Mathematical Methods for Physics and Engineering - Matematica.NET

COMPLEX VARIABLESCC ′C(a)(b)Figure 24.13 The contours used to prove the residue theorem: (a) the originalcontour; (b) the contracted contour encircling each of the poles.to the integral, because f is analytic between C and C ′ . Now the contribution tothe C ′ integral from the polygon (a triangle for the case illustrated) joining thesmall circles is zero, since f is also analytic inside C ′ . Hence the whole value ofthe integral comes from the circles and, by result (24.60), each of these contributes2πi times the residue at the pole it encloses. All the circles are traversed in theirpositive sense if C is thus traversed and so the residue theorem follows. Formally,Cauchy’s theorem (24.40) is a particular case of (24.61) in which C encloses nopoles.Finally we prove another important result, for later use. Suppose that f(z) hasa simple pole at z = z 0 and so may be expanded as the Laurent seriesf(z) =φ(z)+a −1 (z − z 0 ) −1 ,where φ(z) is analytic within some neighbourhood surrounding z 0 . We wish tofind an expression for the integral I of f(z) along an open contour C, whichisthe arc of a circle of radius ρ centred on z = z 0 given by|z − z 0 | = ρ, θ 1 ≤ arg(z − z 0 ) ≤ θ 2 , (24.62)where ρ is chosen small enough that no singularity of f, other than z = z 0 , lieswithin the circle. Then I is given by∫∫∫I = f(z) dz = φ(z) dz + a −1 (z − z 0 ) −1 dz.CCIf the radius of the arc C is now allowed to tend to zero, then the first integraltends to zero, since the path becomes of zero length and φ is analytic andtherefore continuous along it. On C, z = ρe iθ and hence the required expressionfor I is( ∫ θ2)1I = lim f(z) dz = lim a −1 ρ→0∫Cρ→0 θ 1ρe iθ iρeiθ dθ = ia −1 (θ 2 − θ 1 ). (24.63)860C