Part of complex analysis. Prereqs: Laurent series , Cauchy's Integral Theorem , Singularities .
A pole is a place where a complex function "blows up" in a controlled, polynomial way (like 1 / z n 1/z^n 1/ z n residue is the single most important number hiding at that pole — it's the coefficient of the 1 / ( z − z 0 ) 1/(z-z_0) 1/ ( z − z 0 )
WHY do we care? Because Cauchy's theorem says closed-loop integrals of analytic functions are zero, the only thing that survives an integral around a singularity is this 1 / ( z − z 0 ) 1/(z-z_0) 1/ ( z − z 0 ) is the integral (up to 2 π i 2\pi i 2 π i
Definition Isolated singularityz 0 z_0 z 0 isolated singularity of f f f f f f 0 < ∣ z − z 0 ∣ < R 0<|z-z_0|<R 0 < ∣ z − z 0 ∣ < R z 0 z_0 z 0 z 0 z_0 z 0 Laurent series :
f ( z ) = ∑ n = − ∞ ∞ a n ( z − z 0 ) n f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n f ( z ) = ∑ n = − ∞ ∞ a n ( z − z 0 ) n
The n < 0 n<0 n < 0 principal part ) tell us the type of singularity:
Definition Classification
Removable : principal part is zero (no negative powers). e.g. sin z z \frac{\sin z}{z} z s i n z 0 0 0 Pole of order m m m : lowest term is a − m ( z − z 0 ) − m a_{-m}(z-z_0)^{-m} a − m ( z − z 0 ) − m a − m ≠ 0 a_{-m}\neq 0 a − m = 0 1 1 1 simple .Essential : infinitely many negative powers. e.g. e 1 / z e^{1/z} e 1/ z 0 0 0
The residue of f f f z 0 z_0 z 0 a − 1 a_{-1} a − 1 Res ( f , z 0 ) = a − 1 \boxed{\operatorname{Res}(f,z_0)=a_{-1}} Res ( f , z 0 ) = a − 1
We want ∮ C f ( z ) d z \oint_C f(z)\,dz ∮ C f ( z ) d z z 0 z_0 z 0 z = z 0 + ρ e i θ z=z_0+\rho e^{i\theta} z = z 0 + ρ e i θ θ ∈ [ 0 , 2 π ] \theta\in[0,2\pi] θ ∈ [ 0 , 2 π ] d z = i ρ e i θ d θ dz=i\rho e^{i\theta}d\theta d z = i ρ e i θ d θ
∮ C ( z − z 0 ) n d z = ∫ 0 2 π ρ n e i n θ i ρ e i θ d θ = i ρ n + 1 ∫ 0 2 π e i ( n + 1 ) θ d θ \oint_C (z-z_0)^n\,dz=\int_0^{2\pi}\rho^n e^{in\theta}\, i\rho e^{i\theta}\,d\theta = i\rho^{n+1}\int_0^{2\pi} e^{i(n+1)\theta}\,d\theta ∮ 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
Why this step? We parametrize the loop so the integral becomes an ordinary θ \theta θ
Now ∫ 0 2 π e i k θ d θ = 0 \int_0^{2\pi}e^{ik\theta}d\theta = 0 ∫ 0 2 π e ik θ d θ = 0 k ≠ 0 k\neq 0 k = 0 = 2 π =2\pi = 2 π k = 0 k=0 k = 0 k = n + 1 k=n+1 k = n + 1
∮ C ( z − z 0 ) n d z = { 2 π i n = − 1 0 n ≠ − 1 \oint_C (z-z_0)^n\,dz=\begin{cases}2\pi i & n=-1\\ 0 & n\neq -1\end{cases} ∮ C ( z − z 0 ) n d z = { 2 π i 0 n = − 1 n = − 1
Why this step? The exponentials average to zero over a full circle unless the exponent vanishes. The exponent vanishes precisely when n = − 1 n=-1 n = − 1
Therefore every term dies except n = − 1 n=-1 n = − 1 :
∮ C f ( z ) d z = 2 π i a − 1 = 2 π i Res ( f , z 0 ) \oint_C f(z)\,dz = 2\pi i\, a_{-1}=2\pi i\operatorname{Res}(f,z_0) ∮ C f ( z ) d z = 2 π i a − 1 = 2 π i Res ( f , z 0 )
Worked example Simple polef ( z ) = z ( z − 1 ) ( z + 2 ) f(z)=\dfrac{z}{(z-1)(z+2)} f ( z ) = ( z − 1 ) ( z + 2 ) z z 0 = 1 z_0=1 z 0 = 1 Res = lim z → 1 ( z − 1 ) z ( z − 1 ) ( z + 2 ) = 1 1 + 2 = 1 3 \operatorname{Res}=\lim_{z\to1}(z-1)\frac{z}{(z-1)(z+2)}=\frac{1}{1+2}=\frac13 Res = lim z → 1 ( z − 1 ) ( z − 1 ) ( z + 2 ) z = 1 + 2 1 = 3 1 Why this step? The ( z − 1 ) (z-1) ( z − 1 )
Worked example Quotient shortcutf ( z ) = e z z 2 + 1 f(z)=\dfrac{e^z}{z^2+1} f ( z ) = z 2 + 1 e z z 0 = i z_0=i z 0 = i p = e z p=e^z p = e z q = z 2 + 1 q=z^2+1 q = z 2 + 1 q ′ = 2 z q'=2z q ′ = 2 z Res = e i 2 i \operatorname{Res}=\frac{e^i}{2i} Res = 2 i e i Why this step? q ( i ) = 0 q(i)=0 q ( i ) = 0 q ′ ( i ) = 2 i ≠ 0 q'(i)=2i\neq0 q ′ ( i ) = 2 i = 0 p ( i ) ≠ 0 p(i)\neq0 p ( i ) = 0 p / q ′ p/q' p / q ′
Worked example Order-2 polef ( z ) = e z z 2 f(z)=\dfrac{e^z}{z^2} f ( z ) = z 2 e z z 0 = 0 z_0=0 z 0 = 0 m = 2 m=2 m = 2 Res = 1 1 ! lim z → 0 d d z [ z 2 ⋅ e z z 2 ] = lim z → 0 d d z e z = e 0 = 1 \operatorname{Res}=\frac{1}{1!}\lim_{z\to0}\frac{d}{dz}\Big[z^2\cdot\frac{e^z}{z^2}\Big]=\lim_{z\to0}\frac{d}{dz}e^z=e^0=1 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 Why this step? Multiply by z 2 z^2 z 2 m − 1 = 1 m-1=1 m − 1 = 1
Worked example Real integral via residuesCompute ∫ − ∞ ∞ d x x 2 + 1 \displaystyle\int_{-\infty}^{\infty}\frac{dx}{x^2+1} ∫ − ∞ ∞ x 2 + 1 d x z = i z=i z = i Res ( 1 z 2 + 1 , i ) = 1 2 i ⇒ Integral = 2 π i ⋅ 1 2 i = π \operatorname{Res}\Big(\tfrac{1}{z^2+1},i\Big)=\frac{1}{2i}\Rightarrow \text{Integral}=2\pi i\cdot\frac{1}{2i}=\pi Res ( z 2 + 1 1 , i ) = 2 i 1 ⇒ Integral = 2 π i ⋅ 2 i 1 = π Why this step? The big semicircle contribution → 0 \to0 → 0 2 π i × 2\pi i\times 2 π i × π \pi π
Common mistake "Just plug inf ( z 0 ) f(z_0) f ( z 0 ) Why it feels right: residue is "the value at the pole", so you instinctively substitute. Why it's wrong: f ( z 0 ) f(z_0) f ( z 0 ) infinite ; the residue is the coefficient a − 1 a_{-1} a − 1 Fix: always multiply by ( z − z 0 ) m (z-z_0)^m ( z − z 0 ) m then take a limit/derivative.
Common mistake Using the simple-pole formula on a higher-order pole.Why it feels right: lim ( z − z 0 ) f \lim(z-z_0)f lim ( z − z 0 ) f Why wrong: for an order-2 pole, ( z − z 0 ) f (z-z_0)f ( z − z 0 ) f ∞ \infty ∞ m m m Fix: find the true order m m m
Common mistake Forgetting to check the pole is actuallyinside the contour.
Why it feels right: you computed a valid residue. Why wrong: the residue theorem only sums singularities enclosed by C C C Fix: locate every pole, test which lie inside, include only those.
Common mistake Treating an essential singularity with the pole formulas.Why wrong: e 1 / z e^{1/z} e 1/ z m m m z m f z^m f z m f Fix: read a − 1 a_{-1} a − 1 a − 1 = 1 a_{-1}=1 a − 1 = 1 e 1 / z e^{1/z} e 1/ z e 1 / z = ∑ 1 n ! z − n e^{1/z}=\sum \frac{1}{n!}z^{-n} e 1/ z = ∑ n ! 1 z − n
Recall Quick self-test (cover the answers!)
Q: What number is the residue? A: a − 1 a_{-1} a − 1 ( z − z 0 ) − 1 (z-z_0)^{-1} ( z − z 0 ) − 1 Q: Why do all other Laurent terms vanish on integration? A: ∫ 0 2 π e i k θ d θ = 0 \int_0^{2\pi}e^{ik\theta}d\theta=0 ∫ 0 2 π e ik θ d θ = 0 k = 0 k=0 k = 0 n = − 1 n=-1 n = − 1 Q: Residue formula for order-m m m A: 1 ( m − 1 ) ! lim d m − 1 d z m − 1 [ ( z − z 0 ) m f ] \frac{1}{(m-1)!}\lim \frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f] ( m − 1 )! 1 lim d z m − 1 d m − 1 [( z − z 0 ) m f ]
Recall Feynman: explain to a 12-year-oldImagine a whirlpool in a pond. If you paddle in a circle around calm water, you come back with nothing — the water's pushes cancel out. But circle around the whirlpool and it gives you a spin you can't cancel. The residue is the strength of that one whirlpool, and going around it adds exactly 2 π i × 2\pi i\times 2 π i ×
"Residue = Real Important Survivor." Only a − 1 a_{-1} a − 1 ( z − z 0 ) (z-z_0) ( z − z 0 ) m m m "Multiply, Differentiate m − 1 m{-}1 m − 1 ( m − 1 ) ! (m{-}1)! ( m − 1 )!
Laurent series — residue is just one of its coefficients.Cauchy's Integral Theorem — explains why everything but a − 1 a_{-1} a − 1 Cauchy's Integral Formula — special case (residue of f ( z ) z − z 0 \frac{f(z)}{z-z_0} z − z 0 f ( z ) Contour integration — main application: real definite integrals.Argument principle / Rouché's theorem — built on residues counting zeros/poles.
What is the residue of f f f z 0 z_0 z 0 The coefficient
a − 1 a_{-1} a − 1 of
( z − z 0 ) − 1 (z-z_0)^{-1} ( z − z 0 ) − 1 in the Laurent series.
Why does ∮ ( z − z 0 ) n d z = 0 \oint (z-z_0)^n dz=0 ∮ ( z − z 0 ) n d z = 0 n ≠ − 1 n\neq -1 n = − 1 On a circle the integral becomes
i ρ n + 1 ∫ 0 2 π e i ( n + 1 ) θ d θ i\rho^{n+1}\int_0^{2\pi}e^{i(n+1)\theta}d\theta i ρ n + 1 ∫ 0 2 π e i ( n + 1 ) θ d θ , which is
0 0 0 unless
n + 1 = 0 n+1=0 n + 1 = 0 .
Value of ∮ C ( z − z 0 ) − 1 d z \oint_C (z-z_0)^{-1} dz ∮ C ( z − z 0 ) − 1 d z State the Residue Theorem. ∮ C f d z = 2 π i ∑ j Res ( f , z j ) \oint_C f\,dz = 2\pi i\sum_j \operatorname{Res}(f,z_j) ∮ C f d z = 2 π i ∑ j Res ( f , z j ) over poles inside
C C C .
Residue at a simple pole formula? lim z → z 0 ( z − z 0 ) f ( z ) \lim_{z\to z_0}(z-z_0)f(z) lim z → z 0 ( z − z 0 ) f ( z ) .
Residue at an order-m m m 1 ( m − 1 ) ! lim z → z 0 d m − 1 d z m − 1 [ ( z − z 0 ) m f ( z ) ] \frac{1}{(m-1)!}\lim_{z\to z_0}\frac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)] ( 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/q p / q p ( z 0 ) / q ′ ( z 0 ) p(z_0)/q'(z_0) p ( z 0 ) / q ′ ( z 0 ) when
q ( z 0 ) = 0 , q ′ ( z 0 ) ≠ 0 , p ( z 0 ) ≠ 0 q(z_0)=0,\ q'(z_0)\neq0,\ p(z_0)\neq0 q ( z 0 ) = 0 , q ′ ( z 0 ) = 0 , p ( z 0 ) = 0 .
Residue of e z z 2 \frac{e^z}{z^2} z 2 e z 0 0 0 1 1 1 (derivative of
e z e^z e z at
0 0 0 ).
Residue of 1 z 2 + 1 \frac{1}{z^2+1} z 2 + 1 1 i i i What distinguishes a pole from an essential singularity? Pole has finitely many negative Laurent terms; essential has infinitely many.
∫ − ∞ ∞ d x x 2 + 1 \int_{-\infty}^\infty \frac{dx}{x^2+1} ∫ − ∞ ∞ x 2 + 1 d x 2 π i ⋅ 1 2 i = π 2\pi i\cdot\frac{1}{2i}=\pi 2 π i ⋅ 2 i 1 = π .
Cauchy's Integral Theorem
Only 1 over z-z0 term survives
Intuition Hinglish mein samjhoDekho, complex analysis mein pole ka matlab hai woh point jahan function "phat" jaata hai, jaise 1 / z 1/z 1/ z 1 / z 2 1/z^2 1/ z 2 residue us pole pe chhupa ek single number hai: Laurent series mein jo 1 / ( z − z 0 ) 1/(z-z_0) 1/ ( z − z 0 ) a − 1 a_{-1} a − 1
Sabse important baat: agar tum kisi closed loop ke around integral karo, toh Cauchy theorem ke wajah se baaki sab terms cancel ho jaate hain. Sirf a − 1 a_{-1} a − 1 ∫ 0 2 π e i k θ d θ \int_0^{2\pi} e^{ik\theta} d\theta ∫ 0 2 π e ik θ d θ k = 0 k=0 k = 0 − 1 -1 − 1 ∮ C f d z = 2 π i × ( residues ka sum ) \oint_C f\,dz = 2\pi i \times (\text{residues ka sum}) ∮ C f d z = 2 π i × ( residues ka sum ) Residue Theorem hai — bahut powerful, kyunki tough real integrals (jaise ∫ d x x 2 + 1 \int \frac{dx}{x^2+1} ∫ x 2 + 1 d x
Residue nikaalne ka shortcut: simple pole ke liye ( z − z 0 ) f ( z ) (z-z_0)f(z) ( z − z 0 ) f ( z ) Order m m m pole ke liye ( z − z 0 ) m f (z-z_0)^m f ( z − z 0 ) m f ( m − 1 ) (m-1) ( m − 1 ) ( m − 1 ) ! (m-1)! ( m − 1 )! f = p / q f = p/q f = p / q p ( z 0 ) / q ′ ( z 0 ) p(z_0)/q'(z_0) p ( z 0 ) / q ′ ( z 0 )
Galti se bachna: residue ka matlab f ( z 0 ) f(z_0) f ( z 0 ) nahi hai (woh toh infinite hai!), aur sirf woh poles count karo jo contour ke andar ho. Yaad rakho: "Residue = Real Important Survivor" — loop mein bas a − 1 a_{-1} a − 1
