Worked examples — Cauchy's integral theorem and formula

4.10.3 · D3 · Maths › Advanced Topics (Elite Level) › Cauchy's integral theorem and formula

Do tools hi poora toolbox hain. Maano $\gamma$ ek closed loop hai, $f$ ek function hai, $a$ ek point hai.

Jo reader "holomorphic" ya "$\oint$" ka matlab bhool gaya ho, woh pehle parent ka §1 dobara padhe.

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Scenario matrix

Har contour integral jo tum dekhoge woh in cells mein se exactly ek mein aata hai. Har row ek cheez hai jo galat ya special ho sakti hai. Neeche ke examples cell ke hisaab se label kiye gaye hain.

Cell	Description	Kaun sa weapon	Example
A	$\gamma$ ke andar koi pole nahi	Theorem $\to 0$	Ex 1
B	Andar ek simple pole ($n=0$)	Formula, $f(a)$	Ex 2
C	Higher-order pole ($n\ge 1$)	Formula, $f^{(n)}(a)$	Ex 3
D	Pole $\gamma$ ke bahar (degenerate: kuch enclosed nahi)	Theorem $\to 0$	Ex 4
E	Andar do poles → partial fractions se split karo	Formula do baar	Ex 5
F	Pole contour par (illegal input)	Dono nahi — principal value	Ex 6
G	Limiting case: radius $\to$ pole ko swallow/exclude karna	A vs B ki boundary	Ex 7
H	Real-world word problem (physics/AC circuit flavour)	Formula disguise mein	Ex 8
I	Exam twist: same integral, do contours, alag answers	A vs B side by side	Ex 9
J	Contour reversed (clockwise)	Sign flip	Ex 10
K	Non-pole singularities: branch points aur essential singularities	In do weapons se bahar	Ex 11

Ek convention jo har jagah use hogi: $|z|=R$ ka matlab hai origin par centered radius $R$ ka circle, counter-clockwise traverse kiya gaya (positive direction). Figure dekho — counter-clockwise woh arrow direction hai jo enclosed disk ko tumhari left mein rakhta hai.

(Figure s01 — ek blue unit circle jisme yellow arrows sab counter-clockwise ghoom rahe hain; shaded disk har arrow ke left side mein hai. Agar image load na ho: socho ek clock face jisme haath ulte chal rahe hain, aur poora clock face shaded hai — yeh shaded face woh region hai jo loop enclose karta hai.)

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Example 1 — Cell A: andar kuch nahi, toh zero

Steps.

1. $f(z)=z^3-4z+7$ ki singularities identify karo. Yeh step kyun? Theorem tabhi fire karta hai jab $f$ poore enclosed disk par holomorphic ho. Ek polynomial $\mathbb{C}$ ke har point par differentiable hota hai — koi denominator nahi, divide karne ke liye koi root nahi — isliye kahi bhi koi singularity nahi hai.
2. Unme se saari (yani koi nahi) andar hain... kahin nahi, matlab koi $|z|=3$ ke andar nahi. Cell A confirm. Yeh step kyun? Hume confirm karna hai ki "no holes" condition hold karti hai; yahan yeh trivially hold karti hai.
3. Theorem apply karo: ${\oint }_{|z|=3}fdz=0.$

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Example 2 — Cell B: ek simple pole, $f(a)$ use karo

Steps.

1. Shape $\frac{f(z)}{z-a}$ se match karo: yahan $f(z)=\sin z$ aur $a=\frac{\pi }{2}$ hai (toh $n=0$). Yeh step kyun? Formula ko denominator exactly $z-a$ ki tarah likhna chahta hai taaki hum $a$ padh sakein.
2. Kya $a=\pi /2\approx 1.571$, $|z|=2$ ke andar hai? Haan, kyunki $|\pi /2|<2$. Aur $f(z)=\sin z$ har jagah holomorphic hai. Cell B confirm. Yeh step kyun? Agar $a$ bahar hota toh hum Cell D mein hote aur answer $0$ hota.
3. Apply karo $\oint =2\pi if(a)=2\pi i\sin (\frac{\pi }{2})=2\pi i\cdot 1=2\pi i.$

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Example 3 — Cell C: higher-order pole ke liye derivative chahiye

Steps.

1. Likho $z^4=(z-0{)}^{n+1}$. Toh $n+1=4\Rightarrow n=3$, jisme $a=0$ aur $f(z)=e^{2z}$ hai. Yeh step kyun? Generalized formula pole ka order exponent se padhta hai; $n+1$ ko power se match karna bataata hai kitni derivatives leni hain.
2. Confirm karo ki $a=0$, $|z|=1$ ke andar hai (haan) aur $e^{2z}$ holomorphic hai (haan). Cell C confirm.
3. $f^{(3)}(z)$ nikalo. $e^{2z}$ ki har derivative $2$ se multiply ho jaati hai: $f^{\prime }(z)=2e^{2z}$, $f^{\prime \prime }=4e^{2z}$, $f^{\prime \prime \prime }=8e^{2z}$. Toh $f^{(3)}(0)=8$. Yeh step kyun? Formula ko $f^{(n)}(a)=f^{(3)}(0)$ chahiye, bas yahi.
4. Apply karo $\oint =\frac{2\pi i}{n!}{f}^{(n)}(a)=\frac{2\pi i}{3!}\cdot 8=\frac{2\pi i}{6}\cdot 8=\frac{8\pi i}{3}.$

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Example 4 — Cell D: pole bahar hai → theorem phir bhi rule karta hai, answer zero hai

Steps.

1. Integrand ki singularity: $z=5$ par denominator zero. Yeh step kyun? Pole locate karna zaroori hai pehle yeh decide karne se ki kaun sa weapon fire karega.
2. Kya $z=5$, $|z|=1$ ke andar hai? Nahi — $|5|=5>1$. Toh loop ke andar, poora integrand $\frac{e^z}{z-5}$ holomorphic hai (wahan kahin denominator zero nahi hota). Cell D confirm. Yeh step kyun? Yeh Ex 2 se crucial contrast hai: same shape ka integrand, lekin pole excluded hai, toh woh loop ko dikh hi nahi raha.
3. Theorem apply karo (formula nahi): ${\oint }_{|z|=1}\frac{e^z}{z-5}dz=0.$

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Example 5 — Cell E: andar do poles hain → unhe alag karo

Steps.

1. Factor karo aur split karo: $\frac{1}{z^2-1}=\frac{1}{(z-1)(z+1)}=\frac{1/2}{z-1}-\frac{1/2}{z+1}.$ Yeh step kyun? Formula ek baar mein ek enclosed pole handle karta hai. Partial fractions ek two-pole integrand ko do Cell-B integrals mein badal dete hain jo hum pehle se solve karna jaante hain. Yeh confirm karne ke liye ki split sahi hai, dono pieces ko common denominator par laao: $\frac{1/2(z+1)-1/2(z-1)}{(z-1)(z+1)}=\frac{1}{(z-1)(z+1)}$, jo exactly wahi hai jahan se humne shuru kiya tha — toh split valid hai.
2. Dono $z=1$ aur $z=-1$, $|z|=3$ ke andar hain. Cell E confirm. Yeh step kyun? Sirf enclosed poles contribute karte hain; hum verify karte hain ki dono is bade circle se capture ho rahe hain unhe sum karne se pehle.
3. Har piece integrate karo $f=$ constant $\frac{1}{2}$, $n=0$ ke saath: $\oint \frac{1/2}{z-1}dz=2\pi i\cdot \frac{1}{2}=\pi i,\oint \frac{1/2}{z+1}dz=2\pi i\cdot \frac{1}{2}=\pi i.$ Yeh step kyun? Har ek exactly Cauchy's formula hai jisme $f(z)=\frac{1}{2}$ pole par evaluate hota hai.
4. Split ka minus sign lekar combine karo: $\oint =\pi i-\pi i=0.$ Yeh step kyun? Contour integral linear hai: ek sum (ya difference) ke pieces ka integral, unke integrals ka sum (ya difference) hota hai, $\oint (g-h)=\oint g-\oint h$. Toh do partial fractions ke beech ka algebraic minus sign seedha do contour integrals ke beech minus sign ban jaata hai. Hum kuch nayi cheez derive nahi kar rahe — bas yeh use kar rahe hain ki integration $+$ aur $-$ par distribute hota hai.

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Example 6 — Cell F: contour par baitha hua pole (degenerate/illegal)

Shuru karne se pehle ek honest warning. Jab trouble-spot path par hi hota hai, toh integral ki ek single value nahi hoti; alag-alag sensible tarike alag-alag answers dete hain. Standard repair hai principal value: contour se angular half-width $\delta$ ka ek tiny symmetric arc kaato, exactly pole ke center mein, baaki par integrate karo, phir $\delta \to 0$ karo. "Symmetric" ka matlab hai ki pole ke dono taraf se equal angle remove karo, taaki dono sides ke blow-ups ko cancel hone ka fair chance mile. Step 3 ke liye woh picture dhyan mein rakho.

Steps.

1. Pole locate karo: $z=1$. Kya woh $|z|=1$ ke andar, bahar, ya par hai? Exactly par, kyunki $|1|=1$. Yeh step kyun? Formula ki hypothesis "$a$ inside $\gamma$" ab violate ho rahi hai, aur theorem ki hypothesis "$f$ holomorphic on the contour" bhi violate ho rahi hai.
2. Koi bhi weapon legally apply nahi hota. Jaise likha hai, integrand path par unbounded hai aur integral ordinary contour integral ki tarah converge nahi karta. Yeh step kyun? Hum blindly $z=1$ kisi bhi formula mein nahi daalen — woh fake number dega.
3. Symmetric detour: removed arc ki jagah ek tiny semicircle of radius $\epsilon$ lagao jo pole ke around bulge kare. Ek simple pole ke around full small circle $2\pi i$ contribute karta hai (yeh miniature mein Ex 2 hai); half circle — exactly $\pi$ radians of arc — isliye aadha contribute karta hai, $\pi i$. Kyunki principal-value cut symmetric hai, detour exactly yahi half-circle hai, toh principal value $\pi i$ hai. Agar tum semicircle iss tarah push karo ki pole strictly bahar aaye, toh tum koi bhi residue retain nahi karte aur $0$ milta hai; agar pole strictly andar aaye, toh tum poora residue retain karte ho aur $2\pi i$ milta hai. Yeh step kyun? Yeh mechanism explicit karta hai: answer literally woh fraction hai jo tum pole ke around swing karte ho, aur $2\pi$ radians mein se $\pi$ exactly aadha hai — isliye $\pi i$ aata hai.

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Example 7 — Cell G: limiting knife-edge (radius pole ko cross karta hai)

Steps.

1. $z=2$ par pole. Radius $R$ se compare karo. Yeh step kyun? Pole enclosed hai ya nahi — yeh akela cheez hai jo $R$ vary karne se badalti hai.
2. Case $R<2$ (pole bahar): Cell D → Theorem → $I(R)=0.$ Kyun? $R<2$ ke radius $R$ ke poore disk par integrand holomorphic hai.
3. Case $R>2$ (pole andar): Cell B with $f=1$, $a=2$ → $I(R)=2\pi i\cdot 1=2\pi i.$ Kyun? Ab pole enclosed hai aur formula fire karta hai.
4. Case $R=2$: Cell F — pole contour par hai, integral undefined (Ex 6 dekho).

(Figure s02 — ek step graph: $I(R)$ ki value $R<2$ ke liye green flat line hai $0$ par, phir ek dashed yellow arrow ke saath $2\pi i$ ki height tak jump karta hai $R>2$ ke liye red flat line par; ek blue dotted vertical line $R=2$ par pole mark karti hai. Agar image load na ho: socho ek light switch jo radius $2$ se neeche har radius par OFF hai aur suddenly $2\pi i$ value par ON ho jaata hai radius $2$ se upar har radius ke liye.)

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Example 8 — Cell H: word problem (AC-circuit / averaging flavour)

Steps.

1. Mean-value property pehchano. $a=0$ ke saath $|z|=1$ par Cauchy's formula se shuru karo: $f(0)=\frac{1}{2\pi i}{\oint }_{|z|=1}\frac{f(z)}{z}dz.$ Yeh step kyun? "Ring ke around average" Cauchy's formula disguise mein hai; hum diye gaye angular average ko contour integral mein convert karte hain.
2. Parametrize karo $z=e^{i\theta }$, $dz=i{e}^{i\theta }d\theta$, toh $\frac{dz}{z}=id\theta$: $f(0)=\frac{1}{2\pi i}{\int }_0^{2\pi }f(e^{i\theta })id\theta =\frac{1}{2\pi }{\int }_0^{2\pi }f(e^{i\theta })d\theta =A.$ Yeh step kyun? Yeh dikhata hai ki designer ka $A$ exactly $f(0)$ hai — holomorphic functions ke liye mean-value property.
3. Centre evaluate karo: $f(0)=0^2+3\cdot 0+5=5.$ Yeh step kyun? Poora ring average ek single number mein collapse ho jaata hai, centre par $f(a)$.
4. Toh $A=5.$

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Example 9 — Cell I: exam twist, ek integrand, do contours

Steps.

1. Denominator factor karo: $z^2+4=(z-2i)(z+2i)$; poles $z=2i$ aur $z=-2i$ par. Yeh step kyun? Pole locations jaanna poora game hai jab contours alag hoon; har circle in do points ka ek alag subset enclose karega.
2. Contour (a) $|z-2i|=1$ — $2i$ par centered radius $1$ ka circle. Center $2i$ se pole $2i$ tak ki doori $0<1$ hai, toh $2i$ andar hai; $-2i$ tak ki doori $|2i-(-2i)|=4>1$ hai, toh $-2i$ bahar hai. Yeh Cell B hai: exactly ek enclosed simple pole. Yeh step kyun? Hume confirm karna hai ki one-pole formula apply karne se pehle exactly kaun sa ek pole grab hua hai; doosra pole safely bahar hona chahiye ya method change ho jaata hai.
3. Door wale factor ko $f$ mein fold karo. Integrand likho $\frac{f(z)}{z-2i}$ ke roop mein jisme $f(z)=\frac{z}{z+2i}$ hai, toh $a=2i$. Yeh step kyun? Cauchy's formula enclosed pole ko bare $(z-a)$ ke roop mein isolated chahta hai; bahar waale pole ka factor $(z+2i)$ kabhi ${\gamma }_1$ ke andar zero nahi hota, isliye woh wahan holomorphic hai aur legally $f$ ka part ban sakta hai.
4. Formula apply karo: $J_{{\gamma }_1}=2\pi if(2i)=2\pi i\cdot \frac{2i}{2i+2i}=2\pi i\cdot \frac{2i}{4i}=2\pi i\cdot \frac{1}{2}=\pi i.$ Yeh step kyun? $f$ holomorphic hone ke saath ${\gamma }_1$ par aur andar aur $a=2i$ par ek simple pole ke saath, formula seedha value deta hai.
5. Contour (b) $|z|=1$ — origin ke baare mein radius-$1$ circle. Har pole se doori $|\pm 2i|=2>1$ hai, toh koi bhi pole enclosed nahi hai. Yeh Cell A hai. Yeh step kyun? Agar koi singularity enclosed nahi hai, toh integrand poore disk par holomorphic hai aur theorem — formula nahi — apply hota hai.
6. Theorem apply karo: $J_{{\gamma }_2}=0.$ Yeh step kyun? Koi enclosed pole nahi ⇒ holomorphic function ka loop integral ⇒ zero, chahe integrand kitna bhi complicated lage.

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Example 10 — Cell J: same loop ulta chala (orientation)

Steps.

1. Clockwise loop ko $-\gamma$ naam do, jahan $\gamma$ usual counter-clockwise unit circle hai. Yeh step kyun? Hum sirf counter-clockwise loops evaluate karna jaante hain; clockwise wale ko counter-clockwise ke terms mein express karna hoga.
2. Reversal rule apply karo ${\oint }_{-\gamma }fdz=-{\oint }_{\gamma }fdz$. Yeh step kyun? Direction reverse karne se har $\Delta z$ step negate hota hai, isliye poora sum negate ho jaata hai — yeh wahi sign law hai jo orientation definition mein bataya gaya tha, koi nai machinery nahi chahiye.
3. Known value substitute karo: ${\oint }_{-\gamma }\frac{dz}{z}=-{\oint }_{\gamma }\frac{dz}{z}=-2\pi i.$

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Example 11 — Cell K: jab singularity pole nahi hai

Hamare do weapons assume karte hain ki akele bad points poles hain — woh spots jahan $f$ kisi finite whole number $m$ ke liye $1/(z-a{)}^m$ jaisa blow up kare. Contour calculus mein do aur creatures rehte hain, aur yeh jaanna zaroori hai ki woh Cauchy's formula ki reach se bahar hain jaise bataya gaya hai.

Steps.

1. (a) Branch point. $\sqrt{z}$ ka $0$ par branch point hai: origin ke around ek baar jaane par $\sqrt{z}\to -\sqrt{z}$ ho jaata hai, toh yeh punctured disk par single-valued holomorphic function nahi hai. Cauchy's formula require karta hai ek single-valued holomorphic $f$; woh hypothesis fail ho jaati hai. Yeh step kyun? Formula ka pehla word "holomorphic" hai — aur holomorphic mein single-valued shamil hai. Ek multivalued expression ko pehle branch cut (ek curve jise contour cross karna forbidden hai) se tame karna padta hai, jo problem hi badal deta hai. Toh hum yahan "$2\pi if(a)$" likhne ke haqdar nahi hain.
2. (b) Essential singularity — Laurent series use karo. $e^{1/z}={\sum }_{k=0}^{\infty }\frac{1}{k!}{z}^{-k}=1+\frac{1}{z}+\frac{1}{2!z^2}+\cdots$ expand karo. Yeh step kyun? $|z|=1$ par $z^n$ ka loop integral $n\ne -1$ har integer ke liye $0$ hota hai aur $n=-1$ ke liye $2\pi i$ ke barabar hota hai (yahi $\oint dz/z$ fact hai). Toh akela term jo survive karta hai woh $z^{-1}$ ka coefficient hai.
3. $z^{-1}$ ka coefficient padho ($k=1$ term): yeh $\frac{1}{1!}=1$ hai. Yeh step kyun? Yeh coefficient residue kehlata hai; poora integral $2\pi i$ times iske barabar hota hai.
4. Isliye ${\oint }_{|z|=1}{e}^{1/z}dz=2\pi i\cdot 1=2\pi i.$

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Master checklist (har contour integral par yeh run karo)

Recall Quick self-test

${\oint }_{|z|=1}\frac{\cos z}{z^2}dz=?$ ::: $n=1$, $f=\cos z$, $f^{\prime }(0)=-\sin 0=0$, toh $\frac{2\pi i}{1!}\cdot 0=0$. ${\oint }_{|z|=1}\frac{dz}{z-3}=?$ ::: pole $3$ par bahar hai; theorem ⇒ $0$. ${\oint }_{|z|=4}\frac{dz}{z-3}=?$ ::: pole $3$ ab andar hai; $2\pi i\cdot 1=2\pi i$. ${\oint }_{|z|=1}\frac{dz}{z}$ clockwise chala $=?$ ::: $-2\pi i$ (orientation reverse karne se sign flip hota hai).

Related deeper tools: ek saath bahut saare poles enclose karna Residue theorem hai; Ex 8 mein mean-value idea Maximum modulus principle aur Liouville's theorem tak jaata hai; pole ya essential singularity ke paas integrand expand karna Laurent series use karta hai; derivation khud Cauchy-Riemann equations aur Green's theorem par tiki hai.