Advanced Calculus fi..

4578 Advanced Calculus, Fifth EditionEXAMPLE 1 . t., z4 .. ; v,r F-Y- i - --*f g d z = 2ni(Res[f(z). 11 +Res[f(z), -11).1z1=2Since f (z) has first-order poles at h1, we find, by Rule I,zez zeZ eRes [f (z), 11 = lim (z - 1) . - = lim - = -z-+ 1 22-1 z-+lz+l 2'Res[f(z),-l]= lim(z+l).-= zeZ lim -- zeZ --e-1TYz-+-1 z2- 1 z+-1z- 1 -2 'Accordingly,Rule 111 could also have been used:eRes[f(z),I]=el -2, Res[f(z),-I]=-22 z=1This is simpler than Rule I, since the expression A(z)/B'(z), once computed, servesfor all poles of the prescribed type.EXAMPLE 2 r t ~All poles are of first order. Rule 111 gives A(z)/B1(z) = z/(4z3) = 1/(4z2) as theexpression for the residue at any one of the four points. Moreover, z4 = 1 at eachpole, so thatHenceEXAMPLE 3ez$ z(z-121=2dz = 2ni(Res [f (z), 01 + Res [ f (z), I]}.At the first-order pole z =0, application of Rule I gives the residue 1. At thesecond-order pole z = 1, Rule I1 gives