Cauchy's integral theorem and formula: Exercises — Hinglish deep dive

4.10.3 · D4 · Maths › Advanced Topics (Elite Level) › Cauchy's integral theorem and formula

Definition Is poore page ke liye standing conventions (pehle padho!)

Teen ground rules hain jo har exercise mein assume ki jaati hain — yeh har answer ka sign aur validity fix karti hain:

Positive orientation. Har loop counterclockwise (yaani "positive" direction mein) traverse hota hai. Concrete taur pe, jab hum ek circle ko $z = e^{i θ}$ likhte hain to hum angle $θ$ ko $0$ se $2 π$ tak increasing direction mein chalate hain. Clockwise karne se har integral ka sign palat jaayega.
Contour pe koi pole nahi. Cauchy's Theorem aur Formula require karte hain ki loop saare singularities ko avoid kare: ek bad point andar ya bahar ho sakta hai, lekin path pe bilkul nahi (wahan integrand undefined hai aur integral well-posed hi nahi hai). Hum yeh silently har problem mein check karte hain — har pole strictly andar ya strictly bahar hota hai.
Simply connected region. Loop se enclosed region mein apne koi hole nahi hone chahiye (yeh simply connected hai — iska koi bhi loop ek point tak shrink ho sakta hai). Ek circle se bounded disk hamesha qualify karta hai, isliye hum poore page mein circles use karte hain.

Shuru karne se pehle, ek picture poore page ka saara mental model fix kar deti hai.

Figure — Cauchy's integral theorem and formula

Yahan $2 π i$ bas number $2 π \approx 6.283$ hai jo $i = - 1$ se multiply hua hai; yeh hamesha appear karta hai kyunki ek circle ke around counterclockwise ek full trip $2 π$ radians ka angle sweep karti hai.

Level 1 — Recognition

Goal: teen outcomes mein se kaunsa apply hota hai yeh pehchano. Koi heavy computation nahi.

Recall Solution 1.1

Integrand KYA hai? Ek polynomial. YEH kyun settle karta hai? Ek polynomial har point pe holomorphic (complex-differentiable) hota hai — yeh kabhi kisi cheez se divide nahi karta, isliye koi bhi bad point kahin nahi hai, khas taur pe $∣ z ∣ = 1$ ke andar bhi nahi, aur circle pe bhi nahi. Enclosed disk simply connected hai. Yeh exactly Outcome 1 hai. $\oint_{∣ z ∣ = 1} (3 z^{4} - 5 z + 7) d z = 0.$

Recall Solution 1.2

Bad point KYA hai? Denominator $z$ zero hai $z = 0$ pe, aur $0$ circle ka center hai — definitely andar (path pe nahi). Toh yeh Outcome 1 nahi hai. Outcome 2 kyun? Pattern $\frac{f ( z )}{z - a}$ match karo jahan $a = 0$ aur $f (z) = 1$ (constant function, har jagah holomorphic). Cauchy's Integral Formula deta hai $\oint_{∣ z ∣ = 1} \frac{1}{z} d z = 2 π i f (0) = 2 π i \cdot 1 = 2 π i .$ Is classic ka point: ek single hole answer ko nonzero bana deta hai.

Recall Solution 1.3

Bad point KYA hai? $z = 3$ . Yeh KAHAN hai? Origin se iska distance $3$ hai, lekin loop sirf distance $1$ tak pahunchta hai. Toh bad point loop ke bahar hai (aur uske upar bhi nahi). Yeh zero kyun hota hai: $∣ z ∣ = 1$ ke andar aur uske upar integrand $\frac{1}{z - 3}$ bilkul holomorphic hai (wahan kuch blow up nahi hota), aur enclosed disk simply connected hai. Outcome 1. $\oint_{∣ z ∣ = 1} \frac{1}{z - 3} d z = 0.$

Level 2 — Application

Goal: sahi formula mein plug karo aur $f (a)$ evaluate karo.

Recall Solution 2.1

Bad point: $z = 1$ , origin se distance $1$ , jo $< 2$ hai → $∣ z ∣ = 2$ ke andar (path pe nahi). Outcome 2. Pattern match karo: $\frac{f ( z )}{z - a}$ jahan $a = 1$ , $f (z) = e^{z}$ (har jagah holomorphic). $\oint_{∣ z ∣ = 2} \frac{e ^{z}}{z - 1} d z = 2 π i f (1) = 2 π i e^{1} = 2 π i e .$ Numerically $2 π e \approx 17.08$ , toh value lagbhag $17.08 i$ hai.

Recall Solution 2.2

Bad point: $z = 0$ , andar. Pattern: $a = 0$ , $f (z) = sin z$ . $\oint_{∣ z ∣ = 1} \frac{s i n z}{z} d z = 2 π i sin (0) = 2 π i \cdot 0 = 0.$ Hole hone ke bawajood zero kyun? Formula $2 π i f (a)$ return karta hai; yahan $f (0) = sin 0 = 0$ . Hole real hai, lekin numerator wahan vanish ho jaata hai, isliye answer $0$ ho jaata hai. (Geometrically: $sin z / z$ ka actually ek removable singularity hai — ise $0$ pe $1$ redefine kiya ja sakta hai aur yeh fully holomorphic ban jaata hai.)

Recall Solution 2.3

Bad point: $z = 2 i$ . Origin se iska distance $∣2 i ∣ = 2 < 3$ hai → andar (path pe nahi). Outcome 2. Pattern: $a = 2 i$ , $f (z) = z^{2} + 1$ . $f (2 i) = (2 i)^{2} + 1 = 4 i^{2} + 1 = - 4 + 1 = - 3.$ $\oint_{∣ z ∣ = 3} \frac{z ^{2} + 1}{z - 2 i} d z = 2 π i (- 3) = - 6 π i .$

Level 3 — Analysis

Goal: higher-order poles (repeated bad points) aur $n$ sahi choose karna.

Intuition Hum integral ke

through differentiate kyun kar sakte hain (honest reason) Yeh claim ki "basic formula $f (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{z - a} d z$ ko $a$ ke respect mein differentiate karo" ek real question chhupata hai: $\frac{d}{d a}$ derivative ko $\oint$ sign ke andar move karna legal kyun hai? Yahan justification hai, derivative ki definition se.

$a$ mein derivative ek difference quotient ki limit hai: $\frac{d}{d a} (\frac{1}{z - a}) = lim_{h \to 0} \frac{1}{h} (\frac{1}{z - ( a + h )} - \frac{1}{z - a}) = \frac{1}{( z - a ) ^{2}}$ (bas algebra: do fractions combine karo, $h$ 's cancel ho jaate hain, $h \to 0$ lo).
Kyunki contour $γ$ ek fixed, bounded curve hai aur $a$ strictly andar hone se, $z - a$ $γ$ pe zero se bounded away rehta hai (pole kabhi path ko touch nahi karta — rule 2), fraction $\frac{f ( z )}{( z - a ) ^{2}}$ $γ$ ke along uniformly bounded aur continuous hai.
Ek fixed curve pe uniformly convergent functions ke finite integral ka limit integral ke bahar liya ja sakta hai — yeh standard "interchange of limit and finite integral" theorem hai. Toh limit $h \to 0$ (yaani $a$ -derivative) seedha $\oint$ ke through slide kar jaata hai.

Ise ek baar apply karne se milta hai $f^{'} (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{( z - a ) ^{2}} d z$ ; ise $n$ baar repeat karne se $\frac{d ^{n}}{d a ^{n}} \frac{1}{z - a} = \frac{n !}{( z - a ) ^{n + 1}}$ milta hai, jo exactly wahi jagah hai jahan se $n!$ aur exponent $n + 1$ aate hain.

Recall Solution 3.1

Power padho: denominator $z^{3} = (z - 0)^{3}$ hai. $(z - a)^{n + 1}$ match karo: toh $n + 1 = 3 \Rightarrow n = 2$ , $a = 0$ , $f (z) = cos z$ . Do baar differentiate karo: $f^{'} (z) = - sin z$ , $f^{''} (z) = - cos z$ , toh $f^{''} (0) = - cos 0 = - 1$ . $\oint = \frac{2 π i}{2 !} f^{''} (0) = \frac{2 π i}{2} (- 1) = - π i .$

Recall Solution 3.2

Power padho: $z^{2} = (z - 0)^{2} \Rightarrow n + 1 = 2 \Rightarrow n = 1$ , $a = 0$ , $f (z) = e^{2 z}$ . Ek baar differentiate karo: $f^{'} (z) = 2 e^{2 z}$ , toh $f^{'} (0) = 2 e^{0} = 2$ . $\oint = \frac{2 π i}{1 !} f^{'} (0) = 2 π i \cdot 2 = 4 π i .$

Recall Solution 3.3

Bad point $z = 1$ , distance $1 < 2$ → andar. Power: $(z - 1)^{2} \Rightarrow n + 1 = 2 \Rightarrow n = 1$ , $a = 1$ , $f (z) = z$ . Ek baar differentiate karo: $f^{'} (z) = 1$ , toh $f^{'} (1) = 1$ . $\oint = \frac{2 π i}{1 !} f^{'} (1) = 2 π i \cdot 1 = 2 π i .$

Level 4 — Synthesis

Goal: ek loop ke andar do ya zyada poles — kaam ko split karo.

Common mistake Dhyan rakho — repeated linear factors ko EK se ZYADA terms chahiye

Simple "ek factor per term" split tabhi kaam karta hai jab denominator ke factors distinct hon (jaise $(z - 1) (z - 2)$ ). Agar koi factor repeated ho — maano denominator mein $(z - 1)^{2}$ — toh tumhe har power ke liye ek term dena hoga: $\frac{g ( z )}{( z - 1 ) ^{2} ( z - 2 )} = \frac{A}{z - 1} + \frac{B}{( z - 1 ) ^{2}} + \frac{C}{z - 2} .$ Kyun: root-plugging trick $B$ ( $(z - 1)^{2}$ se multiply karo, $z = 1$ set karo) aur $C$ find kar leti hai, lekin $A$ ke liye ek extra step chahiye (differentiate karo, ya dusra coefficient match karo). Aur khas baat: $\frac{B}{( z - 1 ) ^{2}}$ piece ek order-2 pole hai, isliye tumhe ise generalized formula ( $n = 1$ ) se integrate karna hoga, simple wale se nahi. $\frac{A}{z - 1}$ term miss karna, ya $(z - 1)^{2}$ ko simple treat karna, classic galti hai. Is page ke exercises mein saare factors distinct hain, isliye ek term each kaafi hai — lekin repeated factor dikhte hi use pehchano.

Recall Solution 4.1

Poles kahan hain? $z = 1$ aur $z = 2$ ; distances $1$ aur $2$ , dono $< 3$ → dono andar (koi path pe nahi). Split (partial fractions): hum chahte hain $\frac{1}{( z - 1 ) ( z - 2 )} = \frac{A}{z - 1} + \frac{B}{z - 2}$ . Denominators clear karo, $1 = A (z - 2) + B (z - 1)$ . $z = 1$ set karo ( $B$ term khatam): $1 = A (- 1) \Rightarrow A = - 1$ . $z = 2$ set karo ( $A$ term khatam): $1 = B (1) \Rightarrow B = 1$ . Toh integrand hai $\frac{- 1}{z - 1} + \frac{1}{z - 2}$ . Har ek pe formula apply karo (har ek $\frac{const}{z - a}$ form mein hai, yaani $f =$ woh constant): $\oint \frac{- 1}{z - 1} = 2 π i (- 1) = - 2 π i, \oint \frac{1}{z - 2} = 2 π i (1) = 2 π i .$ Add karo: $- 2 π i + 2 π i = 0.$ Sanity comment: do residues cancel ho gaye — yeh genuine feature hai, galti nahi.

Recall Solution 4.2

Recheck karo ki kaunse poles andar hain: loop ki radius $\frac{3}{2} = 1.5$ hai. Pole $z = 1$ : distance $1 < 1.5$ → andar. Pole $z = 2$ : distance $2 > 1.5$ → bahar. (Koi path pe nahi hai.) Sirf andar wala pole contribute karta hai. Wahi split $\frac{- 1}{z - 1} + \frac{1}{z - 2}$ use karo: $\frac{1}{z - 2}$ term is chhoti loop ke andar holomorphic hai, isliye iska integral $0$ hai (Cauchy's Theorem). Sirf $\frac{- 1}{z - 1}$ term bachta hai: $\oint_{∣ z ∣ = 3/2} = 2 π i (- 1) + 0 = - 2 π i .$ Lesson: radius badlane se answer badal jaata hai, kyunki yeh badalta hai ki kaunse poles enclosed hain.

Recall Solution 4.3

Poles: $z = 0$ aur $z = 1$ , distances $0$ aur $1$ , dono $< 2$ → dono andar. $\frac{1}{z ( z - 1 )}$ part split karo (wahi root-plugging trick): $\frac{1}{z ( z - 1 )} = \frac{A}{z} + \frac{B}{z - 1}$ deta hai $1 = A (z - 1) + B z$ . $z = 0$ set karo ( $B$ khatam): $A = - 1$ ; $z = 1$ set karo ( $A$ khatam): $B = 1$ . Toh $\frac{e ^{z}}{z ( z - 1 )} = e^{z} (\frac{- 1}{z} + \frac{1}{z - 1}) = \frac{- e ^{z}}{z} + \frac{e ^{z}}{z - 1} .$ Formula apply karo (ab $f (z) = e^{z}$ har ek pe): $\oint \frac{- e ^{z}}{z} = 2 π i (- e^{0}) = - 2 π i, \oint \frac{e ^{z}}{z - 1} = 2 π i e^{1} = 2 π i e .$ Add karo: $- 2 π i + 2 π i e = 2 π i (e - 1) .$

Level 5 — Mastery

Goal: Cauchy machinery ko ek downstream consequence ke saath combine karo (bounds, uniqueness, real integrals).

Recall Solution 5.1

Pehle, prefactor ka modulus. Humhe $\frac{1}{2 π i}$ chahiye. Yaad karo $∣ i ∣ = 1$ (number $i$ unit circle pe hai, origin se $1$ distance pe), aur quotient ka modulus moduli ka quotient hai, toh $\frac{1}{2 π i} = \frac{1}{2 π ∣ i ∣} = \frac{1}{2 π}$ . $i$ size mein kuch contribute nahi karta — yeh sirf rotate karta hai. Tool — ML inequality: kisi bhi contour ke liye, $\oint g d z \leq (length) \cdot (max ∣ g ∣)$ . Yeh tool kyun? Hum integral exactly compute nahi kar sakte (unknown $f$ ), lekin hum ise bound kar sakte hain. $∣ z ∣ = R$ pe $a = 0$ ke saath: length $= 2 π R$ , aur $\frac{f ( z )}{z ^{2}} = \frac{∣ f ( z ) ∣}{∣ z ∣ ^{2}} \leq \frac{M}{R ^{2}} .$ Toh $∣ f^{'} (0) ∣ = \frac{1}{2 π} \oint \frac{f ( z )}{z ^{2}} d z \leq \frac{1}{2 π} \cdot 2 π R \cdot \frac{M}{R ^{2}} = \frac{M}{R} .$ Payoff: yeh exactly woh estimate hai jo Liouville's theorem ke peeche hai — ek bounded entire function ( $M$ fixed while $R \to \infty$ ) ke liye $f^{'} (0) = 0$ hota hai, aur isliye woh constant hai.

Recall Solution 5.2

Parametrisation set up karo: $∣ z ∣ = 1$ pe counterclockwise trace karte hue, $z = e^{i θ}$ jahan $θ$ increasing $0$ se $2 π$ tak hai; phir $d z = i e^{i θ} d θ$ aur $z - a = e^{i θ} - 0 = e^{i θ}$ . $f \equiv 1$ ke saath, $\oint_{∣ z ∣ = 1} \frac{1}{z} d z = \int_{0}^{2 π} \frac{1}{e ^{i θ}} i e^{i θ} d θ = i \int_{0}^{2 π} d θ .$ Lekin hum yeh bhi jaante hain ki yeh integral $2 π i f (0) = 2 π i$ ke barabar hai (Exercise 1.2). Equate karo: $i \int_{0}^{2 π} d θ = 2 π i \Rightarrow \int_{0}^{2 π} d θ = 2 π .$ Matlab: Cauchy's formula mein mysterious $2 π i$ literally trip ka angle $2 π$ hai, $i$ ke saath dressed jo $d z$ se aata hai. (Agar hum $θ$ doosri taraf $2 π \to 0$ chalate, toh hume $- 2 π i$ milta — isliye direction fix ki jaati hai.)

Recall Solution 5.3

Formula se shuru karo circle $∣ z - a ∣ = r$ pe: $z = a + r e^{i θ}$ parametrise karo jahan $θ$ increasing $0 \to 2 π$ hai (counterclockwise), toh $z - a = r e^{i θ}$ aur $d z = i r e^{i θ} d θ$ . $f (a) = \frac{1}{2 π i} \oint \frac{f ( z )}{z - a} d z = \frac{1}{2 π i} \int_{0}^{2 π} \frac{f ( a + r e ^{i θ} )}{r e ^{i θ}} i r e^{i θ} d θ .$ Cancel karo $r e^{i θ}$ aur $i$ ko $\frac{1}{2 π i}$ ke saath: $f (a) = \frac{1}{2 π} \int_{0}^{2 π} f (a + r e^{i θ}) d θ .$ Matlab: center pe value kisi bhi circle ke around values ka average hai. Yeh Maximum modulus principle ka seed hai — ek interior point kabhi boundary se exceed nahi kar sakta, kyunki woh boundary values ka average hai.

Recall Poore page ka one-line summary

Har bad point locate karo ::: check karo ki kaunse (counterclockwise) loop ke andar hain aur koi uske upar na ho; koi pole nahi → $0$ ; simple pole → $2 π i f (a)$ ; order- $(n + 1)$ pole → $\frac{2 π i}{n !} f^{(n)} (a)$ ; kai poles → split karo (repeated factors ke liye ek term har power ke liye) aur add karo.