Complex Analysis - Maths KU

262 Chapter 5 Integration in the Complex Plane
C
i
–i
y
C 1
C 2
x
and
�
C
dz
z2 � � �
1 1 1
= − dz.
+1 2i C z − i z + i
We now surround the points z = i and z = −i by circular contours C1 and
C2, respectively, that lie entirely within C. Specifically, the choice |z − i| = 1
2
for C1 and |z + i| = 1
2 for C2 will suffice. See Figure 5.32. FromTheorem5.5
we can write
�
dz
z2 �
1
=
+1 2i
C
= 1
�
2i C1
C1
�
1 1
dz −
z − i 2i C1
� �
1 1
− dz +
z − i z + i
1
� � �
1 1
− dz
2i C2 z − i z + i
�
1 1
dz +
z + i 2i C2
�
1 1
dz −
z − i 2i C2
1
dz. (9)
z + i
Figure 5.32 Contour for Example 5 Because 1/(z+i) is analytic on C1 and at each point in its interior and because
1/(z − i) is analytic on C2 and at each point in its interior, it follows from(4)
that the second and third integrals in (9) are zero. Moreover, it follows from
D
C
Figure 5.33 Contour C is closed but
not simple.
(6), with n = 1, that
�
Thus (9) becomes
Remarks
C1
dz
=2πi and
z − i
�
C
�
C2
dz
z2 = π − π =0.
+1
dz
z + i =2πi.
Throughout the foregoing discussion we assumed that C was a simple
closed contour, in other words, C did not intersect itself. Although we
shall not give the proof, it can be shown that the Cauchy-Goursat theorem
is valid for any closed contour C in a simply connected domain D. As
shown in Figure 5.33, the contour C is closed but not simple. Nevertheless,
if f is analytic in D, then �
f(z) dz =0. See Problem23 in Exercises
C
5.3.
EXERCISES 5.3 Answers to selected odd-numbered problems begin on page ANS-16.
In Problems 1–8, show that �
f(z) dz = 0, where f is the given function and C is
C
the unit circle |z| =1.
1. f(z) =z 3 − 1+3i 2. f(z) =z 2 + 1
3. f(z) =
z − 4
z
2z +3
z − 3
4. f(z) =
z2 +2z +2