1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

226 CHAPTER 6 • COMPLEX INTEGRATION

The points z = ±i./2 lie interior to ct (0), so Corollary 6.1 implies that

--= =21Tt.
1

dz.

c;t(OJ z±iJ2

Substituting these values into Equation ( 6-38) yields

r ~ dz = 21Ti + 21Ti = 47ri.

lctcoJ z + 2

• EXAMPLE 6.15 Show that fctcil ,t~ 2 dz= 2?Ti.

Solution Recall that Ct (i) is the circle { z : lz -i i = 1} having positive ori-

entation. Using partial fractions again, we have

r ~dz= r dz + r dz.

l c:(i) z^2 + 2 l c:(i) z + i./2 l c:(i) z - ivl2

In this case, z = iJ2 lies interior to Ct (i) but z = -iJ2 does not, as shown

in Figure 6.27. By Corollary 6.1, the second integral on the right side of this

equation has the value 27ri. The first integral equals zero by the Cauchy- Goursat

theorem because the function f (z) = •+!,/2 is analytic on a simply connected

domain that contains Ci (i). Thus,

r ~ dz = 0 + 27ri = 21Ti.

lc:ciJ z + 2

• EXAMPLE 6.16 Show that fc ;.,-_^2 • dz = -67ri, where C is the "figure

eight" contour shown in Figure 6.28(a).

y

• i

- ivT

Figure 6.27 The circle Ci (i) and the points z = ±i./2.