Functions of a Complex Variable (S1)

VI. RESIDUE

⊲ Definition: residue of a function f at point z0

⊲

⊲ Relationship between complex integration and power expansion

⊲ Techniques and applications of complex RESIDUE CALCULUS

Complex differentiation, complex integration and power series expansions • provide three approaches to the theory of holomorphic functions.

Cauchy integral formulas can be seen as providing • the relationship between the first two.

Residues serve to formulate the relationship between • complex integration and power series expansions. DEFINITION OF RESIDUE

♦ Let f be holomorphic everywhere within and on a closed curve C except possibly at a point z0 in the interior of C where f may have an .

Define residue of f at z : ♣ 0 1 Resz0 f = f(z) dz 2πi IC

If z is a non-singular point, Res f = 0. Otherwise, Res f may be = 0. 0 z0 z0 6 RESIDUE THEOREM ♦ Let C be closed path within and on which f is holomorphic except for m isolated singularities. Then

m f(z) dz = 2πi Resz f I j C Xj=1 Im z C

z 0 zm zj

Re z

⊲ reformulation of Cauchy theorem via arguments similar to those used for deformation theorem Relationship with Laurent expansion

Consider expansion of f about z0: ∞ n 1 f(ξ) f(z) = cn(z z0) , where cn = dξ − 2πi I (ξ z )n+1 nX=−∞ Γ − 0

⊲ Res f is nothing but the coefficient of (z z )−1 z0 − 0 in the Laurent expansion of f about z0 1 c−1 = dξ f(ξ) = Resz0 f 2πi IΓ

Laurent expansion thus provides a general method to compute residues. • EXAMPLES Compute the residue at the singularity of the function • 1 z f(z)= − (1 2z)2 −

z =1/2 pole of order 2 1 − z 1 1 1 1 1 = − ⇒ Res =1 2f = − (1 − 2z)2 8 (z − 1/2)2 4 z − 1/2 z / 4

residue |{z} Compute the residue at the singularity of the function • 2 f(z)= e1/z

z =0

2 1/z 1 1 1 e =1+ + + . . . ⇒ Res =0f =0 z2 2! z4 z Calculation of residues in the case of poles

⊲ If z0 is pole of order n for f, then h(z) f(z)= , h holomorphic and h(z0) =06 (z − z0)n

Substituting this into the definition of residue gives 1 h(z) Resz0 f = n dz 2πi IC (z − z0) − 1 dn 1h(z) = n−1 by Cauchy formula (n − 1)!  dz z=z0 n−1 1 d n = lim −1 [(z − z0) f(z)] z→z0 (n − 1)! dzn

1 − z Ex.(previous slide) f(z)= (1 − 2z)2 − d − 2 d 1 z − 1 Resz=1/2f = lim [(z 1/2) f(z)]= lim [ ]= z→1/2 dz z→1/2 dz 4 4 METHODS TO CALCULATE RESIDUES General method: from Laurent expansion ♠ ∞ n 1 f(ξ) f(z)= cn(z z0) , where cn = dξ − 2πi I (ξ z )n+1 n=X−∞ Γ − 0

Resz0 f = c−1 1 = dξ f(ξ) 2πi IΓ

Method for z pole of order n: ♠ 0 n−1 1 d n Resz0 f = lim n−1 [(z z0) f(z)] z z0 (n 1)! dz − → −

For n =1: Resz0 f = lim [(z z0)f(z)] z z0 − → sin z = 6 dz I IC z where C is the circle of centre z = 0 and radius 1

z = 0 pole of order 5 sin z 1 z3 z5 1 1 1 1 1 = z + + ... = + + analytic part z6 z6  − 3! 5!  z5 − 6 z3 120 z residue |{z} sin z iπ Res Integrand = = 6 dz = 2πi z=0( ) = ⇒ I IC z 60

cos z Note : 6 dz = 0 IC z z = 0 pole of order 6 with zero residue CALCULATION OF CONTOUR INTEGRALS BY RESIDUE THEOREM

Let C be the circle of centre z = 0 and radius 3.

5z 2 dz − = 2πi [Resz=0(Integrand)+ Resz=1(Integrand)] I z(z 1) C − = 2πi(2 + 3) = 10πi

−1/z dz e = 2πi Resz=0(Integrand) IC = 2πi( 1) = 2πi − −

1 πz/4 Res Integrand Res − Integrand dz 2 e = 2πi [ z=+i( )+ z= i( )] IC z + 1 eiπ/4 e−iπ/4 π = 2πi + = 2πi sin = iπ√2  2i 2i  4 − EVALUATION OF REAL INTEGRALS BY COMPLEX CONTOUR INTEGRATION METHODS ∞ x2 I = dx Z−∞ (x2 + 1)(x2 + 4)

2 z Im z dz Γ 2 2 R 2i Γ (z +1)(z +4) R i R

−i Re z −2i R x2 z2 = dx 2 2 + dz 2 2 Z−R (x + 1)(x + 4) ZSR (z + 1)(z + 4) 1 1 π By residue theorem = 2πi[Resz=+if + Resz=+2if] = 2πi( + )= . • IΓR −6i 3i 3

By Jordan lemma 0 for R because zf(z) 0 . • ZSR → → ∞ | | →

Thus I = π/3 . 2π 1 I = dθ Z0 2+cos θ

z + 1/z z = eiθ ; dz = izdθ ; cos θ = 2 1 1 2 1 So I = dz = dz 2 IC1 iz 2+(z + 1/z)/2 IC1 i z + 4z + 1

z + = − 2 + 3 z z + C1 2π By residue theorem I = 2πi [Resz=z+ f] = . • √3 2π 2π − cos θ − cos θ I1 = dθe cos(θ + sin θ) , I2 = dθe sin(θ + sin θ) Z0 Z0

−iθ z = eiθ ; dz = ieiθdθ ; e−1/z = e−e = e− cos θ+i sin θ 2π −1/z − cos θ i(θ+sin θ) So dz e = i dθ e e = i(I1 + iI2) IC1 Z0

By residue theorem

−1/z dz e = 2 π i Res(z=0) C 1 = − 2 π i C1

Thus 2πi = i(I + iI ) I = 2π , I = 0 . − 1 2 ⇒ 1 − 2 TECHNIQUES OF CONTOUR INTEGRATION: Choice of integrand in the complex z plane

∞ cos x Example : Consider I = dx 2 . Z0 x + 1 cos z dz 2 does not vanish on semicircle SR for R . ZS z + 1 → ∞ Im z R Γ eiz R Take instead f(z)= . z2 + 1 R Re z eiz By Jordan lemma 2 dz 0 for R . • ZSR z + 1 → → ∞ eiz e−1 π Res By residue theorem 2 dz = 2πi z=if = 2πi = . • IΓR z + 1 2i e ∞ ∞ + cos x + sin x π π Thus dx + i dx = = I = . Z−∞ x2 + 1 Z−∞ x2 + 1 e ⇒ 2e TECHNIQUES OF CONTOUR INTEGRATION: Choice of contour in the complex z plane

∞ ex/2 Example : Consider I = dx . Z−∞ cosh x ez/2 f(z)= has infinitely many poles, z = i(π/2+ nπ), n Z . • cosh z ∈ Choose contour so as to enclose only a finite number of poles:

Rectangular contour R encircles one only, z = iπ/2, for any L . • Im z iπ R Re z − L L

z/2 e iπ/4 By residue theorem dz = 2πi Resz=iπ/2[f] = 2πe IR cosh z Let L . Integrals along vertical sides vanish because →∞ − − − − − cosh(L+iy) = eL+iy+e L iy /2 eL+iy e L iy /2=(eL e L)/2 eL/4 , | | | | ≥ || |−| || − ≥ and so, by Darboux inequality,

π e(L+iy)/2 π eL/2 −L/2 for i dy dy L = 4π e 0 L . | Z0 cosh(L + iy)|≤ Z0 e /4 → → ∞

Similarly for the other vertical side.

Because cosh(x + iπ)= cosh x, integrals along horizontal sides are related by − − L e(x+iπ)/2 L ex/2 dx = eiπ/2 dx . ZL cosh(x + iπ) Z−L cosh x

∞ + ex/2 2π eiπ/4 π Taking L = I = dx = = = π√2 . → ∞ ⇒ Z−∞ cosh x 1+ eiπ/2 cos(π/4)