4.10.5
Part of complex analysis. Prereqs: Laurent series , Cauchy's Integral Theorem , Singularities .
Ek pole woh jagah hai jahan ek complex function "blow up" karta hai ek controlled, polynomial tarike se (jaise 1/ z n Residue woh sabse important number hai jo us pole par chupi hoti hai — yeh Laurent expansion mein 1/ ( z − z 0 )
HUM KYUN CARE KARTE HAIN? Kyunki Cauchy's theorem kehta hai ki analytic functions ke closed-loop integrals zero hote hain, kisi singularity ke around integral mein sirf yahi 1/ ( z − z 0 ) hi integral hai (up to 2 π i
Definition Isolated singularityz 0 isolated singularity hai f f 0 < ∣ z − z 0 ∣ < R z 0 z 0 Laurent series likh sakte hain:
f ( z ) = ∑ n = − ∞ ∞ a n ( z − z 0 ) n
n < 0 principal part ) humein singularity ka type batate hain:
Definition Classification
Removable : principal part zero hai (koi negative powers nahi). e.g. z s i n z 0 Pole of order m : sabse chhota term a − m ( z − z 0 ) − m a − m = 0 1 simple kehlaata hai.Essential : infinitely many negative powers. e.g. e 1/ z 0
z 0 f residue coefficient a − 1 Res ( f , z 0 ) = a − 1
Hum chahte hain ∮ C f ( z ) d z z 0 z = z 0 + ρ e i θ θ ∈ [ 0 , 2 π ] d z = i ρ e i θ d θ
∮ C ( z − z 0 ) n d z = ∫ 0 2 π ρ n e in θ i ρ e i θ d θ = i ρ n + 1 ∫ 0 2 π e i ( n + 1 ) θ d θ
Yeh step kyun? Hum loop ko parametrize karte hain taaki integral ek ordinary θ
Ab ∫ 0 2 π e ik θ d θ = 0 k = 0 = 2 π k = 0 k = n + 1
∮ C ( z − z 0 ) n d z = { 2 π i 0 n = − 1 n = − 1
Yeh step kyun? Exponentials ek poore circle par average hokar zero ho jaate hain jab tak exponent vanish na kare. Exponent exactly tab vanish karta hai jab n = − 1
Isliye har term mar jaati hai sivaay n = − 1 :
∮ C f ( z ) d z = 2 π i a − 1 = 2 π i Res ( f , z 0 )
Worked example Simple polef ( z ) = ( z − 1 ) ( z + 2 ) z z 0 = 1 Res = lim z → 1 ( z − 1 ) ( z − 1 ) ( z + 2 ) z = 1 + 2 1 = 3 1 Yeh step kyun? ( z − 1 )
Worked example Quotient shortcutf ( z ) = z 2 + 1 e z z 0 = i p = e z q = z 2 + 1 q ′ = 2 z Res = 2 i e i Yeh step kyun? q ( i ) = 0 q ′ ( i ) = 2 i = 0 p ( i ) = 0 p / q ′
Worked example Order-2 polef ( z ) = z 2 e z z 0 = 0 m = 2 Res = 1 ! 1 lim z → 0 d z d [ z 2 ⋅ z 2 e z ] = lim z → 0 d z d e z = e 0 = 1 Yeh step kyun? z 2 m − 1 = 1
Worked example Real integral via residues∫ − ∞ ∞ x 2 + 1 d x z = i Res ( z 2 + 1 1 , i ) = 2 i 1 ⇒ Integral = 2 π i ⋅ 2 i 1 = π Yeh step kyun? Bade semicircle ka contribution → 0 2 π i × π
f ( z 0 ) Kyun sahi lagta hai: residue "pole par value" hai, toh instinctively substitute karte hain. Kyun galat hai: f ( z 0 ) infinite hai; residue coefficient a − 1 Fix: pehle hamesha ( z − z 0 ) m tab limit/derivative lo.
Common mistake Higher-order pole par simple-pole formula use karna.Kyun sahi lagta hai: lim ( z − z 0 ) f Kyun galat hai: order-2 pole ke liye, ( z − z 0 ) f ∞ m Fix: true order m
Common mistake Yeh check karna bhool jaana ki pole actually contour keandar hai ya nahi.
Kyun sahi lagta hai: tune ek valid residue compute kiya. Kyun galat hai: residue theorem sirf C Fix: har pole locate karo, test karo kaun andar hai, sirf unhe include karo.
Common mistake Essential singularity par pole formulas lagaana.Kyun galat hai: e 1/ z m z m f Fix: Laurent series se seedha a − 1 a − 1 = 1 e 1/ z e 1/ z = ∑ n ! 1 z − n
Recall Quick self-test (answers cover karo!)
Q: Residue actually kaunsa number hai? A: a − 1 ( z − z 0 ) − 1 Q: Integration mein baaki saare Laurent terms kyun vanish ho jaate hain? A: ∫ 0 2 π e ik θ d θ = 0 k = 0 n = − 1 Q: Order-m A: ( m − 1 )! 1 lim d z m − 1 d m − 1 [( z − z 0 ) m f ]
Recall Feynman: ek 12-saal ke bachche ko explain karoEk pond mein ek whirlpool imagine karo. Agar tum shant paani ke around circle mein paddling karo, tum kuch liye bina wapis aa jaate ho — paani ke dhakke ek dusre ko cancel kar dete hain. Lekin whirlpool ke around ghoomne par woh tumhe ek spin deta hai jo cancel nahi hoti. Residue us ek whirlpool ki strength hai, aur us ke around jaane par exactly 2 π i ×
"Residue = Real Important Survivor." Sirf a − 1 ( z − z 0 ) m "Multiply, Differentiate m − 1 ( m − 1 )!
Laurent series — residue sirf iska ek coefficient hai.Cauchy's Integral Theorem — explain karta hai kyun a − 1 Cauchy's Integral Formula — special case (z − z 0 f ( z ) Contour integration — main application: real definite integrals.Argument principle / Rouché's theorem — residues par based hain jo zeros/poles count karte hain.
What is the residue of f z 0 Laurent series mein ( z − z 0 ) − 1 a − 1
Why does ∮ ( z − z 0 ) n d z = 0 n = − 1 Ek circle par integral i ρ n + 1 ∫ 0 2 π e i ( n + 1 ) θ d θ 0 n + 1 = 0
Value of ∮ C ( z − z 0 ) − 1 d z 2 π i
State the Residue Theorem. ∮ C f d z = 2 π i ∑ j Res ( f , z j ) C
Residue at a simple pole formula? lim z → z 0 ( z − z 0 ) f ( z )
Residue at an order-m ( m − 1 )! 1 lim z → z 0 d z m − 1 d m − 1 [( z − z 0 ) m f ( z )]
Quotient shortcut for simple pole of p / q p ( z 0 ) / q ′ ( z 0 ) q ( z 0 ) = 0 , q ′ ( z 0 ) = 0 , p ( z 0 ) = 0
Residue of z 2 e z 0 1 e z 0
Residue of z 2 + 1 1 i 2 i 1
What distinguishes a pole from an essential singularity? Pole mein finitely many negative Laurent terms hote hain; essential mein infinitely many hote hain.
∫ − ∞ ∞ x 2 + 1 d x 2 π i ⋅ 2 i 1 = π
Cauchy's Integral Theorem
Only 1 over z-z0 term survives
