Foundations — Residues and poles

4.10.5 · D1 · Maths › Advanced Topics (Elite Level) › Residues and poles

0. Wo plane jahan sab kuch rehta hai

Kisi bhi formula se pehle, tumhe yeh jaanna hai ki ek complex number kaisa dikhta hai.

Humein yeh kyun chahiye. Residues is plane mein points $z_0$ par rehte hain, aur hum is plane par bani paths ke along integrate karte hain. Agar $z$ sirf ek line par ek number hota, toh kisi cheez ke "around jaane" ka koi matlab nahi hota — loop karne ke liye tumhe do dimensions chahiye.

1. Power series kya hoti hai?

Baad ki kai definitions "power series" phrase par lean karti hain, isliye hum ise use karne se pehle pin down karte hain.

Humein yeh kyun chahiye. Complex exponential (next) ek power series hai, aur "analytic" ka matlab hoga "locally ek power series ke equal". Toh yeh woh vocabulary hai jisme woh definitions bolti hain.

2. Complex exponential aur polar form

Kisi point ke "around jaane" ke liye humein ek angle chahiye. Woh tool jo "angle" ko "circle par ek point" mein convert karta hai woh complex exponential hai, toh hum pehle ise build karte hain.

3. Complex derivative — "analytic" ka asli matlab

Analytic word ek limit par rest karta hai, toh hum woh limit state karte hain us word ko use karne se pehle.

4. Ek complex number ke functions, aur woh kahan toot jaate hain

5. Laurent series — sab kuch ka fuel

Ek Taylor series (ek power series sirf powers $z^0,z^1,z^2,\dots$ ke saath) sirf nice points describe kar sakti hai. Ek blow-up describe karne ke liye hum negative powers bhi allow karte hain. Woh negative powers sirf ek ring-shaped region mein sense banate hain, toh hum pehle us ring ko name karte hain.

6. Do integrals-around-a-loop, aur woh theorems jo inhe tame karti hain

7. Residue Theorem — derivation poori tarah dikhaya gaya

Ab hum ${\oint }_{C}fdz$ compute karte hain radius $\rho$ ke ek small circle ke around jo exactly ek singularity $z_0$ enclose karta hai, step by step.

Step 1 — loop ko parametrize karo. $C$ ko circle $|z-z_0|=\rho$ hone do jo counter-clockwise trace hoti hai: $z=z_0+\rho e^{i\theta },\theta \in [0,2\pi ].$ Tiny step $dz$ paane ke liye ise $\theta$ ke respect mein differentiate karo: $\frac{dz}{d\theta }=\rho \cdot i{e}^{i\theta }⟹dz=i\rho e^{i\theta }d\theta .$ Kyun: yeh abstract loop-integral ko single real variable $\theta$ mein ek ordinary integral mein convert karta hai.

Step 2 — Laurent series insert karo aur sum ko integral ke saath swap karo. $\rho$ choose karo jisme $r<\rho <R$ ho taki circle annulus ke andar ho. Wahan series uniformly converge karti hai, aur — Section 5 mein note kiya gaya — uniform convergence exactly woh theorem hai jo infinite sum aur integral swap karne ki permission deta hai (finite errors tab bhi finite rehte hain jab term by term integrate kiya jaata hai). Isliye: ${\oint }_{C}fdz={\oint }_C{\sum }_{n=-\infty }^{\infty }{a}_n(z-z_0{)}^{n}dz={\sum }_{n=-\infty }^{\infty }{a}_n{\oint }_C(z-z_0{)}^{n}dz.$

Step 3 — ek term evaluate karo. $z-z_0=\rho e^{i\theta }$ aur $dz=i\rho e^{i\theta }d\theta$ ke saath: ${\oint }_C(z-z_0{)}^{n}dz={\int }_0^{2\pi }(\rho e^{i\theta }{)}^{n}i\rho e^{i\theta }d\theta =i{\rho }^{n+1}{\int }_0^{2\pi }{e}^{i(n+1)\theta }d\theta .$ Kyun: hum saare $\rho$'s collect karte hain aur do exponentials $e^{in\theta }\cdot e^{i\theta }=e^{i(n+1)\theta }$ ko merge karte hain.

Step 4 — $k=n+1$ ke saath averaging fact apply karo. Section 2 ke formula mein exponent index $k$ tha; yahan $k=n+1$ hai (aur kyunki $n$ ek integer hai, $k$ bhi hai). Formula do cases mein split hota hai — aur humein dono rakhne hain: ${\int }_0^{2\pi }{e}^{i(n+1)\theta }d\theta =\{\begin{array}{ll}2\pi & n+1=0(\text{i.e. }n=-1)\\ 0& n+1\ne 0(\text{i.e. }n\ne -1).\end{array}$ Har case ko Step 3 mein wapas feed karo: ${\oint }_C(z-z_0{)}^{n}dz=\{\begin{array}{ll}i{\rho }^0\cdot 2\pi =2\pi i& n=-1\\ i{\rho }^{n+1}\cdot 0=0& n\ne -1.\end{array}$ Notice karo ki radius $\rho$ surviving case se vanish ho gaya (${\rho }^{n+1}={\rho }^0=1$) — answer is baat par depend nahi karta ki circle kitni badi hai.

Step 5 — sum ko collapse karo. Step 2 ke sum mein, har ek term $0$ se multiply hota hai sivaay $n=-1$ wale ke, jiska factor $2\pi i$ hai. Woh surviving term coefficient $a_{-1}$ carry karta hai: ${\oint }_{C}f(z)dz={\sum }_{n\ne -1}{a}_n\cdot 0+a_{-1}\cdot 2\pi i=2\pi i{a}_{-1}=2\pi i\mathrm{Res}(f,z_0).$ Toh sirf $a_{-1}$ term bachta hai, aur poora loop-integral us ek number times $2\pi i$ ke barabar hai — residue. Yahi poora punchline hai.

Foundations topic ko kaise feed karte hain

Neeche ka map top-to-bottom mein dependence ki chain ki tarah padha jaata hai: har arrow ka matlab hai "upper box ko own karo pehle, tab lower wala meaningful hoga". Do theorem boxes (Cauchy, Residue) bottom mein hain kyunki woh upar ki sab cheez consume karte hain.

Equipment checklist

Parent note padhne se pehle right side cover karo aur khud ko test karo.

Connections

• Laurent series — woh series jiska coefficient $a_{-1}$ residue hai.
• Singularities — bad points (removable / pole / essential) classify karta hai jahan residues rehte hain.
• Cauchy's Integral Theorem — reason ki loop par saare non-$a_{-1}$ terms die kyun karte hain.
• Cauchy's Integral Formula — residue idea $\frac{f(z)}{z-z_0}$ par apply kiya gaya.
• Contour integration — real integrals evaluate karne ke liye ${\oint }_C$ use karna.
• Residues and poles — parent topic jinke liye yeh foundations build karte hain.