1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

6.4 • THE FUNDAMENTAL THEOREMS OF INTEGRATION 233

y

= x

(a) The path C joining z 1 and Z2·

=

(b) The path that is a portion

of the unit circle Id= t.

x

Figure 6.32 The simply connected domain D shown in Examples 6.19 and 6.20.

• EXAMPLE 6.19 We let D = { z = rei^8 : r > 0 and - 7r < 8 < 7r} be the
  simply connected domain shown in Figure 6.32. We know that f (z) = ~ is
  analytic in D and has an antiderivative F (z) =Log (z), for all z ED. If C is a
  contour in D that joins the point z 1 to the point .zi, then Theorem 6.9 implies
  that

1

dz - = 1·• dz - = Logz2 - Logz1.

c z %1 z

• EXAMPLE 6.20 Show that fctco) ~· = 27ri.

Solution Recall that Ct (0) is the unit circle with positive orientation. We
let C be that circle with the. point - 1 omitted, as shown' in Figure 6.32(b ). The
contour C is contained in the simply connected domain D of Example 6.19. We
know that f (z) = ~ is analytic in D , and has an antiderivative F (z) = Log (z),
for all z E D. Therefore, if we let z2 approach -1 on C through the upper
half-plane and z 1 approach -1 on C through the lower half-plane,

r dz = lim r· dz

lct(O) z ::2--l (::2EC,Imz2>0) 1.1 z

z1- - l (z1EC 1 lmz1 <0)

= lim Log z2 - lim Log z 1

•2--l (z2EC, lmz2>0) •1--l (z1EC, lmz1 <0)

= i7r - ( -i7r) = 27ri.

-------•EXERCISES FOR SECTION 6.4

For Exercises 1-14, find the value of the definite integral using Theorem 6.9, and

explain why you are justified in using it.

1. f c z^2 dz, where C is the line segment from 1 + i to 2 + i.