TAILIEUCHUNG - Ebook Complex analysis for mathematics and engineering (2/E): Part 2
Part 2 book “Complex analysis for mathematics and engineering” has contents: Residue theory, conformal mapping, applications of harmonic functions, fourier series and the laplace transform, undergraduate student research projects. | Residue Theory The Residue Theorem The Cauchy integral formulae in Section are useful in evaluating contour integrals over a simple closed contour C where the integrand has the form f(z)/(z — Zo)k and/is an analytic function. In this case, the singularity of the integrand is at worst a pole of order k at zo- In this section we extend this result to integrals that have a finite number of isolated singularities and lie inside the contour C. This new method can be used in cases where the integrand has an essential singularity at z0 and is an important extension of the previous method. Let /have a nonremovable isolated singularity at the point zo- Then/has the Laurent series representation (1) f(z) = 2 an(z - zo)" valid for 0 0 such that the positively oriented circles C* = C^(Zk) (for k = 1, 2, . . . , rc) are mutually disjoint and all lie inside C. Using Theorem , the extended Cauchy-Goursat theorem, it follows that (6) \f(z)dz= JC 2 f f{z)dz. JQ. Since / is analytic in a punctured disk with center Zk that contains the circle C*, equation (4) can be used to obtain (7) f(z) dz = 2ni Res[/, z*] for k = 1, 2, . . . , n. Using equation (7) in equation (6) results in [ f(z) dz = 2ni J Res[/, zkl and the theorem is proven. Calculation of Residues The calculation of a Laurent series expansion is tedious in most circumstances. Since the residue at zo involves only the coefficient a^\ in the Laurent expansion, we seek a method to calculate the residue from special information about the nature of the singularity at zoI f / h a s a removable singularity at zo, then a_n = 0 for n = 1, 2, . . . . Therefore if zo is a removable singularity, then Res[/, zo] = 0. Calculation of Residues 247 Theorem (Residues at Poles) (i) If f has a simple pole at zo, then (1) Res[/, zo] = lim(z - z0)/(z). Z->ZQ (ii) If f has a pole of order 2 at zo, then (2) Res[/, zo] = lim - (z - z0)2/(z). z->za dZ (iii) Iff has a pole of order k at zo, .
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