Visual walkthrough — Cauchy's integral theorem and formula

4.10.3 · D2 · Maths › Advanced Topics (Elite Level) › Cauchy's integral theorem and formula

---

Step 0 — Complex number kya hota hai, ek tasveer mein?

KYA. Ek complex number $z$ simply ek kaagaz ke seedhe tukde par ek point hai. Hum us tukde ko do axes dete hain: horizontal wala ek number measure karta hai jise hum $x$ kehte hain, aur vertical wala ek number measure karta hai jise hum $y$ kehte hain. Jo point horizontal $x$ aur height $y$ par baitha hai, use likhte hain

$z=\underset{\text{kitna daayein}}{\underbrace{x}}+\underset{\text{"90° ghuma" ka marker}}{\underbrace{i}}\underset{\text{kitna upar}}{\underbrace{y}}$

Symbol $i$ koi rahasya quantity nahi hai — ise ek label samjho jo kehta hai "yeh hissa upar ki taraf point karta hai". Toh $z$ origin se dot tak ek arrow hai.

KYUN. Neeche sab kuch us sheet par kheenche gaye ek loop ke around ek pen chalane ke baare mein hai. Loop ki baat karne ke liye pehle hume yeh agree karna hoga ki pen jahan bhi ho sakta hai woh ek number $z$ hai.

TASVEER. Red dot $z$ hai; uska floor par shadow $x$ hai, wall par shadow $y$ hai.

---

Step 1 — Ek function, ek contour, aur $\oint fdz$ ka matlab

KYA. Ek function $f$ har input point $z$ leta hai aur doosra point $f(z)$ wapas deta hai — ek aur arrow. Ek contour $\gamma$ (Greek letter "gamma") ek path hai jo hum plane par kheenchte hain; ek closed contour woh path hai jo wahan wapas aata hai jahan se chala tha, jaise ek dhage ka loop.

Symbol ${\oint }_{\gamma }f(z)dz$ ka matlab hai: loop $\gamma$ ke saath chote chote steps mein chalte jao; har jagah function ki value $f(z)$ ko agle liye gaye chote step $dz$ se multiply karo; un sab chote products ko jod lo.

• $\oint$ (circle-integral) $=$ "ek closed loop par sum".
• $dz$ $=$ path ke saath ek tiny step arrow (position mein ek chota change, khud ek complex number).
• $f(z)dz$ $=$ do arrows ka product (complex multiplication rotate aur scale karta hai).

AISE DEFINE KARNE KA KYUN. Real integrals "height times tiny width" jodti hain. Complex integral "arrow times tiny step" jodti hai. Kyunki $dz$ ek direction carry karta hai, loop ko ulta chalna har $dz$ ko flip kar deta hai — isliye loop ki orientation matter karti hai. Hum hamesha counter-clockwise chalte hain (jise positive orientation kehte hain) jab tak kuch aur na kaha jaye.

TASVEER. Loop ko tiny step arrows $dz$ mein kaata gaya hai; unme se ek red mein highlight hai.

---

Step 2 — Loop integral ko do real loop integrals mein split karo

KYA. Function ko parts mein likho, $f=u+iv$, jahan $u(x,y)$ aur $v(x,y)$ ordinary real-valued functions hain ($f$ ki "floor value" aur "wall value"). Step ko $dz=dx+idy$ likho. Multiply karo:

${\oint }_{\gamma }fdz={\oint }_{\gamma }(u+iv)(dx+idy)=\underset{\text{real part}}{\underbrace{{\oint }_{\gamma }(udx-vdy)}}+i\underset{\text{imaginary part}}{\underbrace{{\oint }_{\gamma }(vdx+udy)}}$

Term by term: $u,v$ point ke real functions hain; $dx$ ek tiny sideways move hai, $dy$ ek tiny upar ka move; woh $i^2=-1$ hi hai jo $iv\cdot idy$ ko negative $-vdy$ mein badal diya.

KYUN. Hum abhi tak nahi jaante ki ek complex loop integral kaise attack karein, lekin hum real loop integrals ke liye plane par pehle se ek tool jaante hain: Green's theorem. Isliye hum ek mushkil complex loop ko do jaane-pehchaane real loops mein translate karte hain.

TASVEER. Wohi loop, ab do ordinary "circulation" loops ke roop mein socha gaya — ek $(u,-v)$ track kar raha hai, ek $(v,u)$.

---

Step 3 — Har real loop ko Green's theorem se daalo

KYA. Green's theorem loop ke around circulation ko bhare huye area $R$ par sum mein convert karta hai:

${\oint }_{\gamma }(Pdx+Qdy)={\iint }_R(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dA$

Yahan $P,Q$ do real functions hain jo $dx,dy$ par chal rahi hain; $\frac{\partial Q}{\partial x}$ ka matlab hai "daayein nudge karne par $Q$ kitni tezi se badalta hai"; ${\iint }_R\dots dA$ ka matlab hai "loop ke andar har tiny area patch par jodo".

Apply karo do baar:

• Real part ($P=u,Q=-v$): ${\iint }_R(-v_x-u_y)dA$.
• Imag part ($P=v,Q=u$): ${\iint }_R(u_x-v_y)dA$.

(Subscripts shorthand hain: $v_x=\partial v/\partial x$, etc.)

KYUN. Green's theorem bilkul wahi bridge hai "boundary par kya hota hai" se "andar kya hota hai" tak — Cauchy ki bilkul wohi spirit. Yeh tabhi kaam karta hai jab $u,v$ pure filled region $R$ par smooth hon: koi hole allowed nahi. Yeh baat Step 6 ke liye yaad rakhna.

TASVEER. Boundary loop andar ki taraf collapse hokar little area patches ki ek mesh mein badal jaata hai, har ek ek "curl" value carry karta hai.

---

Step 4 — Collapse: Cauchy–Riemann dono integrands ko khatam kar deta hai

KYA. Ek complex-differentiable function wahi same derivative deta hai chahe tum kis bhi direction se approach karo. Real-axis approach ko imaginary-axis approach ke barabar force karne se Cauchy-Riemann equations milte hain:

$u_x=v_y\text{aur}{u}_y=-v_x$

Inhe seedha Step 3 ke do area integrands mein plug karo:

$-v_x-u_y=-v_x-(-v_x)=0,u_x-v_y=v_y-v_y=0$

Dono integrands har point par exactly zero hain — "small" nahi, zero. Toh dono area sums $0$ hain, aur isliye

$\boxed{{\oint }_{\gamma }f(z)dz=0}\text{(Cauchy’s Integral Theorem)}$

YEH KYUN KAAM KARTA HAI. Complex differentiability real differentiability se kahin zyada strong demand hai, aur Cauchy–Riemann equations wahi extra strength hain jo likhkar dikhaayi gayi hain. Wohi strength curl ko har jagah vanish karati hai.

TASVEER. Har little patch ka "curl" arrow kuch nahi ban jaata — poori mesh blank ho jaati hai.

Recall

Simply connected kyun zaroori tha Green's theorem ke liye $f$ ko poore filled region par smooth hona tha. Ek domain no holes ke saath (simply connected) guarantee karta hai ki loop ko achhe patches se bhara ja sakta hai. Andar ek bhi bura point aur yeh step illegal hai — yahi agli steps ki poori kahaani hai.

---

Step 5 — Agla karishma: boundary andar ka haal jaanti hai

KYA. Loop ke andar ek point $a$ chuno. Purane function se bana ek naya function dekho: $g(z)=\frac{f(z)}{z-a}$ Denominator $z-a$ woh arrow hai $a$ se current point $z$ tak. Yeh har jagah theek hai siwaay $z=a$ ke, jahan yeh zero hota hai aur $g$ blow up ho jaata hai. Toh $g$ mein exactly ek bura point hai — ek pole — $a$ par baitha hua.

YEH SPECIFIC $g$ KYUN. Hum chahte hain ek formula jo boundary data se $f(a)$ pad sake. $z-a$ se divide karna exactly $a$ par ek controlled blow-up banata hai; wohi blow-up hamare liye $f(a)$ ki value "pakdega".

TASVEER. Bada loop $\gamma$ jisme hole $a$ mark hai; ek boundary point $z$ ke liye arrow $z-a$ red mein drawn hai.

---

Step 6 — Loop ko $a$ ke around ek tiny circle par shrink karo

KYA. Bade loop $\gamma$ aur $a$ ke centre par radius $\epsilon$ ke tiny circle $C_{\epsilon }$ ke beech, function $g$ ke koi bure points nahi hain (sirf ek pole, $a$, chote circle ke andar band hai). Step 4 se, kisi bhi hole-free region ke loop par $g$ ka integral zero hota hai — toh outer aur inner loops same integral dene chahiye:

${\oint }_{\gamma }\frac{f(z)}{z-a}dz={\oint }_{C_{\epsilon }}\frac{f(z)}{z-a}dz$

Ab tiny circle ko parametrise karo. Uske har point par $z=a+\epsilon e^{i\theta }$ hai:

• $\epsilon$ chota radius hai, $\theta$ angle hai jo $0$ se $2\pi$ tak sweep hota hai,
• $e^{i\theta }$ angle $\theta$ par point karne wala unit arrow hai (Euler ka spin marker),
• tiny step hai $dz=i\epsilon e^{i\theta }d\theta$.

Substitute karo: ${\oint }_{C_{\epsilon }}\frac{f(z)}{z-a}dz={\int }_0^{2\pi }\frac{f(a+\epsilon e^{i\theta })}{\epsilon e^{i\theta }}i\epsilon e^{i\theta }d\theta =i{\int }_0^{2\pi }f(a+\epsilon e^{i\theta })d\theta$

Upar aur neeche $\epsilon e^{i\theta }$ exactly cancel ho jaate hain — blow-up aur tiny step ek doosre ko khatam karne ke liye bane the. Yehi poora trick hai.

SHRINK KARNA KYUN. Ek tiny circle par, $f$ muskil se badalta hai: yeh almost constant $f(a)$ ke barabar hota hai. Shrink karna ek mushkil integral ko almost-constant integral se replace karta hai.

TASVEER. Bada loop andar ki taraf deform hota hai (dashed) jab tak woh $a$ ke around small red circle $C_{\epsilon }$ ko hug na kare; unke beech shaded ring pole-free hai.

---

Step 7 — $\epsilon \to 0$ karo aur jawab collect karo

KYA. Jaise jaise radius $\epsilon$ shrink hokar $0$ hota hai, har point $a+\epsilon e^{i\theta }$ $a$ mein slide ho jaata hai, toh continuity se $f(a+\epsilon e^{i\theta })\to f(a)$, ek constant. Ek constant sum se bahar aa jaata hai:

$i{\int }_0^{2\pi }f(a+\epsilon e^{i\theta })d\theta \stackrel{\epsilon \to 0}{\to }if(a){\int }_0^{2\pi }d\theta =if(a)\cdot 2\pi =2\pi if(a)$

Akela $2\pi$ simply circle ke ek chakkar ka total angle hai. Dono sides ko $2\pi i$ se divide karo:

$\boxed{f(a)=\frac{1}{2\pi i}{\oint }_{\gamma }\frac{f(z)}{z-a}dz}\text{(Cauchy’s Integral Formula)}$

SHRINK LEGAL KYUN HAI. Value $\epsilon$ par depend nahi karti thi (Step 6 ne ise ek baar ke liye fix kar diya tha), toh limit lena sirf woh jawab zahir karta hai jo har radius ke liye pehle se sach tha. Boundary $\gamma$ sach mein $f(a)$ encode karti hai.

TASVEER. Circle ek dot mein shrink ho jaata hai; uske upar $f$ ek akeli height $f(a)$ par flat ho jaati hai, aur $2\pi$ angle jawab stamp kar deta hai.

---

Step 8 — Degenerate cases jo tumne zaroor dekhe honge

KYA. Teen edge scenarios, har ek ki apni tasveer-fact:

1. Andar koi pole nahi ($a$ bahar $\gamma$ ke): toh $f/(z-a)$ pure filled loop par smooth hai, toh Step 4 apply hota hai aur integral $0$ hai. Formula quietly theorem ban jaata hai.
2. Pure hole ${\oint }_{|z|=1}\frac{1}{z}dz$: yahan $f=1$, $a=0$ andar hai, toh formula deta hai $2\pi i\cdot f(0)=2\pi i$. Zero nahi — kyunki region mein $0$ par ek hole hai, toh theorem kabhi allowed hi nahi tha.
3. Higher-order pole (denominator $(z-a{)}^{n+1}$): $a$ mein formula ko $n$ baar differentiate karna (har derivative $(z-a{)}^{-1}$ se ek aur factor nikaalti hai, $n!$ produce karti hai) deta hai $f^{(n)}(a)=\frac{n!}{2\pi i}{\oint }_{\gamma }\frac{f(z)}{(z-a{)}^{n+1}}dz$ Isliye ek complex derivative secretly infinite saare derivatives grant karta hai.

TEENO KYUN DIKHAO. Reader ko kabhi loop dekhke ruk nahi jaana chahiye: pole bahar → $0$; simple pole andar → $2\pi if(a)$; repeated pole → $n!$ wala version.

TASVEER. Teen mini-panels: pole bahar (integral $0$), pole andar (integral $2\pi if(a)$), double pole ($n!$ formula).

---

Ek-tasveer summary

Upar ki sab kuch ek single diagram mein: bada loop $\gamma$, smooth (hole-free) interior jahan Cauchy–Riemann force karta hai $\oint fdz=0$, aur punctured case jahan $a$ par shrink karna boundary se $f(a)$ padhta hai.

Recall Poore walkthrough ki Feynman-style retelling

Ek smooth sheet par dhage ka ek loop kheencho. Uske around chalte hue "value times step" jodne se zero milta hai — kyunki smoothness rule (Cauchy–Riemann) andar har tiny patch ko perfectly balanced banata hai, toh Green's theorem kuch bhi sum nahi karta (Steps 1–4). Ab ek special point $a$ dabao aur apne function ko "arrow from $a$" se divide karo, wahin ek single blow-up create karte hue (Step 5). Outer loop ko andar slide karo jab tak woh $a$ ke around ek tiny ring ko hug na kare — legal hai, kyunki unke beech ka doughnut smooth hai (Step 6). Us tiny ring par blow-up aur tiny step exactly cancel ho jaate hain, function basically constant $f(a)$ hota hai, aur ek full trip ek $2\pi$ ka angle contribute karta hai; arithmetic $2\pi if(a)$ ugalti hai (Step 7). $2\pi i$ se divide karo: edge ne andar ki baat bata di. Finally, agar special point bahar hai toh phir zero milta hai, aur agar blow-up kai baar stack hai toh differentiate karo $n!$ wala version paane ke liye (Step 8).

Recall Quick self-test

$\oint fdz=0$ sirf enclosed singularity ke bina kyun hota hai? ::: Green's theorem ko $f$ ki poore filled region par smoothness chahiye; ek pole woh tod deta hai. Formula mein $2\pi$ kahan se aata hai? ::: Shrinking circle ke ek chakkar ka total angle ${\int }_0^{2\pi }d\theta$. Step 6 mein blow-up kya cancel karta hai? ::: Tiny step $dz=i\epsilon e^{i\theta }d\theta$ denominator mein $\epsilon e^{i\theta }$ ko khatam karta hai.

Related: Cauchy-Riemann equations · Green's theorem · Residue theorem · Laurent series · Liouville's theorem · Maximum modulus principle