Complex Analysis - Maths KU

3i
–3i
y
Figure 5.44 Contour for Example 2
C
x
5.5 Cauchy’s Integral Formulas and Their Consequences 275
Solution By factoring the denominator as z2 +9=(z− 3i)(z +3i) wesee
that 3i is the only point within the closed contour C at which the integrand
fails to be analytic. See Figure 5.44. Then by rewriting the integrand as
z
z2 +9 =
�
z
f(z)
z +3i
,
z − 3i
we can identify f(z) =z/(z +3i). The function f is analytic at all points
within and on the contour C. Hence, fromCauchy’s integral formula (1) we
have
�
C
z
z2 �
dz =
+9 C
z
z +3i
dz =2πi f(3i) =2πi3i = πi.
z − 3i 6i
Second Formula We shall now build on Theorem5.9 by using it to
prove that the values of the derivatives f (n) (z0),n= 1, 2, 3, ... of an analytic
function are also given by a integral formula. This second integral formula is
similar to (1) and is known by the name Cauchy’s integral formula for
derivatives.
Theorem 5.10 Cauchy’s Integral Formula for Derivatives
Suppose that f is analytic in a simply connected domain D and C is
any simple closed contour lying entirely within D. Then for any point z0
within C,
f (n) (z0) = n!
�
2πi C
f(z)
dz. (6)
(z − z0) n+1
Partial Proof We will prove (6) only for the case n = 1. The remainder
of the proof can be completed using the principle of mathematical induction.
We begin with the definition of the derivative and (1):
f ′ f(z0 +∆z) − f(z0)
(z0) = lim
∆z→0 ∆z
��
�
1
f(z)
f(z)
= lim
dz −
∆z→02πi
∆z C z − (z0 +∆z) C z − z0
�
1
f(z)
= lim
∆z→02πi
(z − z0 − ∆z)(z − z0) dz.
C
�
dz
Before continuing, let us set out some preliminaries. Continuity of f on the
contour C guarantees that f is bounded (see page 124 of Section 2.6), that
is, there exists a real number M such that |f(z)| ≤M for all points z on C.