1550251515-Classical_Complex_Analysis__Gonzalez_

Integration 427

so that

j f(z) dz= j f(z) dz

'l'l '/'2

Conversely, suppose that 1 is a closed rectifiable curve. Then 1 and -1

have the same endpoints. If the integral depends only on the endpoints,

we have

j f(z) dz= j J(z) dz= - j J(z) dz

')' -')' ')'

so that

j J(z) dz= 0

')'

Theorem 7.6 Let f(z) be continuous along the smooth arc 1: z = z(t),
a ::.::; t ::.::; /3, and let

F(z) =('I') 1: J(() d(

be the function defined by the integral from the initial point zo E 1 to a
variable point z E 1. Then F( z) has a directional derivative at each point
of 1 in the direction of the arc, and

F~(z) = f(z)

Proof Let z + 6.z = z(t + 6.t), where t + 6.t E [a, /3]. Then it follows that

6.F = F(z + 6.z) - F(z) = ('/') 1:+Az J(() d( -('I') 1: J(() d(

r+Az r+Az

=('/') Jz f(()d(=f(z)D.z+('I') Jz [f(()-J(z)]d(

Hence we have
6.F l
1

z+Az

- =f(z)+ _(.,,) [f(()-f(z)]d(

6.z 6.z z

(7.8-7)

Letting

M(D.t) = max lf(()-f(z)i = max IJ(((r))-f(z(t))I

. (EN(z) /r-t/:s;/At/

we get, on applying Theorem 7.2-6.

l

.2:_('1')1z+Az[f(()-f(z)]d(I:::; _l_M(D.t)D.s

6.z z 16.zl