Complex integration — contour integrals: Exercises

4.10.2 · D4 · Maths › Advanced Topics (Elite Level) › Complex integration — contour integrals

Shuru karne se pehle, ek shared picture samajh lo is "playground" ki. Ek contour woh curve hai jiske saath hum complex numbers ke flat plane mein chalte hain; ek pole woh point hai jahan function "blow up" ho jaata hai (ek "spike"). Pura khel yeh hai: kya woh spike mere loop ke andar hai ya nahi?

Formula Chaar tools (poore detail mein, yaheen)

1. The $d z$ rule. Kisi path $z (t)$ , $t \in [a, b]$ ke liye, ek contour integral ek ordinary real integral hota hai: $\int_{γ} f (z) d z = \int_{a}^{b} f (z (t)) z^{'} (t) d t$ , yaani $d z = z^{'} (t) d t$ .

2. The $z^{n}$ -loop lemma. Unit circle par, ek baar counterclockwise chalte hue, $\oint_{∣ z ∣ = 1} z^{n} d z = {2 π i, 0, n = - 1 n \in Z, n \neq = - 1.$

3. Cauchy's Integral Theorem & Formula. Agar $f$ ek closed contour $γ$ par aur usके andar analytic hai, toh $\oint_{γ} f (z) d z = 0$ ; aur agar $a$ , $γ$ ke andar hai, toh $f (a) = \frac{1}{2 π i} \oint_{γ} \frac{f ( z )}{z - a} d z .$

4. The Residue Theorem. Agar $f$ , $γ$ ke andar analytic hai siwaaye isolated poles $z_{k}$ ke jo $γ$ se enclosed hain, toh $\oint_{γ} f (z) d z = 2 π i k \sum Res_{z = z_{k}} f$ jahan $Res_{z_{0}} f$ woh coefficient $c_{- 1}$ hai jo $f$ ki Laurent series mein $z_{0}$ ke aas-paas $(z - z_{0})^{- 1}$ ka coefficient hota hai.

Figure 1 — aise padhna. Kaala circle contour $∣ z ∣ = 2$ hai jisme counterclockwise arrow hai (positive direction). Laal cross $z = 1 + \frac{1}{2} i$ par andar baitha hai: wahan ka ek pole $2 π i Res$ contribute karta hai. Kaala cross $z = 3$ par bahar baitha hai: yeh $0$ contribute karta hai. Yeh akela "inside vs outside" test neeche ki har problem ko drive karta hai.

Figure — Complex integration — contour integrals

L1 — Recognition

Goal: tool ka naam batao. Koi bhaari computation nahi.

Recall Solution 1.1

Kaun sa tool? Tool 2, the $z^{n}$ -loop lemma: $\oint_{∣ z ∣ = 1} z^{n} d z = 2 π i$ sirf tab jab $n = - 1$ ho, warna $0$ .

$z^{4}$ mein $n = 4 \neq = - 1$ hai, toh $\oint z^{4} d z = 0$ .
$z^{- 1}$ mein $n = - 1$ hai, toh $\oint z^{- 1} d z = 2 π i$ .

Kyun: unit circle par $z = e^{i t}$ hai, toh integrand $e^{i (n + 1) t}$ ban jaata hai. Ek pura loop kisi bhi "hilte" exponential ka average zero kar deta hai — jab tak $n + 1 = 0$ use constant $1$ nahi bana deta, jo bach jaata hai.

Recall Solution 1.2

Kaun sa tool? Tool 3, Cauchy's Integral Theorem. Integrand ek polynomial hai, jo har jagah analytic hai — loop ke andar toh koi "spike" hai hi nahi. $\oint_{∣ z ∣ = 1} (3 z^{2} - 5 z + 7) d z = 0.$ Kyun, Green's theorem ke zariye spell out karein. $f = u + i v$ , $z = x + i y$ likhne par, $\oint f d z$ ke real aur imaginary parts do line integrals $\oint (u d x - v d y)$ aur $\oint (v d x + u d y)$ ban jaate hain. Green's theorem har closed line integral ko enclosed region $D$ par ek area integral mein badal deta hai: "Green's-theorem integrands" hain $(- v_{x} - u_{y})$ aur $(u_{x} - v_{y})$ . Kisi analytic $f$ ke liye Cauchy–Riemann equations $u_{x} = v_{y}, u_{y} = - v_{x}$ dono ko exactly zero bana dete hain, toh dono area integrals vanish ho jaate hain aur loop $0$ hota hai.

Recall Solution 1.3

Kaun sa tool? "Kya spike andar hai?" Pole $z = 3$ par hai. Contour radius $2$ ka circle hai. Kyunki $∣3∣ = 3 > 2$ hai, pole bahar hai (Figure 1 ke kaale cross ki tarah). Loop ke andar function analytic hai, toh $\oint_{∣ z ∣ = 2} \frac{d z}{z - 3} = 0.$

L2 — Application

Goal: ek tool cleanly chalao, shuru se ant tak.

Recall Solution 2.1

Parametrise (Tool 1): seedhi line linear hoti hai, $z (t) = t (2 + i)$ for $t \in [0, 1]$ , toh $z^{'} (t) = 2 + i$ aur $d z = (2 + i) d t$ . $\int_{0}^{1} t (2 + i) \cdot (2 + i) d t = (2 + i)^{2} \int_{0}^{1} t d t = (2 + i)^{2} \cdot \frac{1}{2} .$ Ab $(2 + i)^{2} = 4 + 4 i + i^{2} = 3 + 4 i$ hai, toh answer hai $\frac{1}{2} (3 + 4 i) = \frac{3}{2} + 2 i$ . Sanity check: $z$ analytic hai, toh answer antiderivative $\frac{z ^{2}}{2}$ ke endpoints par evaluate hone ke barabar hona chahiye: $\frac{( 2 + i ) ^{2}}{2} - 0 = \frac{3 + 4 i}{2} = \frac{3}{2} + 2 i$ . ✔

Recall Solution 2.2

Kaun sa tool? Tool 3, CIF: $f (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{z - a} d z$ . Match karo: $f (z) = e^{z}$ , $a = 2$ . Pole $z = 2$ , $∣2∣ = 2 < 3$ hai, toh yeh $∣ z ∣ = 3$ ke andar hai, aur $e^{z}$ wahan analytic hai. $\oint_{∣ z ∣ = 3} \frac{e ^{z}}{z - 2} d z = 2 π i f (2) = 2 π i e^{2} .$

Recall Solution 2.3

Pehle factor karo: $z^{2} - 1 = (z - 1) (z + 1)$ , jisse $z = 1$ aur $z = - 1$ par simple poles milte hain; dono ka modulus $1 < 2$ hai, toh dono andar hain. Cover-up method (upar ki definition dekho): $z = 1$ ke liye $(z - 1)$ cover karo, bache hue mein $z = 1$ substitute karo $\frac{1}{z + 1}$ ; $z = - 1$ ke liye $(z + 1)$ cover karo, bache mein $z = - 1$ substitute karo $\frac{1}{z - 1}$ : $Res_{z = 1} = \frac{1}{1 + 1} = \frac{1}{2}, Res_{z = - 1} = \frac{1}{- 1 - 1} = - \frac{1}{2} .$ Sum $= \frac{1}{2} - \frac{1}{2} = 0$ hai, toh Residue Theorem se $\oint_{∣ z ∣ = 2} \frac{d z}{z ^{2} - 1} = 2 π i \cdot 0 = 0.$

L3 — Analysis

Goal: sahi tool chunna aur justify karna ki baaki kyun fail karte hain.

Recall Solution 3.1

Poles locate karo: $z = 1$ (modulus $1 < 2$ , andar) aur $z = 3$ (modulus $3 > 2$ , bahar). Sirf $z = 1$ count karta hai (exactly Figure 1 ka "inside vs outside" test). Simple pole $z = 1$ par cover-up: factor $(z - 1)$ cover karo, baki mein $z = 1$ substitute karo: $Res_{z = 1} = \frac{z}{z - 3}_{z = 1} = \frac{1}{1 - 3} = - \frac{1}{2} .$ $\oint_{∣ z ∣ = 2} \frac{z d z}{( z - 1 ) ( z - 3 )} = 2 π i (- \frac{1}{2}) = - π i .$

Recall Solution 3.2

Residues kyun nahi? $\overset{z}{ˉ}$ (conjugate) analytic nahi hai — yeh Cauchy–Riemann equations fail karta hai — toh koi Cauchy/residue machinery nahi lagti. Hume haath se parametrise karna hoga (Tool 1). Parametrise karo: $z (t) = e^{i t}$ for $t \in [0, π]$ ; tab $\overset{z}{ˉ} = e^{- i t}$ aur $z^{'} (t) = i e^{i t}$ . $\int_{0}^{π} e^{- i t} \cdot i e^{i t} d t = i \int_{0}^{π} d t = iπ .$ Path-dependence check: $1$ se $- 1$ tak seedhi line par, $z (t) = 1 - 2 t$ lo ( $t \in [0, 1]$ ), jo real hai toh $\overset{z}{ˉ} = z = 1 - 2 t$ aur $z^{'} = - 2$ : $\int_{0}^{1} (1 - 2 t) (- 2) d t = - 2 [t - t^{2}]_{0}^{1} = - 2 (0) = 0.$ Do paths, same endpoints, alag answers ( $iπ$ vs $0$ ) — path-dependence confirm ho gayi, exactly isliye kyunki $\overset{z}{ˉ}$ analytic nahi hai.

Figure 2 — aise padhna. Dono paths $z = 1$ se shuru hote hain aur $z = - 1$ par khatam. Laal upar wala semicircle $iπ$ deta hai; kaala seedha segment real axis ke saath $0$ deta hai. Same endpoints, alag answers — yeh visual proof hai ki $\int \overset{z}{ˉ} d z$ path-dependent hai kyunki $\overset{z}{ˉ}$ analytic nahi hai. (Kisi analytic $f$ se tulna karo, jahan Figure 1 ki logic dono paths ko agree karne par majboor kar deti.)

L4 — Synthesis

Goal: real integral crack karne ke liye contour tricks combine karo.

Definition Bada arc kab vanish hota hai? (Jordan's lemma aur degree test)

Real line ko semicircle se close karna tab hi kaam karta hai jab arc ka contribution $R \to \infty$ hone par khatam ho jaata hai. Do saaf sufficient conditions hain:

Degree test (rational functions). Agar $f = \frac{P ( z )}{Q ( z )}$ ho aur $de g Q \geq de g P + 2$ , toh arc par $∣ f ∣ \leq \frac{c}{R ^{2}}$ aur arc length $π R$ hai, toh arc integral zyada se zyada $\frac{c π R}{R ^{2}} = \frac{c π}{R} \to 0$ hai. Yeh " $∣ f ∣ \sim 1/ R^{2}$ " heuristic precisely kiya gaya hai: tumhe denominator ko numerator se kam se kam do degrees aage hona chahiye.
Jordan's lemma (oscillatory factor). $e^{im z}$ wale integrals mein jab $m > 0$ ho, arc upper half-plane mein vanish hota hai jab tak $f (z) \to 0$ uniformly arc par ho — chahe sirf $de g Q \geq de g P + 1$ ho — kyunki $∣ e^{im z} ∣ = e^{- m Im z}$ upper arc par decay karta hai.

Figure 3 mein dono ingredients dikhte hain: shrinking bound $c π / R$ aur growing arc length $π R$ , jinका product phir bhi $\to 0$ hota hai.

Recall Solution 4.1

Plan: $x$ ki jagah complex $z$ lo, $\frac{1}{z ^{2} + 4}$ ko real axis ke aas-paas integrate karo jo upper half-plane mein radius $R$ ke bade semicircle se close ho (counterclockwise, positive orientation), phir $R \to \infty$ karo. Arc check (degree test): yahaan $P = 1$ , $Q = z^{2} + 4$ hai, aur $de g Q = 2 \geq de g P + 2 = 2$ , toh arc vanish ho jaata hai. Isliye closed loop real-line integral ke barabar hai. Poles: $z^{2} + 4 = 0 \Rightarrow z = \pm 2 i$ . Sirf $z = 2 i$ upper half-plane mein hai. Cover-up / derivative residue at $z = 2 i$ ( $q = z^{2} + 4$ , $q^{'} = 2 z$ ke saath): $Res_{z = 2 i} = \frac{1}{2 z}_{z = 2 i} = \frac{1}{4 i} .$ $\int_{- \infty}^{\infty} \frac{d x}{x ^{2} + 4} = 2 π i \cdot \frac{1}{4 i} = \frac{π}{2} .$ Check: $\frac{1}{a} arctan \frac{x}{a}$ mein $a = 2$ dalo toh $\frac{1}{2} \cdot π = \frac{π}{2}$ milta hai. ✔

Edge case — LOWER half-plane mein close karna. Tum real axis ko neeche semicircle se bhi close kar sakte ho. Woh loop clockwise trace hota hai, toh woh $- 2 π i \sum (lower residues)$ ke barabar hai. Neeche ka akela pole $z = - 2 i$ hai jiska $Res_{- 2 i} = \frac{1}{2 ( - 2 i )} = - \frac{1}{4 i}$ hai, jisse $- 2 π i \cdot (- \frac{1}{4 i}) = \frac{π}{2}$ milta hai — wahi answer, jaisa hona chahiye. Do orientations, ek sachch: orientation reverse karne ka $- 1$ doosre residue ke sign flip ko cancel kar deta hai.

Recall Solution 4.2

Poles: $z^{4} = - 1$ ki roots hain $z = e^{iπ /4}, e^{i 3 π /4}, e^{i 5 π /4}, e^{i 7 π /4}$ . Upper half-plane mein do hain: $z_{1} = e^{iπ /4}$ aur $z_{2} = e^{i 3 π /4}$ . Arc check (degree test): $P = z^{2}$ , $Q = z^{4} + 1$ , $de g Q = 4 \geq de g P + 2 = 4$ , toh arc contribution $\to 0$ hai. Cover-up / derivative residues of $\frac{z ^{2}}{z ^{4} + 1}$ , $q^{'} = 4 z^{3}$ ke saath, toh $Res = \frac{z ^{2}}{4 z ^{3}} = \frac{1}{4 z}$ :

\operatorname{Res}_{z_2}=\frac{1}{4e^{i3\pi/4}}=\tfrac14 e^{-i3\pi/4}.$$ Sum $=\tfrac14\big(e^{-i\pi/4}+e^{-i3\pi/4}\big)=\tfrac14\big(\cos\tfrac\pi4 - i\sin\tfrac\pi4 - \cos\tfrac\pi4 - i\sin\tfrac\pi4\big)=\tfrac14\big(-2i\sin\tfrac\pi4\big)=\tfrac14\left(-2i\cdot\tfrac{\sqrt2}{2}\right)=-\tfrac{i\sqrt2}{4}.$ $$\int_{-\infty}^{\infty}\frac{x^2}{x^4+1}\,dx = 2\pi i\left(-\frac{i\sqrt2}{4}\right)=\frac{\pi\sqrt2}{2}=\frac{\pi}{\sqrt2}.$$

L5 — Mastery

Goal: higher-order poles, Laurent series, aur full case-handling.

Definition Laurent series aur order-

$m$ residue formula Ek isolated pole $z_{0}$ ke paas, ek function jo wahaan analytic nahi hai phir bhi ek Laurent series ke roop mein likha ja sakta hai — ek ordinary power series plus negative powers ka ek finite tail:

+\ c_0 + c_1(z-z_0)+c_2(z-z_0)^2+\cdots$$ Pole ki **order $m$** hoti hai agar $c_{-m}\neq 0$ ho lekin sare deeper coefficients vanish ho jaayein. **Residue** hamesha woh akela coefficient $c_{-1}$ hota hai — sirf wahi term ek loop mein survive karti hai (Tool 2). - **Practice mein series kaise banaayein:** koi jaani-pehchaani power series lo (jaise $e^z=1+z+\tfrac{z^2}{2}+\cdots$) aur pole factor se divide karo. $e^z$ ko $z^2$ se divide karne par har power do se neeche shift ho jaata hai, aur $c_{-1}$ term turant saamne aa jaati hai. - **Order-$m$ residue formula (jab series mushkil ho):** $$\operatorname{Res}_{z_0}f=\frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{\,m-1}}{dz^{\,m-1}}\Big[(z-z_0)^m f(z)\Big].$$ **Simple** pole ke liye ($m=1$) yeh cover-up rule tak reduce ho jaata hai (koi derivative nahi, bas substitute karo). **Double** pole ke liye ($m=2$) yeh kehta hai: $(z-z_0)^2$ se multiply karo, **ek baar** differentiate karo, phir limit lo.

Recall Solution 5.1

Method A — Laurent series ( $c_{- 1}$ padho). $e^{z} = 1 + z + \frac{z ^{2}}{2} + \dots$ expand karo aur $z^{2}$ se divide karo (har power do se neeche aati hai): $\frac{e ^{z}}{z ^{2}} = \frac{1}{z ^{2}} + \frac{1}{z} + \frac{1}{2} + \frac{z}{6} + \dots$ $\frac{1}{z}$ ka coefficient $c_{- 1} = 1$ hai. Woh akela term hi ek loop ka survivor hai (Tool 2), toh $\oint_{∣ z ∣ = 1} \frac{e ^{z}}{z ^{2}} d z = 2 π i c_{- 1} = 2 π i \cdot 1 = 2 π i .$ Method B — order- $m$ formula (cross-check). Yahaan $m = 2$ , $z_{0} = 0$ hai. $z^{2}$ se multiply karo: $(z - 0)^{2} f = e^{z}$ . Ek baar differentiate karo: $\frac{d}{d z} e^{z} = e^{z}$ . Limit $z \to 0$ lo aur $(m - 1)! = 1! = 1$ se divide karo: $Res_{0} \frac{e ^{z}}{z ^{2}} = \frac{1}{1 !} lim_{z \to 0} \frac{d}{d z} [e^{z}] = e^{0} = 1.$ Dono methods agree karte hain: $c_{- 1} = 1$ , integral $= 2 π i$ . ✔

Recall Solution 5.2

Poles: $z = 0$ ( $∣0∣ = 0 < 3$ , andar), $z = 1$ ( $< 3$ , andar), $z = 4$ ( $> 3$ , bahar). $0$ aur $1$ rakho. Cover-up method (har pole par factor cover karo, baki plug in karo): $Res_{0} = \frac{2 z + 1}{( z - 1 ) ( z - 4 )}_{z = 0} = \frac{1}{( - 1 ) ( - 4 )} = \frac{1}{4},$ $Res_{1} = \frac{2 z + 1}{z ( z - 4 )}_{z = 1} = \frac{3}{( 1 ) ( - 3 )} = - 1.$ Sum $= \frac{1}{4} - 1 = - \frac{3}{4}$ hai, toh Residue Theorem se $\oint_{∣ z ∣ = 3} \frac{( 2 z + 1 ) d z}{z ( z - 1 ) ( z - 4 )} = 2 π i (- \frac{3}{4}) = - \frac{3 π i}{2} .$

Recall Solution 5.3

Substitution trick: $z = e^{i θ}$ set karo toh jab $θ$ , $0 \to 2 π$ run karta hai, $z$ unit circle ko ek baar counterclockwise traverse karta hai (positive orientation). Tab $cos θ = \frac{z + z ^{- 1}}{2}, d θ = \frac{d z}{i z} .$ Kyun: $d z = i e^{i θ} d θ = i z d θ$ hai, toh $d θ = d z / (i z)$ ; yeh trig integral ko ek rational contour integral mein convert kar deta hai. Substitution ka poora algebra (step by step): $5 + 4 cos θ = 5 + 4 \cdot \frac{z + z ^{- 1}}{2} = 5 + 2 z + 2 z^{- 1} .$ Toh integrand times $d θ$ hai

= \frac{1}{iz\,(5+2z+2z^{-1})}\,dz = \frac{1}{i\,(5z+2z^{2}+2)}\,dz,$$ jahan aakhiri step mein $iz$ ka $z$, bracket mein multiply hua: $z(5+2z+2z^{-1})=5z+2z^2+2$. Isliye $$\int_0^{2\pi}\frac{d\theta}{5+4\cos\theta}=\oint_{|z|=1}\frac{dz}{i\,(2z^2+5z+2)}.$$ **Denominator factor karo:** $2z^2+5z+2=(2z+1)(z+2)$, toh poles hain $z=-\tfrac12$ (andar, $|{-\tfrac12}|<1$) aur $z=-2$ (bahar). Sirf $z=-\tfrac12$ count karta hai. **Cover-up residue** of $\dfrac{1}{i(2z+1)(z+2)}$ at $z=-\tfrac12$ ($2z+1=2(z+\tfrac12)$ likho, $(z+\tfrac12)$ cover karo): $$\operatorname{Res}_{-1/2}=\frac{1}{i\cdot 2\,(z+2)}\bigg|_{z=-1/2}=\frac{1}{i\cdot 2\cdot \tfrac32}=\frac{1}{3i}.$$ $$\int_0^{2\pi}\frac{d\theta}{5+4\cos\theta}=2\pi i\cdot\frac{1}{3i}=\frac{2\pi}{3}.$$

Recall Master check — ek-line reasons

Ex 3.1 mein $z = 3$ ne contribute kyun nahi kiya? ::: Yeh $∣ z ∣ = 2$ ke bahar hai, toh enclosed nahi hai. $\int \overset{z}{ˉ} d z$ ko kaun sa tool handle karta hai aur kyun? ::: Direct parametrisation, kyunki $\overset{z}{ˉ}$ analytic nahi hai (koi residues nahi). $\int_{0}^{2 π} (\dots cos θ) d θ$ ko contour integral mein kaise badloge? ::: $z = e^{i θ}$ , $cos θ = \frac{z + z ^{- 1}}{2}$ , $d θ = \frac{d z}{i z}$ set karo. $e^{z} / z^{2}$ ka $0$ par residue? ::: $c_{- 1} = 1$ , kyunki $e^{z} / z^{2} = z^{- 2} + z^{- 1} + \frac{1}{2} + \dots$ . Order- $m$ formula plain cover-up tak kab reduce hota hai? ::: Jab $m = 1$ (simple pole): koi derivative nahi, bas substitute karo. Contour ki orientation reverse karne par kya hota hai? ::: Integral $- 1$ se multiply ho jaata hai (clockwise = negative).

Connections

Residue theorem applications — L4 aur L5 exactly yahi techniques hain.
Laurent series — Ex 5.1 mein double-pole residue ek coefficient read-off hai.
Cauchy–Riemann equations — isliye Ex 3.2 ka $\overset{z}{ˉ}$ path-dependent hai.
Analytic functions — "loop $= 0$ " aur "loop counts spikes" ke beech ki dividing line.
Green's theorem — woh engine jo Cauchy's theorem ko pehli jagah sach banata hai.

Concept Map

Neeche ka flowchart woh decision tree hai jo tumhe har contour integral par chalana chahiye: pehle poochho ki $f$ , andar analytic hai ya nahi; agar haan toh loop $0$ hai; agar nahi, toh enclosed poles dhoondo, har residue compute karo (simple $\Rightarrow$ cover-up; higher-order $\Rightarrow$ Laurent $c_{- 1}$ ya order- $m$ formula), aur unhe $2 π i$ se multiply karke sum karo. Real integrals semicircle close karke ya $z = e^{i θ}$ substitute karke aate hain.