1550251515-Classical_Complex_Analysis__Gonzalez_

430 Chapter 7

that F'(z) = f(z) for every z E R. More generally, the property holds if
'Y is a rectifiable arc in R.
Proof. (a) Smooth arc. First suppose that the integral is independent
of the path 'Y in R. Let z 0 be a fixed point, z a variable point in R, and
'Y any smooth arc with graph contained in R connecting z 0 and z. Then
the function

F(z) ={I) r f(()d(

lzo

is single-·valued, since the value of the integral is uniquely determined by

z. We wish to prove that F(z) is analytic and that F'(z) = f(z).

Let 'Yi be a smooth arc connecting z 0 and z, and let 01 be the inclina-
tion angle of the semi tangent to 'Yi at z. Let 12 be another smooth arc
connecting z 0 and z, having at z a semitangent with an inclination angle
02 =F 01 + br. By Theorem 7.6 we have

F8 1 (z) = f(z) and

so that F 01 (z) = F 02 (z), and it follows from Theorem 6.13 that F(z) is
monogenic at z. Since z E R is arbitrary, we conclude that F'( z) is analytic

in R and that F'(z) = f(z).

Next, suppose that there exists in Ra primitive F(z) of f(z), and let 1:

z = z(t), a :::; t :::; /3, Then, by Theorem 7.1-9, we have

1

z1 1{J 1{J d

(^7 ) f(()d( = f(z(t))z'(t)dt = dF(z(t))dt

zo °' °' t

= F(z(/3)) - F(z(a)) = F(z1) - F(zo)

Thus the value of the integral is independent of the path joining the points

zo and z1.

(b) Rectifiable arc. As before, let

F(z) = (')') r f(() d(
lzo
and assume that the integral is independent of the rectifiable arc 'Y joining
Zo and z. Consider a neighborhood N(z) about z contained in R, and let
z + h E N(z). Call 1' the line segment joining z and z + h. Since the
integral is independent of the path, it is immaterial what path is taken to
connect z 0 to z + h. Hence we have