Complex Analysis - Maths KU

y
i
0
C 2
C 1
Figure 5.45 Contour for Example 4
x
5.5 Cauchy’s Integral Formulas and Their Consequences 277
EXAMPLE 4 Using Cauchy’s Integral Formula for Derivatives
�
Evaluate
5.45.
C
z3 +3
dz, where C is the figure-eight contour shown in Figure
z(z − i) 2
Solution Although C is not a simple closed contour, we can think of it as
the union of two simple closed contours C1 and C2 as indicated in Figure
5.45. Since the arrows on C1 flow clockwise or in the negative direction, the
opposite curve –C1 has positive orientation. Hence, we write
�
C
z3 +3
dz =
z(z − i) 2
�
C1
�
= −
z3 +3
dz +
z(z − i) 2
−C1
z 3 +3
(z − i) 2
z
�
C2
�
dz +
z3 +3
dz
z(z − i) 2
C2
z 3 +3
z
(z − i) 2 dz = −I1 + I2,
and we are in a position to use both formulas (1) and (6).
To evaluate I1 we identify z0 =0,f(z) =(z 3 +3)/(z −i) 2 , and f(0) = −3.
By (1) it follows that
�
I1 =
−C1
z 3 +3
(z − i) 2
z
dz =2πi f(0) = 2πi(−3) = −6πi.
To evaluate I2 we now identify z0 = i, n =1,f(z) =(z 3 +3)/z, f ′ (z) =
(2z 3 − 3)/z 2 , and f ′ (i)=3+2i. From(6) we obtain
�
I2 =
Finally, we get
�
C
C2
z3 +3
z 2πi
dz =
(z − i) 2 1! f ′ (i) =2πi(3+2i) =−4π +6πi.
z 3 +3
z(z − i) 2 dz = −I1 + I2 =6πi +(−4π +6πi) =−4π +12πi.
5.5.2 Some Consequences of the Integral Formulas
An immediate and important corollary to Theorem 5.10 is summarized next.
Theorem 5.11 Derivative of an Analytic Function Is Analytic
Suppose that f is analytic in a simply connected domain D. Then f
possesses derivatives of all orders at every point z in D. The derivatives
f ′ ,f ′′ ,f ′′′ , ... are analytic functions in D.