residue $c_{- 1} = 1$ ( $\frac{1}{z}$ term). Dono seedha series se padh lo. ✔

Real-world aur exam-twist

Worked example

Ex 8 — Cell: Word problem. Ek stable signal amplifier ka transfer function hai $H (z) = \frac{1}{z - \frac{1}{2}}$ . Region $∣ z ∣ > \frac{1}{2}$ (pole ke bahar "causal" region) ke liye, $H$ ko $1/ z$ ki powers mein expand karo aur impulse-response coefficients $h_{n}$ (coefficient of $z^{- n - 1}$ ) padho. Forecast: kya coefficients $n$ badhne ke saath grow karenge ya decay? (Ek stable amplifier decay karna chahiye.)

Steps.

Hum $∣ z ∣ > \frac{1}{2}$ mein hain, toh $\frac{1/2}{z} < 1$ . Badi quantity $z$ factor out karo: $H (z) = \frac{1}{z} \cdot \frac{1}{1 - \frac{1}{2 z}} = \frac{1}{z} \sum_{n \geq 0} (\frac{1}{2 z})^{n} = \sum_{n \geq 0} \frac{1}{2 ^{n}} z^{- n - 1} .$ Yeh step kyun? Pole ke bahar, $z$ dominate karta hai, toh $\frac{1}{2 z}$ chhota ratio hai — yeh woh ring hai jo singularity encircle karti hai, exactly wahan physical (causal) expansion rehti hai.
$h_{n} = \frac{1}{2 ^{n}}$ padho: $h_{0} = 1, h_{1} = \frac{1}{2}, h_{2} = \frac{1}{4}, \dots$ . Yeh step kyun? $z^{- n - 1}$ ka coefficient hai hi is convention mein $n$ -th impulse-response sample.

Result: $h_{n} = (1/2)^{n}$ — ek decaying response ⇒ stable. Residue $c_{- 1} = h_{0} = 1$ hai $Res_{z = 1/2} H = 1$ .

Verify: $\sum_{n \geq 0} h_{n} = \sum (1/2)^{n} = \frac{1}{1 - 1/2} = 2 = H (1)$ ? Check karo: $H (1) = \frac{1}{1 - 1/2} = 2$ . ✔ Series ko $z = 1$ par sum karne se function ki value reproduce hoti hai — aur finite total energy stability confirm karti hai.

Worked example

Ex 9 — Cell: Exam twist, product of two blow-ups. $f (z) = \frac{e ^{z}}{z ^{2} ( z - 1 )}$ about $0$ , ring $0 < ∣ z ∣ < 1$ mein. $0$ par residue nikalo. Forecast: $0$ par pole order $2$ hai; grind karne se pehle residue guess karo.

Steps.

$0 < ∣ z ∣ < 1$ mein do analytic ingredients expand karo: $e^{z} = 1 + z + \frac{z ^{2}}{2} + \dots$ aur $\frac{1}{z - 1} = \frac{- 1}{1 - z} = - \sum_{n \geq 0} z^{n} = - 1 - z - z^{2} - \dots$ (valid kyunki $∣ z ∣ < 1$ ). Yeh step kyun? Dono yahan honest Taylor series hain; sirf singular factor $\frac{1}{z ^{2}}$ hai, jise hum alag rakhte hain aur last mein apply karte hain.
Do power series multiply karo, $z^{1}$ tak terms rakhte hue (hume $z^{2}$ se divide karne ke baad $z^{- 1}$ tak pohonchne ke liye enough chahiye): $e^{z} \cdot \frac{1}{z - 1} = (1 + z + \dots) (- 1 - z - \dots) = - 1 - 2 z + O (z^{2}) .$ Yeh step kyun? Residue $f$ mein $z^{- 1}$ ka coefficient hai; $z^{2}$ se divide karna powers 2 se shift karta hai, toh hume is product mein $z^{1}$ ka coefficient chahiye.
$z^{2}$ se divide karo: $f (z) = \frac{- 1 - 2 z + O ( z ^{2} )}{z ^{2}} = - \frac{1}{z ^{2}} - \frac{2}{z} + O (1) .$ Yeh step kyun? Ab principal part visible hai: order-2 pole, aur $c_{- 1} = - 2$ .

Result: $0$ par residue hai $c_{- 1} = - 2$ .

Verify: Order-2 formula: $Res = lim_{z \to 0} \frac{d}{d z} (z^{2} f) = lim_{z \to 0} \frac{d}{d z} \frac{e ^{z}}{z - 1} = lim \frac{e ^{z} ( z - 1 ) - e ^{z}}{( z - 1 ) ^{2}} = \frac{1 \cdot ( - 1 ) - 1}{1} = - 2$ . ✔ (Yeh Cauchy integral formula machinery disguise mein hai.)

Recall Poori drill ki one-line summary

Do sawaal ::: (1) removable / pole-of-order- $m$ / essential — dekhte hain kahan negative tail rukti hai; (2) kaun si ring — $u$ aisa chunein ki $∣ u ∣ < 1$ wahan, phir expand. Ek function ke kai series kyun ho sakte hain ::: Alag rings, singularities se separated, alag valid geometric expansions force karti hain. Jab pole ho toh fast residue ::: $Res = \frac{1}{( m - 1 )!} lim_{z \to a} \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - a)^{m} f (z)]$ .

Active recall

0

ke baare mein

\frac{s i n z}{z}

mein principal part kya hai?

Zero — yeh removable singularity hai, sabse chhota power

z^{0}

hai.

0

par

\frac{c o s z}{z ^{3}}

ka residue?

c_{- 1} = - \frac{1}{2}

0

ke baare mein

\frac{1}{z ( z - 3 )}

ke kitne distinct Laurent series hain?

Do — ek

0 < ∣ z ∣ < 3

mein, ek

∣ z ∣ > 3

mein.

0 < ∣ z - 1∣ < 1

mein

1

ke baare mein

\frac{1}{z - 2}

expand karne par, kya principal part hai?

Nahi — yeh pure Taylor series hai; singularity boundary par hai, encircle nahi hoti.

e^{1/ z}

z^{2}

se multiply karne se essential singularity remove kyun nahi hoti?

Negative tail infinite hai; ek finite power shift use truncate nahi kar sakta.

∣ z ∣ > \frac{1}{2}

mein

\frac{1}{z - 1/2}

ke impulse coefficients?

h_{n} = (1/2)^{n}

— decaying, hence stable.

0

par

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

ka residue?

- 2