Complex integration — contour integrals: Worked examples

4.10.2 · D3 · Maths › Advanced Topics (Elite Level) › Complex integration — contour integrals

Kuch bhi karne se pehle: teen facts, ek convention, aur ek piece of notation jis par hum constantly lean karenge. Taaki tumhe kabhi parent note chase na karna pade, yahan hain, poori tarah se restated.

Recall Teen parent facts, yahan restated (taaki tum yeh page kabhi na chhodon)

Parent fact 1 — power lemma. Unit circle $z = e^{i t}$ par, har integer power obey karta hai $\oint_{∣ z ∣ = 1} z^{n} d z = {2 π i, 0, n = - 1 n \in Z, n \neq = - 1.$ Kyun: $z = e^{i t}$ substitute karne se integral $i \int_{0}^{2 π} e^{i (n + 1) t} d t$ ban jaata hai; woh exponential poore loops complete karta hai aur $0$ pe cancel hota hai jab tak $n + 1 = 0$ na ho, jahan woh constant $i$ hai, $2 π i$ deta hai. Sirf $\frac{1}{z}$ term ek loop mein survive karta hai — yeh sab kuch ka seed hai.

Parent fact 2 — Cauchy's theorem. Agar $f$ ek closed loop $γ$ par aur andar analytic hai, toh $\oint_{γ} f d z = 0$ . Kyun: $f = u + i v$ likhna aur Green's theorem apply karna loop integral ko area integrals mein convert karta hai jinke integrands exactly Cauchy–Riemann equations $u_{x} = v_{y}, u_{y} = - v_{x}$ hain — dono vanish ho jaate hain.

Parent fact 3 — deform-to-a-tiny-circle argument (the "inside" test). Agar $f$ ek bade loop aur pole $a$ ke around ek chhote circle ke beech ring mein analytic hai, toh dono loop integrals equal hain (ring koi singularity enclose nahi karta, isliye fact 2 se difference $0$ hai). Chhote circle $z = a + ε e^{i t}$ ko shrink karne par $\oint_{γ} \frac{f ( z )}{z - a} d z = 2 π i f (a)$ milta hai. Consequence: loop ke bahar ka pole kuch nahi contribute karta aur $0$ deta hai; loop ke andar ka pole hamesha $2 π i$ times uska residue contribute karta hai.

The scenario matrix

Har contour problem in cells mein se kisi ek mein rehti hai. Neeche ke examples cell ke label se mark hain.

Cell	Kya cheez ise distinct banati hai	Kaun sa tool jeetta hai
A — non-analytic integrand	$f$ like $\overset{z}{ˉ}$ ; Cauchy fails	brute parametrise $d z = z^{'} d t$
B — analytic, no pole inside	$f$ loop mein har jagah smooth	Cauchy's theorem $\Rightarrow 0$
C — one simple pole inside	one spike	$2 π i \cdot \frac{p}{q ^{'}}$
D — several poles, some in some out	"is it inside?" se selection	sum only the enclosed residues
E — higher-order (repeated) pole	spike counted twice	derivative residue formula
F — pole exactly ON the contour	degenerate/limiting	undefined (or principal value)
G — real integral $\int_{- \infty}^{\infty}$	close with a big arc	arc $\to 0$ , then residues
H — real word/exam twist	phrased in disguise	translate, then use C–G
I — essential singularity	$e^{1/ z}$ -type, infinitely sharp	read $c_{- 1}$ off the Laurent series
J — branch point / branch cut	$z, lo g z, x^{s}$	keyhole contour, cannot encircle freely

Ab hum cells A → J cover karte hain. Sign, "inside?" test, aur degenerate cases dekho.

Cell A — Non-analytic: brute force hi ek raasta hai

Worked example A1 —

$\int_{γ} \overset{z}{ˉ}^{2} d z$ along $z (t) = e^{i t}$ , $t : 0 \to π$ (unit circle ka upper half, counterclockwise) Forecast: $\overset{z}{ˉ}$ analytic nahi hai, isliye path matters aur answer $0$ ya $2 π i$ ka multiple hona zaruri nahi. Guess: koi plain complex number. Apna guess likh lo.

Parametrise. $z = e^{i t} \Rightarrow \overset{z}{ˉ} = e^{- i t}$ , isliye $\overset{z}{ˉ}^{2} = e^{- 2 i t}$ , aur $z^{'} (t) = i e^{i t}$ . Yeh step kyun? Cauchy's theorem ko analyticity chahiye; $\overset{z}{ˉ}$ mein $u_{x} \neq = v_{y}$ hai, isliye hum shortcut nahi le sakte — hume literally clock $t$ ke saath path par chalna hoga.
Integrand assemble karo. $\overset{z}{ˉ}^{2} z^{'} = e^{- 2 i t} \cdot i e^{i t} = i e^{- i t}$ . Kyun? Yeh sirf definition se $f (z (t)) z^{'} (t)$ hai.
$t$ mein integrate karo. $\int_{0}^{π} i e^{- i t} d t = i \cdot \frac{e ^{- i t}}{- i}_{0}^{π} = - e^{- i t}_{0}^{π} = - (e^{- iπ} - e^{0}) = - (- 1 - 1) = 2.$

Verify: answer $2$ ek plain real number hai — exactly wahi jo "non-analytic, path matters" ne predict kiya tha. Yeh $0$ nahi hai, aur $2 π i$ ka multiple bhi nahi. Agar tumhe $0$ mila toh tumne shayad $\overset{z}{ˉ}$ ko analytic treat kiya — Analytic functions warning dekho.

Cell B — Analytic, koi pole nahi: answer $0$ hai

Worked example B1 —

$\oint_{∣ z ∣ = 1} cos z d z$ Forecast: $cos z$ poore plane par analytic hai (ek entire function — kahin koi spike nahi). Computing se pehle loop value predict karo.

Loop ke andar analyticity check karo. $cos z$ har jagah Cauchy–Riemann equations satisfy karta hai; koi denominator vanish hone wala nahi, isliye koi pole enclosed nahi hai. Yeh step kyun? Cauchy's theorem ki ek hi hypothesis hai "analytic on and inside $γ$ " (parent fact 2).
Cauchy's Integral Theorem apply karo. $\oint_{∣ z ∣ = 1} cos z d z = 0$ . Yeh step kyun? Loop par aur andar analytic function ka wahan ek single-valued antiderivative hota hai ( $cos z$ ke liye $sin z$ ); ek closed loop apne starting point par wapas aata hai, isliye end par antiderivative minus start par antiderivative $= 0$ . Yeh exactly Cauchy's theorem hai, aur isliye "analytic inside" loop ko vanish karne par majboor karta hai.

Verify (brute check ki woh sach mein $0$ hai). $z = e^{i t}$ parametrise karo: $\int_{0}^{2 π} cos (e^{i t}) i e^{i t} d t$ . $cos z$ ka antiderivative $sin z$ hai, aur $sin z$ ek poore loop ke baad apne start par wapas aata hai ( $z$ wahin khatam hota hai jahan shuru hua), isliye integral $sin z$ hai same point minus itself pe evaluate hota hai $= 0$ . ✔

Cell C — Andar ek simple pole

Worked example C1 —

$\oint_{∣ z ∣ = 1} \frac{e ^{z}}{2 z - 1} d z$ Forecast: ek denominator zero $z = \frac{1}{2}$ par. Kya yeh $∣ z ∣ = 1$ ke andar hai? $∣ \frac{1}{2} ∣ = \frac{1}{2} < 1$ — haan. Toh expect karo ek nonzero $2 π i \times (kuch)$ .

$p$ aur $q$ name karo; pole locate karo. $f = \frac{p ( z )}{q ( z )}$ likho jahan upar $p (z) = e^{z}$ aur neeche $q (z) = 2 z - 1$ ho. Phir $q (z) = 0 \Rightarrow z = \frac{1}{2}$ , unit circle ke andar. Kyun? Sirf loop ke andar ke spikes count karte hain (parent fact 3: bahar ka pole $0$ contribute karta hai).
Simple-pole residue $p / q^{'}$ se. $q^{'} (z) = 2$ , isliye $Res_{z = 1/2} = \frac{p ( 1/2 )}{q ^{'} ( 1/2 )} = \frac{e ^{1/2}}{2}$ . $p / q^{'}$ kyun? Upar shortcut definition dekho — $\frac{1}{z - 1/2}$ ka coefficient hi residue hai.
Residue theorem. $\oint = 2 π i \cdot \frac{e ^{1/2}}{2} = π i e^{1/2}$ .

Verify: $\frac{e ^{z}}{2 z - 1} = \frac{1}{2} \cdot \frac{e ^{z}}{z - 1/2}$ rewrite karo. Cauchy's Integral Formula se yeh $\frac{1}{2} \cdot 2 π i e^{z}_{z = 1/2} = π i e^{1/2}$ hai. Same answer. ✔

Cell D — Kai poles: sirf enclosed wale count karte hain

Figure — Complex integration — contour integrals

Figure s01 — inside/outside test. Lavender loop counterclockwise unit circle $∣ z ∣ = 1$ hai. $z = \frac{1}{2}$ par ek coral star loop ke andar hai (uska residue count karta hai); $z = 2$ par ek grey star bahar hai (uska residue ignore hota hai — parent fact 3 use nothing pe push karta hai). Circle par lavender arrow tumhe yaad dilata hai ki walk counterclockwise hai. Cell D ka poora trick yeh picture padhna hai: circle, phir sirf andar ke stars ka sum karo.

Worked example D1 —

$\oint_{∣ z ∣ = 1} \frac{1}{( z - \frac{1}{2} ) ( z - 2 )} d z$ Forecast: do poles, $z = \frac{1}{2}$ aur $z = 2$ . Sirf $z = \frac{1}{2}$ andar hai. Predict karo: ek single residue survive karta hai.

Inside/outside test. $∣ \frac{1}{2} ∣ = \frac{1}{2} < 1$ → andar. $∣2∣ = 2 > 1$ → bahar, ignored. Kyun? Contour ko ek aise pole ke paas deform karna jo tum enclose nahi karte use count se bahir rakhta hai.
$z = \frac{1}{2}$ par Residue. Simple pole; $(z - \frac{1}{2})$ factor ko cover karo aur baaki mein plug in karo: $Res_{1/2} = \frac{1}{( \frac{1}{2} - 2 )} = \frac{1}{- \frac{3}{2}} = - \frac{2}{3}$ . Cover-up kyun? $Res = lim_{z \to 1/2} (z - \frac{1}{2}) f (z)$ singular factor remove karta hai, baaki ki finite value chhodta hai.
Sirf enclosed residues sum karo. $\oint = 2 π i (- \frac{2}{3}) = - \frac{4 π i}{3}$ .

Verify: partial fractions $\frac{1}{( z - \frac{1}{2} ) ( z - 2 )} = \frac{- 2/3}{z - \frac{1}{2}} + \frac{2/3}{z - 2}$ . Doosra term $∣ z ∣ = 1$ ke andar analytic hai (uska pole bahar hai) → uska loop integral $0$ hai. Pehla $- \frac{2}{3} \cdot 2 π i = - \frac{4 π i}{3}$ deta hai. ✔ Sirf inside pole bola.

Cell E — Higher-order (repeated) pole

Cell F — Pole bilkul contour PAR (degenerate case)

Worked example F1 —

$\oint_{∣ z ∣ = 1} \frac{d z}{z - 1}$ Forecast: $z = 1$ par pole. Lekin $∣1∣ = 1$ — yeh circle par hi hai. Predict karo: dikkat.

Inside/outside test fail karta hai. $z = 1$ na andar hai na bahar; woh $γ$ par hi hai. Yeh kyun matter karta hai? Deformation argument (parent fact 3) pole ke around clear space assume karta hai; yahan koi space nahi hai.
Diagnose. $t = 0$ ke paas integrand $\frac{1}{e ^{i t} - 1}$ $\frac{1}{i t}$ jaisa behave karta hai, jiska singularity par integral diverge karta hai. Isliye ordinary contour integral exist nahi karta.
Repair (upar diye indentation rule se): $z = 1$ ke aas-paas ek tiny semicircle notch karo. Woh detour $π i \cdot Res_{z = 1} \frac{1}{z - 1} = π i \cdot 1 = π i$ contribute karta hai — ek poore loop ka aadha — aur remaining principal-value arc analysis ka doosra aadha supply karta hai. Poori principal-value bookkeeping ke liye Residue theorem applications dekho.

Verify (kyun "no value"): compare karo $z = 1$ thoda andar moved se ( $\oint = 2 π i$ ) versus thoda bahar ( $\oint = 0$ ). Dono limits disagree karte hain ( $2 π i \neq = 0$ ), isliye on-contour case mein genuinely koi single value nahi hai. ✔ Disagreement hi proof hai ki yeh degenerate hai — aur note karo ki half-residue $π i$ exactly $0$ aur $2 π i$ ke beechon beech hai, jaisa indentation rule predict karta hai.

Cell G — Ek real integral, ek bade arc se closed

Figure s02 — semicircular (Jordan) contour. Real axis par mint segment woh integral hai jo hum actually chahte hain, $[- R, R]$ ; radius $R$ ka lavender arc loop ko upper half-plane ke through close karta hai. Coral stars un poles ko mark karte hain jo loop enclose karta hai (upper half); grey stars unhe mark karte hain jo bahar reh jaate hain (lower half). Do arrows walk dikhate hain: real axis ke saath daayein, phir arc ke saath counterclockwise wapas. Jaise $R \to \infty$ arc ka contribution $0$ tak shrink hota hai, real integral $2 π i$ times sirf coral residues ke barabar hota hai.

Worked example G1 —

$\int_{- \infty}^{\infty} \frac{x ^{2}}{x ^{4} + 1} d x$ Forecast: integrand positive hai aur $1/ x^{2}$ ki tarah decay karta hai, isliye answer ek positive real number hai. $2$ ke paas ek value guess karo (spoiler: yeh $\frac{π}{2} \approx 2.22$ hai).

Contour choose karo. Real axis $[- R, R]$ + upper semicircular arc $z = R e^{i θ}$ , $θ : 0 \to π$ . Loop ko $γ_{R}$ bolo. Upper kyun? Hume arc vanish hona chahiye taaki closed loop limit mein real-line integral ke barabar ho.
Dikhao ki arc contribution $\to 0$ (the key WHY). Arc par, $∣ z ∣ = R$ hai, isliye $z^{4} + 1 \geq R^{4} - 1$ aur $∣ z^{2} ∣ = R^{2}$ . Isliye integrand $\frac{R ^{2}}{R ^{4} - 1}$ se bound hai. ML-estimate se (integral $\leq$ max-modulus $M$ times arc-length $L$ ), $L = π R$ ke saath, $=\frac{\pi R^3}{R^4-1}\xrightarrow{R\to\infty}0.$$ *Yeh bound kyun?* $R^3/R^4=1/R\to0$: denominator ki degree numerator ko do powers se beat karta hai *plus* arc length ka extra $R$, isliye arc genuinely mar jaata hai. (Yeh Jordan's lemma ka elementary version hai.)$
Enclosed poles dhundho. $z^{4} + 1 = 0 \Rightarrow z = e^{iπ /4}, e^{i 3 π /4}, e^{i 5 π /4}, e^{i 7 π /4}$ . Upper half-plane mein do hain $z_{1} = e^{iπ /4}$ aur $z_{2} = e^{i 3 π /4}$ . Sirf upper kyun? Hamara loop sirf upar enclose karta hai; neeche ke do bahar hain.
$p / q^{'}$ se Residues. Upar $p (z) = z^{2}$ aur neeche $q (z) = z^{4} + 1$ name karo, isliye $q^{'} (z) = 4 z^{3}$ aur $Res = \frac{z ^{2}}{4 z ^{3}} = \frac{1}{4 z}$ .
- $z_{1} = e^{iπ /4}$ par: $\frac{1}{4} e^{- iπ /4}$ .
- $z_{2} = e^{i 3 π /4}$ par: $\frac{1}{4} e^{- i 3 π /4}$ .
Sum karo, multiply karo, $R \to \infty$ lo. Sum $= \frac{1}{4} (e^{- iπ /4} + e^{- i 3 π /4}) = \frac{1}{4} (\frac{2}{2} - i \frac{2}{2} - \frac{2}{2} - i \frac{2}{2}) = \frac{1}{4} (- i 2) = - \frac{i}{2 2} .$ Isliye $\oint_{γ_{R}} = 2 π i \cdot (- \frac{i}{2 2}) = \frac{π}{2}$ . Jaise $R \to \infty$ arc drop hota hai (Step 2), bacha rehta hai $\int_{- \infty}^{\infty} \frac{x ^{2}}{x ^{4} + 1} d x = \frac{π}{2} .$

Verify: numerically $\frac{π}{2} \approx 2.2214$ , ek positive real — forecast aur yeh fact match karta hai ki ek real function ke real integral ka answer real hona chahiye ( $i$ 's cancel ho gaye, jaise hone chahiye). ✔

Cell H — Real-world / exam twist (pehle translate karo)

Worked example H1 — Ek signal-processing average:

$\frac{1}{2 π} \int_{0}^{2 π} \frac{d θ}{5 + 4 cos θ}$ Forecast: yeh ek real trig integral hai, lekin $2 π$ aur $cos θ$ chilla rahe hain "unit circle disguise mein." Ek chhota sa positive number guess karo (yeh $\frac{1}{5 + 4 c o s θ}$ ka average hai, jo $\frac{1}{9}$ se $1$ tak range karta hai).

$z = e^{i θ}$ substitute karo. Phir $cos θ = \frac{1}{2} (z + \frac{1}{z})$ aur $d θ = \frac{d z}{i z}$ , aur jaise $θ : 0 \to 2 π$ , $z$ ek baar $∣ z ∣ = 1$ counterclockwise trace karta hai. Yeh step kyun? Trig integral ko ek contour integral mein convert karta hai jo hum residues se attack kar sakte hain — yeh poora reason hai ki contour methods "real" problems ke liye matter karte hain.
Rewrite karo. $5 + 4 cos θ = 5 + 2 (z + \frac{1}{z}) = \frac{2 z ^{2} + 5 z + 2}{z}$ . Isliye $=\frac{1}{2\pi i}\oint_{|z|=1}\frac{dz}{2z^2+5z+2}.$$$
Poles andar. $2 z^{2} + 5 z + 2 = (2 z + 1) (z + 2) = 0 \Rightarrow z = - \frac{1}{2}$ (andar) ya $z = - 2$ (bahar). Check kyun? Sirf enclosed spikes count karte hain (Cell D logic).
$z = - \frac{1}{2}$ par Residue. Upar $p (z) = 1$ aur neeche $q (z) = 2 z^{2} + 5 z + 2$ name karo, isliye $q^{'} (z) = 4 z + 5$ ; $z = - \frac{1}{2}$ par, $q^{'} = - 2 + 5 = 3$ , isliye $Res = \frac{1}{3}$ milta hai.
Assemble karo. $\frac{1}{2 π i} \cdot 2 π i \cdot \frac{1}{3} = \frac{1}{3}$ .

Verify: $\frac{1}{3} \approx 0.333$ $\frac{1}{9} \approx 0.111$ aur $1$ ke beech hai — $\frac{1}{5 + 4 c o s θ}$ ka ek legitimate average. Units: dimensionless average, match karta hai. ✔

Cell I — Essential singularity (koi finite pole order nahi)

Worked example I1 —

$\oint_{∣ z ∣ = 1} e^{1/ z} d z$ Forecast: $0$ par infinitely sharp spike. Tum $p / q^{'}$ ya derivative formula use nahi kar sakte. Guess karo: answer phir bhi $2 π i \times (woh 1/ z coefficient)$ hai.

Laurent series expand karo. $e^{1/ z} = \sum_{k \geq 0} \frac{1}{k ! z ^{k}} = 1 + \frac{1}{z} + \frac{1}{2 ! z ^{2}} + \dots$ . Series kyun? Koi pole order exist nahi karta, isliye ek hi handle coefficient list khud hai (parent fact 1: sirf $\frac{1}{z}$ survive karta hai).
$c_{- 1}$ padho. $\frac{1}{z}$ ka coefficient $k = 1$ term hai: $\frac{1}{1 !} = 1$ . Isliye $Res_{0} e^{1/ z} = 1$ .
Residue theorem. $\oint = 2 π i \cdot 1 = 2 π i$ .

Verify: term by term integrate karo. Har term $\frac{1}{k !} z^{- k}$ ek power $z^{n}$ hai jahan $n = - k$ ; parent fact 1 se sab $0$ dete hain except $n = - 1$ (yaani $k = 1$ ), jo $2 π i \cdot 1$ deta hai. Isliye $\oint = 2 π i$ . ✔ Infinitely many doosre terms $0$ integrate karte hain — spike ki shape irrelevant hai, sirf uska $c_{- 1}$ matter karta hai.

Cell J — Branch point / branch cut (tum freely encircle nahi kar sakte)

Figure s03 — keyhole contour. $z$ ya $x^{s}$ jaisi functions multivalued hoti hain: $z = 0$ ke around ek baar chalo aur value badal jaati hai (e.g. $z$ sign flip karta hai). $0$ par ek branch point ek branch cut force karta hai (positive real axis ke saath coral ray) jo contour cross karne ke liye forbidden hai. Lavender keyhole cut ke saath chipkti hai: ek bada circle bahar, wapas cut ke bilkul upar, $0$ ke around ek tiny circle, aur wapas cut ke bilkul neeche. Cut ke "upar" aur "neeche" ke beech ka gap exactly wahi hai jo method kaam karta hai. Isliye tum naively ek branch point par residue drop nahi kar sakte — padne ke liye koi single-valued $c_{- 1}$ nahi hai.

Worked example J1 —

$\int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x$ keyhole se Forecast: $x^{- 1/2}$ ka $0$ par branch point hai; standard result $π$ hai. $3$ ke paas ek value guess karo.

Keyhole set up karo. $f (z) = \frac{z ^{- 1/2}}{z + 1}$ use karo positive real axis ke saath branch cut ke saath, aur $z^{- 1/2} = e^{- \frac{1}{2} l o g z}$ define karo $ar g z \in [0, 2 π)$ ke saath. Keyhole kyun? $0$ ke around ek plain loop illegal hai — value cut ke across jump karti hai; keyhole cut ke dono sides ke saath chalataa hai cut cross karne ki bajaay.
Do edges ek known factor se disagree karte hain. Cut ke upar ( $ar g z = 0$ ) integrand $\frac{x ^{- 1/2}}{x + 1}$ hai. Cut ke neeche ( $ar g z = 2 π$ ), $z^{- 1/2} = e^{- \frac{1}{2} (2 π i)} x^{- 1/2} = e^{- iπ} x^{- 1/2} = - x^{- 1/2}$ . Doosri taraf wapas chalna direction bhi flip karta hai, isliye do straight edges add ho jaate hain $(1 - e^{- iπ}) \int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x = 2 \int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x$ . Yeh step kyun? Poora trick yeh hai ki multivaluedness usi integral ko ek nonzero difference mein badal deta hai jise hum solve kar sakte hain.
Enclosed pole $z = - 1$ par Residue. Hamare branch mein $z = - 1 = e^{iπ}$ hai, isliye $z^{- 1/2} = e^{- iπ /2} = - i$ ; simple pole $z = - 1$ par $\frac{z ^{- 1/2}}{z + 1}$ ka residue $z^{- 1/2}_{z = - 1} = - i$ hai. Chhote aur bade circles radii $\to 0, \infty$ hone par $0$ contribute karte hain.
Assemble karo. $2 \int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x = 2 π i \cdot (- i) = 2 π$ hai, isliye integral $π$ hai.

Verify: $x = u^{2}$ substitute karo: $\int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x = \int_{0}^{\infty} \frac{2 d u}{u ^{2} + 1} = 2 \cdot \frac{π}{2} = π$ . ✔ Keyhole answer se match karta hai — aur dikhata hai ki branch point ko apna contour kyun chahiye residue drop ki bajaay.

Sign & orientation sweep — poora clockwise case

Worked example H2 — Same integrand, reversed loop:

$\oint_{∣ z ∣ = 1, clockwise} \frac{d z}{z}$ Forecast: counterclockwise ne $+ 2 π i$ diya (parent fact 1 with $n = - 1$ ). Walk reverse karne par sign flip hokar $- 2 π i$ hona chahiye. Computing se pehle predict karo.

Parametrisation reverse karo. Clockwise matlab angle decrease karta hai, isliye $z = e^{- i t}$ lo, $t : 0 \to 2 π$ . Phir $z^{'} (t) = - i e^{- i t}$ . Kyun? Exponent ke derivative ka sign hi "kis taraf" encode karta hai — $- i$ flip carry karta hai.
Integrand assemble karo. $\frac{1}{z} z^{'} = \frac{1}{e ^{- i t}} \cdot (- i e^{- i t}) = - i$ . Kyun? Same $f (z (t)) z^{'} (t)$ recipe; $e^{- i t}$ factors cancel ho jaate hain, ek constant chhodta hai.
$t$ mein integrate karo. $\int_{0}^{2 π} (- i) d t = - i \cdot 2 π = - 2 π i .$

Verify: yeh exactly counterclockwise value $2 π i$ ka $- 1$ times hai. ✔ Orientation ek single global sign hai, kuch aur nahi. General rule (hamare convention se): counterclockwise $+$ hai, clockwise $-$ hai; $γ$ reverse karna integral ko $- 1$ se multiply karta hai. Har pehle example ne $+$ convention assume kiya; ek clockwise chalaana sirf result negate karta hai.

Recall checkpoints

Recall Kaun sa cell, aur pehla move kya hai?

$\oint_{∣ z ∣ = 1} \overset{z}{ˉ} d z$ ::: Cell A (non-analytic) — brute parametrise karo, residues use nahi kar sakte. $\oint_{∣ z ∣ = 1} sin z d z$ ::: Cell B (analytic, no pole) — Cauchy's theorem se $0$ ke barabar hai. $\oint_{∣ z ∣ = 1} \frac{e ^{z}}{z ^{3}} d z$ ::: Cell E (order-3 pole) — do derivatives chahiye; answer $2 π i \cdot \frac{1}{2} = π i$ . $z = 1$ par pole $∣ z ∣ = 1$ par hai ::: Cell F — integral undefined hai; principal value / indentation chahiye. $\oint_{∣ z ∣ = 1} e^{1/ z} d z$ ::: Cell I (essential singularity) — series se $c_{- 1} = 1$ padho; answer $2 π i$ . $\int_{0}^{\infty} \frac{x ^{- 1/2}}{x + 1} d x$ ::: Cell J (branch point) — keyhole contour, plain loop nahi. Loop clockwise reverse karo ::: Poore integral ko $- 1$ se multiply karo.

Connections

Analytic functions — Cell A vs B: analyticity decide karta hai ki paths matter karte hain ya nahi.
Cauchy–Riemann equations — "kya yeh andar analytic hai?" ke peeche ka test.
Laurent series — $c_{- 1}$ se residues padhna (Cells E, I, aur sab kuch ka origin).
Green's theorem — woh engine jo Cell B ka answer $0$ banata hai.
Residue theorem applications — Cells G, H, J aur Cell F ka principal-value repair.

Worked examples — Complex integration — contour integrals