Complex Analysis - Maths KU

5.4 Independence of Path 267
and z(b) =z1. Using (11) of Section 5.2 and the fact that F ′ (z) =f(z) for
each z in D, we then have
�
� b
f(z) dz = f(z(t))z ′ � b
(t) dt = F ′ (z(t))z ′ (t) dt
C
� b
=
a
a
a
d
F (z(t)) dt
dt
← chain rule
= F (z(t)) | b
a
= F (z(b)) − F (z(a)) = F (z1) − F (z0).
EXAMPLE 2 Applying Theorem 5.7
In Example 1 we saw that the integral �
2zdz, where C is shown in Figure
C
5.39, is independent of the path. Now since the f(z) =2z is an entire function,
it is continuous. Moreover, F (z) =z2 is an antiderivative of f since F ′ (z) =
2z = f(z). Hence, by (4) of Theorem5.7 we have
� −1+i
−1
2zdz= z 2� � −1+i
−1
=(−1+i)2 − (−1) 2 = −1 − 2i.
EXAMPLE 3 Applying Theorem 5.7
Evaluate �
C cos zdz, where C is any contour with initial point z0 = 0 and
terminal point z1 =2+i.
Solution F (z) = sin z is an antiderivative of f(z) = cos z since F ′ (z) =
cos z = f(z). Therefore, from(4) we have
�
� 2+i
cos zdz=
= sin(2 + i) − sin 0 = sin(2 + i).
C
0
cos zdz= sin z | 2+i
0
If we desire a complex number of the form a + ib for an answer, we can use
sin(2+i) ≈ 1.4031−0.4891i (see part (b) of Example 1 in Section 4.3). Hence,
�
� 2+i
cos zdz= cos zdz≈ 1.4031 – 0.4891i.
C
0
Some Conclusions We can draw several immediate conclusions from
Theorem5.7. First, observe that if the contour C is closed, then z0 = z1 and,
consequently,
�
f(z) dz =0. (5)
C
✎