428 Chapter^7
where
r+D.z
.6.s = (-y) Jz Id(!Since J(((r)) is a continuous function of r, it follows that M(.6.t) ---t 0
as .6.t -~ 0. Also, since 'Y is smooth, we have .6.s/l.6.zl ---t 1 as .6.t ---t 0.
Therefore, taking limits in (7.8-7) as .6.t ---t 0 [i.e., as .6.z = z( t+.6.t)-z(t) ---t
O], we obtain
F~(z) = J(z)Corollairy 7.3 If f(z) is continuous along the smooth arc"(, and if
('Y) r tco d( = o
lzofor every z E 'Y*, then f ( z) = 0 along I.
Proof 'l'he assumption implies that F( z) = 0 for all z E 1. Hence f ( z) =
F~(z) = 0 for all z E 1.
Note {'Y) J:a^1 f ( () d( = 0 for a fixed upper limit z1 E 1* does not imply
that J(z) vanishes somewhere on 'Y*·
Example J:-n: eizdz = 0, yet eiz does not vanish anywhere on (0, 27r].
Theorem 7. 7 If ¢( z) is such that ¢ 0 ( z) = f ( z) for every z on the smooth
arc I, the directional derivative being taken in the direction of the arc, then
{'Y) 1z J(() d( = 1/J(z) + C
zofor some constant C, and
1
z1
(-y) !(() d( = 'l/;(z1)-'!f;(zo)
zoProof From Theorem 7.6 we know that
F(z) = (-y) r J(() d(
lzois such that F~(z) = J(z) for z E 1. Also, 'l/;9(z) = J(z) for f E 1.
Hence, by Corollary 6.2 we have
F(z) = 'l/;(z) + C (7.8-8)
where C is a constant. We may evaluate C by letting t = a or z = z 0
[with L( 1) = O]. Thus we get
'!f;(zo) + C = 0 or C = -'lf;(zo)