Complex integration — contour integrals

4.10.2 · Maths › Advanced Topics (Elite Level)

---

Contour integral HAI kya?

---

Key results ko scratch se derive karna

Step 1 — Fundamental example $\oint \frac{dz}{z}$

Unit circle lo $z(t)=e^{it}$, $t\in [0,2\pi ]$, counterclockwise. Toh $z^{\prime }(t)=i{e}^{it}$.

${\oint }_{|z|=1}\frac{dz}{z}={\int }_0^{2\pi }\frac{1}{e^{it}}(i{e}^{it})dt={\int }_0^{2\pi }idt=2\pi i.$

Yeh step kyun? Humne parametrisation substitute ki aur $e^{it}$ ko cancel hone diya. Bacha hua $\int idt$ trivial hai. Answer $2\pi i\ne 0$ — yahi saari residue theory ki seed hai.

Step 2 — Power $z^n$, $n\ne -1$

Usi circle ke saath, $\oint z^{n}dz={\int }_0^{2\pi }{e}^{int}i{e}^{it}dt=i{\int }_0^{2\pi }{e}^{i(n+1)t}dt.$

Integer $n\ne -1$ ke liye, $e^{i(n+1)t}$ $[0,2\pi ]$ pe periodic hai, isliye uska integral $0$ hai.

$\boxed{{\oint }_{|z|=1}{z}^{n}dz=\{\begin{array}{ll}2\pi i,& n=-1\\ 0,& n\in \mathbb{Z},n\ne -1\end{array}}$

Yeh step kyun? Sirf $1/z$ term ek loop mein survive karta hai. Har residue formula isi fact ka disguised roop hai.

Step 3 — Analytic $f$ ke liye path independence (Cauchy's theorem ka idea)

$f=u+iv$, $z=x+iy$, $dz=dx+idy$ likho: ${\int }_{\gamma }fdz={\int }_{\gamma }(udx-vdy)+i{\int }_{\gamma }(vdx+udy).$ Green's theorem apply karo har piece pe closed $\gamma$ se enclosed region $D$ ke upar:

+i\iint_D\!\Big(\tfrac{\partial u}{\partial x}-\tfrac{\partial v}{\partial y}\Big)dA.$$ Agar $f$ **analytic** hai, toh ==Cauchy–Riemann equations== $u_x=v_y,\ u_y=-v_x$ **dono** integrands ko zero kar dete hain. > [!formula] Cauchy's Integral Theorem > Agar $f$ ek closed contour $\gamma$ par aur uske andar analytic hai, toh > $$\oint_\gamma f(z)\,dz = 0.$$ > **KYUN:** analyticity = Cauchy–Riemann = dono Green's-theorem integrands zero hain. ### Step 4 — Cauchy's Integral Formula $f$ ko analytic lo, $a$ ko $\gamma$ ke andar lo. $\dfrac{f(z)}{z-a}$ consider karo, jiska $a$ par ek bura point hai. $\gamma$ ko ek tiny circle $z=a+\varepsilon e^{it}$ par deform karo (allowed hai, kyunki $f/(z-a)$ unke beech ki ring mein analytic hai, isliye wahan loop integral $0$ hai). Toh $$\oint_\gamma\frac{f(z)}{z-a}dz=\int_0^{2\pi}\frac{f(a+\varepsilon e^{it})}{\varepsilon e^{it}}\, i\varepsilon e^{it}\,dt = i\int_0^{2\pi}f(a+\varepsilon e^{it})\,dt \xrightarrow{\varepsilon\to0} 2\pi i\,f(a).$$ > [!formula] Cauchy Integral Formula & Residue Theorem > $$f(a)=\frac{1}{2\pi i}\oint_\gamma\frac{f(z)}{z-a}\,dz,\qquad > \oint_\gamma f\,dz = 2\pi i\sum_k \operatorname{Res}_{z=z_k} f.$$ > Ek **residue** $\operatorname{Res}_{z_0}f$ coefficient $c_{-1}$ hai $f$ ki Laurent series mein > $z_0$ ke baare mein — exactly wahi ek term jo Step 2 mein survive ki. ![[4.10.02-Complex-integration-—-contour-integrals.png]] --- ## Worked examples > [!example] Example 1 — Direct parametrisation > $\displaystyle\int_\gamma \bar z\,dz$ evaluate karo, $\gamma$ = $0$ se $1+i$ tak seedhi line. > - Parametrise karo $z(t)=t(1+i)$, $t\in[0,1]$, isliye $\bar z = t(1-i)$, $z'=1+i$. **Kyun?** seedhi line = linear $z(t)$. > - $\int_0^1 t(1-i)(1+i)\,dt = \int_0^1 t\cdot 2\,dt = 1.$ **Kyun?** $(1-i)(1+i)=2$. > - Note: $\bar z$ **analytic nahi** hai, isliye answer path pe depend karta hai. $z$ se compare karo. > [!example] Example 2 — Residue theorem use karke > $\displaystyle\oint_{|z|=2}\frac{z}{z^2+1}\,dz$. > - Poles jahan $z^2+1=0\Rightarrow z=\pm i$; dono $|z|=2$ ke andar hain. **Kyun?** $|{\pm i}|=1<2$. > - $\operatorname{Res}_{z=i}=\dfrac{z}{2z}\Big|_{i}=\tfrac12$, usi tarah $-i$ par bhi $\tfrac12$ milta hai. > **Kyun?** simple pole rule $\operatorname{Res}=\dfrac{p(z)}{q'(z)}$ jahan $q'=2z$. > - Sum $=1$, isliye integral $=2\pi i\cdot 1 = 2\pi i.$ > [!example] Example 3 — Ek real integral contour se > $\displaystyle\int_{-\infty}^{\infty}\frac{dx}{x^2+1}$. > - Upper half-plane mein ek bade semicircle se close karo (radius $R\to\infty$). **Kyun?** arc ka > contribution $\to 0$ kyunki $|f|\sim 1/R^2$ aur arc length $\sim \pi R$. > - Andar sirf pole $z=i$ hai: $\operatorname{Res}_{i}\frac{1}{z^2+1}=\frac{1}{2i}$. > - Integral $=2\pi i\cdot\frac{1}{2i}=\pi.$ $\arctan x\big|_{-\infty}^{\infty}=\pi$ se match karta hai. ✔ --- ## Forecast-then-Verify > [!recall]- $\oint_{|z|=1}\frac{e^z}{z}dz$ compute karne se pehle forecast karo > Pehle guess karo: $e^z/z$ ki Laurent series ka kaunsa term ise control karta hai? … $1/z$ coefficient > $e^0=1$ hai, isliye CIF se $f(0)=1$ aur integral $=2\pi i\cdot 1 = 2\pi i$. Kya tumne $2\pi i$ predict kiya tha? --- ## Common mistakes (Steel-man + fix) > [!mistake] "Saare closed loop integrals zero hote hain." > **Kyun sahi lagta hai:** Cauchy's theorem kehta hai $\oint f=0$. **Catch yeh hai:** iske liye $f$ ko > **andar analytic** hona chahiye. $\oint dz/z = 2\pi i \neq 0$ kyunki $z=0$ andar ek singularity hai. > **Fix:** $0$ declare karne se pehle poles check karo. > [!mistake] "$\int_\gamma f\,dz$ hamesha sirf endpoints pe depend karta hai." > **Kyun sahi lagta hai:** real integrals ko sirf endpoints chahiye; aur analytic $f$ ke liye yeh *sach* hai. > **Fix:** path-independence ke liye analyticity chahiye. $\int_\gamma \bar z\,dz$ (Example 1) path-dependent hai. > [!mistake] $z'(t)$ ($dz$ factor) bhool jaana. > **Kyun sahi lagta hai:** real calculus mein $dx$ "bas wahan hota hai". **Fix:** $dz=z'(t)\,dt$ hamesha; > $ie^{it}$ drop karne se answer khatam ho jaata hai. > [!mistake] Orientation sign. > Counterclockwise $+2\pi i$ deta hai; clockwise sign flip kar deta hai. **Fix:** standard positive orientation > counterclockwise hai; $\gamma$ reverse karne se integral $-1$ se multiply ho jaata hai. --- ## 80/20 — woh 20% jo 80% deta hai 1. $dz=z'(t)\,dt$, phir yeh $t$ mein ek real integral hai. 2. $\oint z^n dz = 2\pi i$ sirf tab jab $n=-1$, warna $0$. 3. Andar har jagah analytic $\Rightarrow \oint=0$. 4. Warna $\oint = 2\pi i\sum\text{Res}$, aur simple-pole residue $=\frac{p}{q'}$. --- ## #flashcards/maths Parametrisation $z(t)$ ke terms mein $dz$ kya hai? ::: $dz=z'(t)\,dt$. $\oint_{|z|=1} z^n\,dz$ ki value? ::: $2\pi i$ agar $n=-1$, warna $0$ (integer $n$). Cauchy's Integral Theorem batao. ::: Agar $f$ closed $\gamma$ par aur uske andar analytic hai, toh $\oint_\gamma f\,dz=0$. $f$ ki kaunsi property dono Green's-theorem integrands ko vanish karati hai? ::: Cauchy–Riemann equations $u_x=v_y,\ u_y=-v_x$. Cauchy's Integral Formula batao. ::: $f(a)=\frac{1}{2\pi i}\oint_\gamma \frac{f(z)}{z-a}\,dz$, $a$ $\gamma$ ke andar. Residue Theorem ka statement. ::: $\oint_\gamma f\,dz = 2\pi i\sum_k \operatorname{Res}_{z_k} f$. $p/q$ ka simple pole $z_0$ par residue (jahan $q(z_0)=0$)? ::: $\operatorname{Res}=p(z_0)/q'(z_0)$. $\int_\gamma \bar z\,dz$ path-dependent kyun hai? ::: Kyunki $\bar z$ analytic nahi hai, isliye Cauchy's theorem apply nahi hota. $\oint_{|z|=1} e^z/z\,dz$ = ? ::: $2\pi i$ (kyunki $f(0)=e^0=1$ CIF se). Contour orientation ulta karne ka effect? ::: Integral $-1$ se multiply ho jaata hai. --- > [!recall]- Feynman: ek 12-saal ke bachche ko samjhao > Socho tum ek flat maidan mein chal rahe ho aur "arrows" collect kar rahe ho (har point tumhe ek chhota arrow deta hai). > Agar maidan bilkul smooth hai, toh loop ke saath poora chakkar lagaane par arrows cancel ho jaate hain aur tumhare paas > **kuch nahi** bachta ($0$). Lekin agar tumhare loop ke andar ek magic spike (ek "pole") hai, toh tum use kabhi cancel nahi kar sakte — uske around har loop tumhe **wahi fixed amount** deta hai, $2\pi i$ uski strength ka. Isliye ek mushkil sum bass loop ke andar spikes ginne mein badal jaata hai! > [!mnemonic] > **"Sirf $1/z$ loop mein survive karta hai; jo bhi analytic hai woh zero ho jaata hai."** > Aur: **R**esidue **T**heorem = **R**ound **T**rip spikes count karta hai ($2\pi i\sum$Res). ## Connections - [[Cauchy–Riemann equations]] — analyticity loop integral ko kyun khatam kar deti hai. - [[Laurent series]] — residue = coefficient $c_{-1}$. - [[Green's theorem]] — Cauchy's theorem ke peeche ka real-analysis engine. - [[Residue theorem applications]] — real definite integrals evaluate karna. - [[Analytic functions]] — woh class jiske liye paths matter nahi karte. ## 🖼️ Concept Map ```mermaid flowchart TD P[Parametrise contour z of t] -->|converts 2D walk to| I[Contour integral in t] I -->|integrate 1/z on circle| S[Seed result 2 pi i] I -->|integrate z^n n not -1| Z[Result 0] S -->|only 1/z survives loop| R[Residue idea] Z -->|only 1/z survives loop| R A[Analytic f] -->|satisfies| CR[Cauchy-Riemann equations] CR -->|kills Green integrands| CT[Cauchy Integral Theorem loop = 0] I -->|split u+iv and Green| CT CT -->|deform around bad point a| CIF[Cauchy Integral Formula] S -->|supplies 2 pi i factor| CIF A -->|f over z minus a| CIF ```