Complex Analysis - Maths KU

268 Chapter 5 Integration in the Complex Plane
z 0
z
s
D
z + ∆z
Important
Figure 5.40 Contour used in proof
of (7)
☞
Next, since the value of �
C f(z) dz depends only on the points z0 and z1, this
value is the same for any contour C in D connecting these points. In other
words:
If a continuous function f has an antiderivative F in D, then
�
f(z) dz is independent of the path.
C
In addition, we have the following sufficient condition for the existence of an
antiderivative.
If f is continuous and �
f(z) dz is independent of the path C in a
C (7)
domain D, then f has an antiderivative everywhere in D.
The last statement is important and deserves a proof.
Proof of (7) Assume f is continuous and �
f(z) dz is independent of the
C
path in a domain D and that F is a function defined by F (z) = � z
f(s) ds,
z0
where s denotes a complex variable, z0 is a fixed point in D, and z represents
any point in D. We wish to show that F ′ (z) =f(z), that is,
� z
F (z) = f(s) ds (8)
z0
is an antiderivative of f in D. Now,
� z+∆z � z � z+∆z
F (z +∆z) − F (z) = f(s) ds− f(s) ds = f(s) ds. (9)
z
z
z0
Because D is a domain, we can choose ∆z so that z +∆z is in D. Moreover,
z and z +∆z can be joined by a straight segment as shown in Figure 5.40.
This is the contour we use in the last integral in (9). With z fixed, we can
write
� z+∆z � z+∆z
f(z)∆z = f(z) ds = f(z) ds or f(z) = 1
� z+∆z
f(z) ds. (10)
∆z
From(9) and the last result in (10) we have
F (z +∆z) − F (z)
− f(z) =
∆z
1
� z+∆z
[f(s) − f(z)] ds.
∆z
Now f is continuous at the point z. This means for any ε>0 there exists
a δ>0 so that |f(s) − f(z)|