Advanced Calculus fi..

RULE V If f (z) has a zero of first order at oo, thenChapter 8 Functions of a Complex Variable 581Res [ f (z), oo] = - lim z f (2).Z'W:~MT ti: , \ r -6If f (z) has a zero of second or higher order at oo, the residve at at is zero. • .RULE VIRes [ f (z), ool = -Res - f (llz), 0 .[z:The proof of Rule V is left as an exercise (Problem 8 following Section 8.18).To prove Rule VI, we writea-1 a-2f(z)= ...+ a, zn+...+a~z+ao+ - +-+...ZThen for 0 < lzl < R-',z2I(lzl > R).Henceand the rule follows. This result reduces the problem to evaluation of a residue atzero, to which Rules I through IV are applicable.EXAMPLE 1 We consider the integralof Example 2 in the preceding section. There is no singularity outside the path otherthan oo, and at oo the function has a zero of order 3; hence the integral is zero.EXAMPLE 2715 "-lzl=21(Z + 1)4(~2 - 9)(z - 4) dz.Here there is a fourth-order pole inside the path, at which evaluation of the residueis tedious. Outside the path there are first-order poles at f 3 and 4 and a zero of order7 at oo. Hence by Rule I the integral equals