Methods of Applied Mathematics Lecture Notes

58 CHAPTER 4. COMPLEX INTEGRATION4.3.2 The residue calculusSay that f(z) has an isolated singularity at z 0 . Let C δ (z 0 ) be a circle about z 0that contains no other singularity. Then the residue of f(z) at z 0 is the integralres(z 0 ) = 1 ∫f(z) dz. (4.35)2πi C δ (z 0 )Theorem. (Residue Theorem) Say that C ∼ 0 in R, so that C = ∂S withthe bounded region S contained in R. Suppose that f(z) is analytic in R exceptfor isolated singularities z 1 , . . . , z k in S. Then∫k∑f(z) dz = 2πiCj=1res(z j ). (4.36)Proof: Let R ′ be R with the singularities omitted. Consider small circlesC 1 , . . . , C k around these singularities such that C ∼ C 1 + · · · + C k in R ′ . Applythe Cauchy integral theorem to C − C 1 − . . . − C k .If f(z) = g(z)/(z − z 0 ) with g(z) analytic near z 0 and g(z 0 ) ≠ 0, then f(z)is said to have a pole of order 1 at z 0 .Theorem. If f(z) = g(z)/(z − z 0 ) has a pole of order 1, then its residue atthat pole isres(z 0 ) = g(z 0 ) = lim (z − z 0 )f(z). (4.37)z→z 0Proof. By the Cauchy integral formula for each sufficiently small circle Cabout z 0 the function g(z) satisfiesThis is the residue.g(z 0 ) = 1 ∫g(z)dξ = 1 ∫f(z) dz. (4.38)2πi C z − z 0 2πi C4.3.3 EstimatesRecall that for complex numbers we have |zw| = |z||w| and |z/w| = |z|/|w|.Furthermore, we haveWhen |z| > |w| this allows us to estimate||z| − |w|| ≤ |z ± w| ≤ |z| + |w|. (4.39)1|z ± w| ≤ 1|z| − |w| . (4.40)Finally, we have|e z | = e Rz . (4.41)