1550251515-Classical_Complex_Analysis__Gonzalez_

Singularities/Residues/ Applications where P1(z) = A-m(z - ai)-m + · · · + A-1(z - ai)-^1 fi(z) =Ao+ Ai(z - ai) + A2(z -ai)^2 + ...

666 Chapter 9 f. Thus oo is a pole of f iff the degree of the denominator is smaller than the degree of the numerator. In any ca ...

Singularities/Residues/ Applications 667 Thus the residue of f at a is equal to the coefficient A_ 1 in the Laurent expansion of ...

668 Chapter9 R Fig. 9.5 so that B_1 = 2 ~i J J(() d( c+ Hence we obtain Res J(z) = - 2 1 z=oo 7rZ. J J(() d( = -B-1 (9.7-5) c- i ...

Singularities/Residues/ Applications 9.8 Some Special Rules for the Computation of Residues 669 (a) Residue at a finite simple p ...

670 Chapter9 (b) Residue at a finite pole of order m (m > 1). If a# oo is a pole of order m of f(z), we have the Laurent repr ...

Singularities/Residues/Applications 671 or _ (1/2!)h"(a)g'(a)-(1/3!)h'"(a)g(a) ¥-~:J(z) - [h"(a)/2!]2 g'(a) 2 h"'(a)g(a) = 2 h"( ...

672 Chapter^9 that g(O) = g'(O) = 1, h(O) = h'(O) = h"'(O) = 0, h"(O) = 1. Hence by using (9.8-4) we obtain ez Res =2 z=O 1-COSZ ...

Singularities/Residues/ Applications 673 Example Let f(z) = (z + 1)/(z^2 + 4). This function is analytic for 2 < lzl < oo ...

674 Chapter9 where g(z) = f(a + (1/z)). Example f(z) = e^1 1z has z = 0 as an isolated essential singularity, and it is its only ...

Singularities/Residues/ Applications z=oo sin3z Res z=l ( Z -- 1 ) 2 10. Resez + z -1 z::iO 12. Res z^4 sin ~ z=O Z 675 z2m 13 ...

676 Chapter9 tyt a, rc;;f~ a2 / / / QC~ /' an c+ Fig. 9.6 which is equivalent to (9.9-1). Remark In view of Theorem 7.13 (strong ...

Singularities/Residues/ Applications 677 which gives an alternative method (often advantageous) for the evaluation of an integra ...

678 Chapter 9 as before. Theorem. 9.10 (General Form of the Residue Theorem). Suppose that the function f is analytic in some op ...

Singularities/Residues/Applications Exercises 9.3 Show that j ( 3 ~ 2 ~ 2 1 ~~~ ~ 4 ) = ~ 7ri, c+: z = 3eit, O $ t $ 271". c+ ...

680 Chapter9 Show that j (z 2 z~:) 2 dz= 7ritsint, c+: z = 2ei^8 , o::::; ()::::; 27r. c+ 12. Let f be analytic in C except fo ...

Singularities/Residues/Applicatio11s y y=0 y=20hr (^0) e Fig. 9.9 Then if a > 0, we have ' lim j eiaz f(z) dz= 0 R--+oo 'Y (F ...

682 Chapter9 where Lemma 9.1 has been used in passing from the fourth to the fifth integral. Since e > 0 is arbitrary the con ...

Singularities/Residues/ Applications 683 where m ~ 2 and h(O) =/:-0. Then and Letting z = Rei^8 it follows that lh(e-i^8 /RI R-+ ...

684 Chapter9 But Hence lim [p 'f/d() = 0 R-+0 la and we obtain lim J f(z) dz= iL((J - a) R-+0 Remark If the arc 'Y has its cente ...