Functions of a Complex Variable (S1) Definition: residue of a function f at point z 0. Residue theorem

Transcription

1 Functions of a omplex Variable (S) VI. RESIDUE ALULUS Definition: residue of a function f at point z 0 Residue theorem Relationship between complex integration and power series expansion Techniques and applications of complex contour integration

2 RESIDUE ALULUS omplex differentiation, complex integration and power series expansions provide three approaches to the theory of holomorphic functions. auchy 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.

3 DEFINITION OF RESIDUE Let f be holomorphic everywhere within and on a closed curve except possibly at a point z 0 in the interior of where f may have an isolated singularity. Define residue of f at z 0 : Res z0 f = f(z) dz 2πi If z 0 is a non-singular point, Res z0 f = 0. Otherwise, Res z0 f may be 0.

4 RESIDUE THEOREM Let be closed path within and on which f is holomorphic except for m isolated singularities. Then f(z) dz = 2πi Im z m Res zj f j= z 0 z j zm Re z reformulation of auchy theorem via arguments similar to those used for deformation theorem

5 Relationship with Laurent expansion onsider Laurent series expansion of f about z 0 : f(z) = c n (z z 0 ) n, where c n = f(ξ) dξ 2πi (ξ z 0 ) n+ n= Γ Res z0 f is nothing but the coefficient of (z z 0 ) in the Laurent expansion of f about z 0 c = dξ f(ξ) = Res z0 f 2πi Γ Laurent expansion thus provides a general method to compute residues.

6 EXAMPLES ompute the residue at the singularity of the function f(z) = z ( 2z) 2 z ( 2z) 2 = 8 z = /2 pole of order 2 (z /2) 2 4 }{{} residue z /2 Res z=/2f = 4 ompute the residue at the singularity of the function f(z) = e /z2 e /z2 = + z 2 + 2! z = 0 essential singularity z Res z=0f = 0

7 alculation of residues in the case of poles f(z) = If z 0 is pole of order n for f, then h(z) (z z 0 ) n, h holomorphic and h(z 0) 0 Substituting this into the definition of residue gives Res z0 f = h(z) 2πi (z z 0 ) dz n [ ] d n h(z) = by auchy formula (n )! dz n = lim z z 0 (n )! z=z 0 d n dz n [(z z 0) n f(z)] Res z=/2 f = lim z /2 Ex.(previous slide) f(z) = z ( 2z) 2 d dz [(z /2)2 f(z)] = lim z /2 d dz [ z ] = 4 4

8 f(z) = METHODS TO ALULATE RESIDUES General method: from Laurent expansion n= c n (z z 0 ) n, where c n = 2πi Γ f(ξ) dξ (ξ z 0 ) n+ Res z0 f = c = 2πi Γ dξ f(ξ) Method for z 0 pole of order n: Res z0 f = lim z z 0 (n )! d n dz n [(z z 0) n f(z)] For n = : Res z0 f = lim z z 0 [(z z 0 )f(z)]

9 sinz z 6 = z 6 I = sinz z 6 dz where is the circle of centre z = 0 and radius z = 0 pole of order 5 ) (z z3 3! + z5 5! +... = z 5 6z = I = }{{} residue sinz z 6 dz = 2πiRes z=0 (Integrand) = iπ 60 z + analytic part Note : cosz z 6 dz = 0 z = 0 pole of order 6 with zero residue

10 ALULATION OF ONTOUR INTEGRALS BY RESIDUE THEOREM Let be the circle of centre z = 0 and radius 3. dz 5z 2 z(z ) = 2πi[Res z=0(integrand)+res z= (Integrand)] = 2πi(2+3) = 0πi dz e /z = 2πi Res z=0 (Integrand) = 2πi( ) = 2πi dz z 2 + eπz/4 = 2πi[Res z=+i (Integrand)+Res z= i (Integrand)] [ ] e iπ/4 = 2πi + e iπ/4 = 2πisin π 2i 2i 4 = iπ 2

11 EVALUATION OF REAL INTEGRALS BY OMPLEX ONTOUR INTEGRATION METHODS I = dx x 2 (x 2 +)(x 2 +4) z 2 z 2 z 2 Γ ( +)( +4) R dz Γ R 2i i Im z R R x 2 = dx R (x 2 +)(x 2 +4) + z 2 dz S R (z 2 +)(z 2 +4) By residue theorem = 2πi[Res z=+i f+res z=+2i f] = 2πi( Γ R 6i + 3i ) = π 3. By Jordan lemma 0 for R because zf(z) 0. S R i 2i Thus I = π/3. Re z

12 I = 2π 0 dθ 2+cosθ So I = z = e iθ ; dz = izdθ ; cosθ = z +/z 2 dz iz 2+(z +/z)/2 = dz 2 i z 2 +4z + z + = z z + By residue theorem I = 2πi [Res z=z+ f] = 2π 3.

13 I = 2π 0 dθe cosθ cos(θ+sinθ), I 2 = 2π 0 dθe cosθ sin(θ+sinθ) z = e iθ ; dz = ie iθ dθ ; e /z = e e iθ = e cosθ+isinθ 2π So dz e /z = i dθ e cosθ e i(θ+sinθ) = i(i +ii 2 ) By residue theorem 0 /z dz e = 2 π i = 2 π i Res(z=0) Thus 2πi = i(i +ii 2 ) I = 2π, I 2 = 0.

14 TEHNIQUES OF ONTOUR INTEGRATION: hoice of integrand in the complex z plane Example : onsider I = 0 dx cosx x 2 +. dz cosz S R z 2 + does not vanish on semicircle S R for R. Take instead f(z) = eiz z 2 +. By Jordan lemma SR e iz z 2 + Γ R dz 0 for R. Im z R Re z By residue theorem ΓR e iz z 2 + dz = 2πi Res z=if = 2πi e 2i = π e. Thus + dx cosx + x 2 + +i dx sinx x 2 + = π e = I = π 2e.

15 TEHNIQUES OF ONTOUR INTEGRATION: hoice of contour in the complex z plane Example : onsider I = dx ex/2 coshx. f(z) = ez/2 has infinitely many poles, z = i(π/2+nπ), n Z. coshz hoose contour so as to enclose only a finite number of poles: Rectangular contour R encircles one only, z = iπ/2, for any L. R iπ Im z L L Re z By residue theorem R dz ez/2 coshz = 2πi Res z=iπ/2[f] = 2πe iπ/4

16 Let L. Integrals along vertical sides vanish because cosh(l+iy) = e L+iy +e L iy /2 e L+iy e L iy /2 = (e L e L )/2 e L /4, and so, by Darboux inequality, π 0 i dy e (L+iy)/2 cosh(l+iy) π 0 dy el/2 e L /4 = 4π e L/2 0 for L. Similarly for the other vertical side. Because cosh(x+iπ) = coshx, integrals along horizontal sides are related by L L dx e (x+iπ)/2 cosh(x+iπ) = eiπ/2 L L dx ex/2 coshx. Taking L = I = + e x/2 coshx dx = 2π eiπ/4 +e iπ/2 = π cos(π/4) = π 2.