Residues and poles: Worked examples — Hinglish deep dive

4.10.5 · D3 · Maths › Advanced Topics (Elite Level) › Residues and poles

Residues and poles ka companion note hai yeh. Yahaan hum nayi rules nahi seekhte — woh rules drill karte hain jo tumhe pehle se pata hain jab tak koi bhi case surprise na kar sake. Topic jo bhi trap laga sakta hai, hum usse purpose se lagate hain aur defuse karte hain.

Shuru karne se pehle: do reminders, phir chaar tools.

Recall Laurent series kya hota hai (cover karke recall karo)

Ek isolated singularity $z_{0}$ ke paas har function ko ek do-taraf ki power series ke roop mein likha ja sakta hai $f (z) = \sum_{n = - \infty}^{\infty} a_{n} (z - z_{0})^{n} = principal (negative) part \dots + \frac{a _{- 2}}{( z - z _{0} ) ^{2}} + \frac{a _{- 1}}{z - z _{0}} + ordinary Taylor part a_{0} + a_{1} (z - z_{0}) + \dots .$ Negative-power wale terms describe karte hain kaise yeh blow up hota hai; residue sirf is series ka ek coefficient hai, yaani $a_{- 1}$ . Poora construction Laurent series mein hai.

Recall Chaar residue tools (cover karke recall karo)

Simple pole: $Res (f, z_{0}) = lim_{z \to z_{0}} (z - z_{0}) f (z)$ — $(z - z_{0})$ factor "blow-up cancel karta hai", phir plug in karo.
Quotient shortcut ( $p / q$ ka simple pole): $\frac{p ( z _{0} )}{q ^{'} ( z _{0} )}$ jab $q (z_{0}) = 0, q^{'} (z_{0}) \neq = 0, p (z_{0}) \neq = 0$ .
Order- $m$ pole: $\frac{1}{( m - 1 )!} lim_{z \to z_{0}} \frac{d ^{m - 1}}{d z ^{m - 1}} [(z - z_{0})^{m} f (z)]$ .
Residue theorem: ek positively-oriented (counter-clockwise) closed contour $C$ ke liye, $\oint_{C} f d z = 2 π i \sum Res (f, z_{j})$ , sirf $C$ ke andar wale poles. (Loop ulta ghumao → $- 1$ se multiply karo.)

Do words jinhe hum zyada use karte hain. ==Pole of order $m$ == matlab Laurent series mein sabse bura term $a_{- m} (z - z_{0})^{- m}$ hai jahan $a_{- m} \neq = 0$ aur isse bura kuch nahi; order $1$ ko simple pole kehte hain. Ek contour $C$ bas ek closed loop hai jiske around integrate karte hain (dekho Contour integration); "inside" matlab woh region jo loop encircle karta hai, aur jab tak bataya na jaye hum hamesha ise counter-clockwise chalate hain — woh sign convention $+ 2 π i$ mein baked hai.

Scenario matrix

Har residue problem jo tumhe milegi woh inhi cells mein se ek mein aati hai. Neeche ke examples cell ke hisaab se label kiye hain taaki tum dekh sako ki poora board cover hai.

Cell	Kya alag banata hai	Kaun sa example
A. Simple pole, cancel-and-plug	ek $(z - z_{0})$ factor, plug in karo	Ex 1
B. Simple pole, quotient shortcut	$p / q$ jahan factoring ugly ho	Ex 2
C. Higher-order pole (derivative)	$m \geq 2$ , $m - 1$ baar differentiate karna hoga	Ex 3
D. Pole hidden by numerator ka zero	order zyada lagta hai, cancel hota hai	Ex 4
E. Degenerate: pole NAHI hai (removable / essential)	formulas silently fail ho jaate hain	Ex 5
F. Sign/location: $C$ ke andar kaun se poles hain?	kuch poles enclosed, kuch nahi	Ex 6
G. Real integral, upper half plane	contour close karo, arc discard karo	Ex 7
H. Word problem (real-world model)	translate karo phir compute karo	Ex 8
I. Exam twist: trig poles ki ek infinite family	$sin z = 0$ har $nπ$ par	Ex 9

Mushkil columns hain D, E, F — wahan students marks khote hain. Hum inhe seedha attack karte hain.

Cell A — Simple pole, cancel and plug

Worked example Ex 1 (Cell A)

$Res (\frac{z + 4}{( z - 2 ) ( z + 3 )}, 2)$ nikalo.

Forecast: aage padhne se pehle ek single number guess karo. (Numerator $z = 2$ par $6$ hai; doosra factor $z = 2$ par $5$ hai…)

Pole confirm karo. $z = 2$ par factor $(z - 2)$ vanish hota hai; baaki finite aur non-zero hai → simple pole. Yeh step kyun? Formula choose karne se pehle order pata hona chahiye. Ek vanishing linear factor = order 1.
$(z - 2)$ se multiply karo: $(z - 2) f = \frac{z + 4}{z + 3}$ . Yeh step kyun? $(z - 2)$ factor blow-up cancel karta hai, kuch finite chodta hai $z = 2$ par.
$z \to 2$ jaane do: $\frac{2 + 4}{2 + 3} = \frac{6}{5}$ . Yeh step kyun? Pole cancel hone ke baad, "the limit" bas substitution hai.

$Res = \frac{6}{5}$

Verify: independently doosre pole $z = - 3$ par residue compute karo: $\frac{- 3 + 4}{- 3 - 2} = - \frac{1}{5}$ . Cross-check ke liye partial fractions karte hain: $\frac{z + 4}{( z - 2 ) ( z + 3 )} = \frac{6/5}{z - 2} + \frac{- 1/5}{z + 3}$ . $\frac{1}{z - 2}$ aur $\frac{1}{z + 3}$ ke coefficients directly padhne se exactly do residues $\frac{6}{5}$ aur $- \frac{1}{5}$ milte hain, jo hamara answer confirm karte hain. ✔

Cell B — Quotient shortcut

Worked example Ex 2 (Cell B)

$Res (\frac{cos z}{z ^{2} + 9}, 3 i)$ nikalo.

Forecast: denominator factor hota hai $(z - 3 i) (z + 3 i)$ ke roop mein, toh yeh simple hai. Kya answer real hoga ya imaginary?

$p = cos z$ , $q = z^{2} + 9$ identify karo. Check karo $q (3 i) = (3 i)^{2} + 9 = - 9 + 9 = 0$ . Achha, yeh hai pole. Yeh step kyun? Shortcut $p / q^{'}$ ko $q (z_{0}) = 0$ chahiye — verify karo, assume mat karo.
$q^{'} (z) = 2 z$ , toh $q^{'} (3 i) = 6 i \neq = 0$ ; aur $p (3 i) = cos (3 i) \neq = 0$ . Yeh step kyun? Dono conditions ( $q^{'} \neq = 0$ , $p \neq = 0$ ) guarantee karte hain simple pole, shortcut use karne ki permission dete hain.
Apply karo: $Res = \frac{p ( 3 i )}{q ^{'} ( 3 i )} = \frac{cos ( 3 i )}{6 i}$ . Yeh step kyun? $cos z$ ko factor karne se zyada fast hai; yahi toh shortcut ka poora point hai.
$cos (3 i) = cosh 3$ se simplify karo: $Res = \frac{cosh 3}{6 i} = - \frac{i cosh 3}{6}$ . Yeh step kyun? $cos (i x) = cosh x$ ek "imaginary-input cosine" ko real number mein badalta hai.

Verify: numerically $cosh 3 \approx 10.0677$ , toh $Res \approx - 1.678 i$ — purely imaginary, jaise expected hai real-even integrand ke complex pole ke liye. ✔

Cell C — Higher-order pole (derivative formula)

Worked example Ex 3 (Cell C)

$Res (\frac{e ^{z}}{( z - 1 ) ^{3}}, 1)$ nikalo.

Forecast: order $m = 3$ hai, toh hum $m - 1 = 2$ baar differentiate karenge. Guess karo: kuch $e^{1}$ se bana hoga.

$(z - 1)^{3}$ se multiply karo: $(z - 1)^{3} f = e^{z}$ . Yeh step kyun? $(z - z_{0})^{m}$ se multiply karne se har negative power kill ho jaata hai, ek clean Taylor series bachta hai.
$m - 1 = 2$ baar differentiate karo: $\frac{d ^{2}}{d z ^{2}} e^{z} = e^{z}$ . Yeh step kyun? $a_{- 3} (z - 1)^{- 3} + a_{- 2} (z - 1)^{- 2} + a_{- 1} (z - 1)^{- 1} + \dots$ mein, $(z - 1)^{3}$ se multiply karne par $a_{- 1}$ $(z - 1)^{2}$ slot mein aa jaata hai; do derivatives + $2!$ use karke ise extract karte hain.
$z = 1$ par evaluate karo aur $(m - 1)! = 2!$ se divide karo: $Res = \frac{1}{2 !} e^{1} = \frac{e}{2}$ . Yeh step kyun? Factorial woh $\frac{1}{2 !}$ undo karta hai jo Taylor coefficients carry karte hain.

$Res = \frac{e}{2}$

Verify: directly Laurent-expand karo: $e^{z} = e \cdot e^{z - 1} = e \sum \frac{( z - 1 ) ^{n}}{n !}$ , toh $\frac{e ^{z}}{( z - 1 ) ^{3}} = e \sum \frac{( z - 1 ) ^{n - 3}}{n !}$ . $(z - 1)^{- 1}$ term $n = 2$ par hai: coefficient $e /2! = e /2$ . ✔

Cell D — Numerator ka ek zero order kam kar deta hai

Yeh wala sneaky hai. Denominator lagta hai order 2 ka, lekin numerator wahan vanish hota hai aur ek factor cancel karta hai.

Worked example Ex 4 (Cell D)

$Res (\frac{z - 1}{( z - 1 ) ^{2} ( z + 2 )}, 1)$ nikalo.

Forecast: $z = 1$ par denominator mein $(z - 1)^{2}$ hai, toh yeh lagging order 2 ka. Lekin numerator $(z - 1)$ hai… sach mein order?

Pehle shared factor cancel karo: $f = \frac{z - 1}{( z - 1 ) ^{2} ( z + 2 )} = \frac{1}{( z - 1 ) ( z + 2 )}$ . Yeh step kyun? Order reduced form se decide hota hai. Common factor pole nahi, woh ek hole hai jo fill ho jaata hai.
Ab sirf ek $(z - 1)$ bacha hai → simple pole. Multiply karo aur plug karo: $(z - 1) f = \frac{1}{z + 2} \to \frac{1}{3}$ . Yeh step kyun? Cell A par reduce ho gaya; sabse simple sahi tool use karo.

$Res = \frac{1}{3}$

Verify: agar tumne galti se order-2 formula unreduced $f$ par use kiya hota: $\frac{1}{1 !} lim \frac{d}{d z} [(z - 1)^{2} f] = lim \frac{d}{d z} \frac{z - 1}{z + 2} = lim \frac{( z + 2 ) - ( z - 1 )}{( z + 2 ) ^{2}} = \frac{3}{9} = \frac{1}{3}$ . Wohi answer — toh yeh formula $m$ overestimate karne ke liye robust hai, lekin sirf tab jab differentiate sahi se karo. ✔

Cell E — Degenerate inputs: jab yeh pole NAHI hota

Formulas tumhe warn nahi karte; removable aur essential singularities khud pehchanni padti hain.

Worked example Ex 5 (Cell E — removable AUR essential)

(a) $Res (\frac{sin z}{z}, 0)$ . (b) $Res (e^{1/ z}, 0)$ .

Forecast: inme se ek $0$ hai, ek $1$ . Kaun sa kaun sa hai?

Part (a) — removable:

Expand karo: $sin z = z - \frac{z ^{3}}{6} + \dots$ , toh $\frac{sin z}{z} = 1 - \frac{z ^{2}}{6} + \dots$ . Yeh step kyun? Laurent series ground truth hai. Koi negative powers nazar nahi aate.
$a_{- 1} = 0$ padho. Koi $1/ z$ term nahi hai. Yeh step kyun? Residue define hota hai $a_{- 1}$ ke roop mein; yahaan clearly $0$ hai.

$Res = 0$

Part (b) — essential:

Expand karo: $e^{1/ z} = \sum_{n = 0}^{\infty} \frac{1}{n !} z^{- n} = 1 + \frac{1}{z} + \frac{1}{2 ! z ^{2}} + \dots$ . Yeh step kyun? Infinitely many negative powers → essential; koi finite $m$ nahi, toh pole formulas use nahi ho sakte ( $z^{m} e^{1/ z}$ kabhi analytic nahi banta).
$z^{- 1}$ coefficient $n = 1$ hai: $a_{- 1} = \frac{1}{1 !} = 1$ . Yeh step kyun? Essential singularities ke liye sirf ek method hai — series se $a_{- 1}$ padho.

$Res = 1$

Verify: (a) $lim_{z \to 0} z \cdot \frac{s i n z}{z} = lim sin z = 0$ — consistent hai removable point ke saath (residue $0$ ). (b) residue theorem same series par apply karo: $\oint_{∣ z ∣ = 1} e^{1/ z} d z$ equals $2 π i$ times coefficient $a_{- 1} = 1$ , kyunki har doosri power $z^{- n}$ (jab $n \neq = 1$ ) circle ke around integrate karke $0$ deta hai. Toh loop integral $2 π i \cdot 1 = 2 π i$ hai, exactly $a_{- 1} = 1$ match karta hai. ✔

Cell F — Contour ke andar kaun se poles hain?

Residue theorem sirf enclosed poles add karta hai. Location (imaginary part ka sign, centre se distance) membership decide karta hai.

Worked example Ex 6 (Cell F)

$\oint_{∣ z ∣ = 2} \frac{z}{( z - 1 ) ( z - 3 )} d z$ compute karo (counter-clockwise).

Forecast: $z = 1$ aur $z = 3$ par poles hain. Circle ka radius $2$ hai. Kaun andar hai?

Neeche ka figure contour aur do poles dikhata hai, taaki tum membership dekh sako sirf assert karne ki jagah. Ise left-to-right padho:

cyan circle loop $∣ z ∣ = 2$ hai; iske upar chota cyan arrow anticlockwise point karta hai — woh anticlockwise direction exactly wahi hai jo theorem ke sign ko $+ 2 π i$ banata hai;
$z = 1$ par filled amber dot circle ke andar baitha hai → uska residue count hota hai;
$z = 3$ par hollow white dot circle ke bahar baitha hai → yeh ignore hota hai.

Poles locate karo: $z = 1$ (origin se distance $1$ ) aur $z = 3$ (distance $3$ ). Yeh step kyun? " $∣ z ∣ = 2$ ke andar" matlab $∣ z_{0} ∣ < 2$ . Toh $z = 1$ in hai (figure mein amber dot), $z = 3$ out hai (cyan circle ke bahar hollow white dot).
Enclosed pole $z = 1$ par residue: $lim_{z \to 1} (z - 1) f = \frac{1}{1 - 3} = - \frac{1}{2}$ . Yeh step kyun? Sirf enclosed poles contribute karte hain; $z = 3$ ko bilkul ignore karo.
Theorem apply karo (contour counter-clockwise hai, toh $+ 2 π i$ use karo): $\oint = 2 π i (- \frac{1}{2}) = - π i$ . Yeh step kyun? $\oint_{C} f = 2 π i \sum_{inside} Res$ positively-oriented loop ke liye.

$\oint_{∣ z ∣ = 2} f d z = - π i$

Verify: contour $∣ z ∣ = 4$ tak widen karo (dono poles andar) aur $Res (f, 3) = \frac{3}{3 - 1} = \frac{3}{2}$ add karo: total $2 π i (- \frac{1}{2} + \frac{3}{2}) = 2 π i$ . Woh bada loop radius-2 loop se alag value dega, aur deta hai — confirm karta hai ki radius $2$ par sirf $z = 1$ count karta hai, $- π i$ deta hai. ✔

Cell G — Contour close karke real integral

Worked example Ex 7 (Cell G)

$\int_{- \infty}^{\infty} \frac{d x}{x ^{4} + 1}$ compute karo.

Forecast: saare chaar poles unit circle par $4 5^{\circ}, 13 5^{\circ}, 22 5^{\circ}, 31 5^{\circ}$ angles par baithe hain. Sirf top two count karte hain. Guess karo answer $π$ ka ek clean multiple hai.

Figure dikhata hai kyun hum curved part throw away kar sakte hain. Ise ek story ki tarah padho:

real axis ke saath flat cyan segment woh integral hai jo hum actually chahte hain, left-to-right traverse kiya;
radius $R$ ka cyan semicircular arc loop ko upper half plane se close karta hai (anticlockwise, $+ 2 π i$ match karta hai);
do filled amber dots ( $z = e^{iπ /4}, e^{i 3 π /4}$ ) enclosed poles hain; axis ke neeche do hollow white dots skip ho jaate hain;
dashed white unit circle sirf ek guide hai jo dikhata hai sare chaar poles ka modulus $1$ hai.

Poles nikalo: $z^{4} = - 1 = e^{iπ}$ , toh $z_{k} = e^{i (π + 2 π k) /4}$ , yaani angles $\frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}$ . Yeh step kyun? $z^{4} + 1 = 0$ solve karne se har pole place hota hai; woh jo positive imaginary part ke saath hain (amber dots, upper half) arc ke andar enclosed hain.
Dikhaao ki arc contribution vanish ho jaata hai jab $R \to \infty$ . Arc par $∣ z ∣ = R$ , toh triangle inequality se $∣ z^{4} + 1∣ \geq ∣ z^{4} ∣ - 1 = R^{4} - 1$ , isliye integrand $\frac{1}{z ^{4} + 1} \leq \frac{1}{R ^{4} - 1}$ satisfy karta hai. Arc ek half-circle hai radius $R$ ka, jiski length $\frac{1}{2} (2 π R) = π R$ hai. Yeh step kyun? Hume prove karna hai ki curved piece gayab ho jaata hai, warna "close the contour" unjustified hai. $R^{4} - 1$ reverse triangle inequality $∣ a + b ∣ \geq ∣ a ∣ - ∣ b ∣$ se aata hai; length $π R$ bas half circle ki circumference $2 π R$ hai.
Arc integral ko (integrand ka max) $\times$ (length) se bound karo: $\int_{arc} \leq \frac{1}{R ^{4} - 1} \cdot π R = \frac{π R}{R ^{4} - 1} \to 0$ jab $R \to \infty$ (numerator $R$ ki tarah grow karta hai, denominator $R^{4}$ ki tarah). Yeh step kyun? Yeh ML-inequality hai: ek integral zyada se zyada apne integrand ki peak size times path length ho sakta hai. Kyunki yeh $0$ par tend karta hai, closed loop akele real-line integral ke barabar hai.
Quotient shortcut use karo $Res = \frac{p}{q ^{'}} = \frac{1}{4 z ^{3}}$ ek pole par. Kyunki wahan $z^{4} = - 1$ hai, $z^{3} = z^{4} / z = - 1/ z$ , toh har pole par $Res = \frac{1}{4 z ^{3}} = \frac{1}{4 ( - 1/ z )} = - \frac{z}{4}$ . Yeh step kyun? $p = 1, q = z^{4} + 1, q^{'} = 4 z^{3}$ ; $z^{3} = - 1/ z$ likhne se cube ek clean linear expression $- z /4$ mein badal jaata hai jo har enclosed pole par evaluate hota hai.
Do enclosed poles $z_{1} = e^{iπ /4}, z_{2} = e^{i 3 π /4}$ par evaluate karo: $- \frac{1}{4} (z_{1} + z_{2}) = - \frac{1}{4} (\frac{2}{2} (1 + i) + \frac{2}{2} (- 1 + i)) = - \frac{1}{4} (2 i) = - \frac{2}{4} i$ . Yeh step kyun? Residue theorem enclosed poles ke residues add karta hai; hum har ek ko $- z /4$ mein plug karte hain aur sum karte hain.
Integral $= 2 π i \cdot (- \frac{2}{4} i) = \frac{π 2}{2} = \frac{π}{2}$ . Yeh step kyun? $2 π i \times (enclosed residues ka sum)$ ; do $i$ multiply hokar $- 1$ dete hain, sign flip karke positive karte hain.

$\int_{- \infty}^{\infty} \frac{d x}{x ^{4} + 1} = \frac{π}{2} \approx 2.2214$

Verify: integrand positive aur even hai; numeric estimate $\approx 2.221$ deta hai, $π / 2$ match karta hai. ✔

Cell H — Word problem (ek real-world model)

Worked example Ex 8 (Cell H)

Ek signal ka frequency response $H (x) = \frac{1}{x ^{2} + a ^{2}}$ ( $a > 0$ , ek bandwidth) se model kiya gaya hai. Uska total "energy weight" $E = \int_{- \infty}^{\infty} \frac{d x}{x ^{2} + a ^{2}}$ hai. $E$ ko $a$ ke terms mein nikalo, phir $a = 2$ ke liye evaluate karo.

Forecast: zyada bandwidth $a$ smaller peak dega lekin area simple reh sakta hai. Guess karo $E \propto 1/ a$ .

$\frac{1}{x ^{2} + a ^{2}}$ ke poles: $z^{2} + a^{2} = 0 \Rightarrow z = \pm ia$ solve karo. Upper half plane mein close karne par sirf $z = ia$ enclosed hota hai. Yeh step kyun? Same closing argument Ex 7 jaise. Arc par $∣ z^{2} + a^{2} ∣ \geq R^{2} - a^{2}$ , toh integrand $\leq \frac{1}{R ^{2} - a ^{2}}$ ; arc length $π R$ se multiply karne par $\frac{π R}{R ^{2} - a ^{2}} \to 0$ milta hai. Toh real-line integral $2 π i \times$ (upper-half residues) ke barabar hai.
$z = ia$ par residue (quotient shortcut, $p = 1$ , $q = z^{2} + a^{2}$ , $q^{'} = 2 z$ ): $\frac{1}{2 ( ia )} = \frac{1}{2 ia}$ . Yeh step kyun? Simple pole ( $q^{'} (ia) = 2 ia \neq = 0$ ), toh $p / q^{'}$ instant hai.
$E = 2 π i \cdot \frac{1}{2 ia} = \frac{π}{a}$ . Yeh step kyun? $i$ cancel ho jaate hain, ek real, positive area bachta hai — jo physical energy honi chahiye.
$a = 2$ ke liye: $E = \frac{π}{2} \approx 1.5708$ . Yeh step kyun? Given bandwidth ko general formula mein substitute karo.

$E = \frac{π}{a}, E (a = 2) = \frac{π}{2}$

Verify: units/sanity: bada $a$ → chhota $E$ (broader, flatter response kam weight concentrate karta hai), intuition match karta hai; $E > 0$ jaise energy honi chahiye; aur symbolic integration confirm karta hai $\int_{- \infty}^{\infty} \frac{d x}{x ^{2} + a ^{2}} = \frac{π}{a}$ . ✔

Cell I — Exam twist: poles ki ek infinite family (trig)

Worked example Ex 9 (Cell I)

Kisi bhi integer $n$ ke liye $Res (\frac{1}{sin z}, z_{0} = nπ)$ nikalo.

Forecast: $sin z = 0$ har $z = nπ$ par — infinitely many simple poles. Answer sirf sign ke through $n$ par depend karna chahiye.

$f = \frac{p}{q}$ likho jahan $p = 1, q = sin z$ . Check karo $q (nπ) = sin (nπ) = 0$ → $π$ ke har multiple par pole. Yeh step kyun? Classify karne se pehle singularities confirm karo; yahaan real axis ke saath countably many hain.
$q^{'} (z) = cos z$ , aur $q^{'} (nπ) = cos (nπ) = (- 1)^{n} \neq = 0$ → har ek simple pole hai. Yeh step kyun? $q^{'} \neq = 0$ order 1 certify karta hai, toh quotient shortcut har pole par ek saath apply hota hai.
Shortcut: $Res = \frac{p ( nπ )}{q ^{'} ( nπ )} = \frac{1}{( - 1 ) ^{n}} = (- 1)^{n}$ . Yeh step kyun? Ek formula poori infinite family cover karta hai — $p / q^{'}$ ki elegance (note karo $1/ (- 1)^{n} = (- 1)^{n}$ kyunki $(- 1)^{n} = \pm 1$ ).

$Res (\frac{1}{sin z}, nπ) = (- 1)^{n}$

Verify: $n = 0$ par: residue $= 1$ ; indeed $\frac{1}{s i n z} = \frac{1}{z} \cdot \frac{1}{1 - z ^{2} /6 + \dots} = \frac{1}{z} (1 + \frac{z ^{2}}{6} + \dots)$ , toh $a_{- 1} = 1 = (- 1)^{0}$ . $n = 1$ ( $z = π$ ) par: $(- 1)^{1} = - 1$ . ✔

Sabhi cells ke liye ek decision tree

Active recall

Recall Answers cover karo

Jab numerator pole par vanish ho, formula choose karne se pehle kya karte hain? ::: Pehle common factor cancel karo — reduced form sach mein order batata hai. Order formula mein bahut bada $m$ use karne par kya hota hai? ::: Safely kaam karta hai — ab analytic factor ke extra derivatives $0$ add karte hain; lekin bahut chhota $m$ $\infty$ deta hai. $\frac{z}{( z - 1 ) ( z - 3 )}$ ke kaun se poles $∣ z ∣ = 2$ ke andar hain? ::: Sirf $z = 1$ ( $∣1∣ < 2$ ); $z = 3$ bahar hai. $C$ ki kya orientation honi chahiye taaki $\oint_{C} f d z = + 2 π i \sum Res$ ? ::: Positive, yaani counter-clockwise; isko reverse karna sign flip karta hai. $\int d x / (x^{4} + 1)$ mein bada arc kyun vanish ho jaata hai? ::: ML-inequality: integrand $\leq 1/ (R^{4} - 1)$ times arc length $π R$ se $π R / (R^{4} - 1) \to 0$ milta hai. $1/ sin z$ ka $z = nπ$ par residue? ::: $(- 1)^{n}$ . $e^{1/ z}$ par $0$ par pole formulas kyun use nahi kar sakte? ::: Yeh essential hai — $z^{m} e^{1/ z}$ kabhi analytic nahi banta; $a_{- 1} = 1$ series se padho.

Connections

Residues and poles — parent note; yeh uske exhausted cases hain.
Laurent series — Cell E (removable/essential) ke liye sirf yehi tool hai.
Cauchy's Integral Formula — Ex 3 ka derivative formula uska higher-order sibling hai.
Cauchy's Integral Theorem — kyun loop ka arc-free part deformation ke under apni value rakhta hai.
Contour integration — Cells G, H yahaan rehte hain.
Argument principle / Rouché's theorem — Ex 9 ke $cot$ -type sums zero-counting power dete hain.
Singularities — upar ka classification decision tree.