Complex Analysis - Maths KU

5.5 Cauchy’s Integral Formulas and Their Consequences 273
After establishing this proposition we shall use it to further show that:
An analytic function f in a simply connected domain possesses derivatives of all
orders.
The ramifications of these two results alone will keep us busy not only for the remainder of
this section but in the next chapter as well.
5.5.1 Cauchy’s Two Integral Formulas
First Formula If f is analytic in a simply connected domain D and
z0 is any point in D, the quotient f(z)/(z − z0) is not defined at z0 and
hence is not analytic in D. Therefore, we cannot conclude that the integral
of f(z)/(z − z0) around a simple closed contour C that contains z0 is zero
by the Cauchy-Goursat theorem. Indeed, as we shall now see, the integral of
f(z)/(z − z0) around C has the value 2πif(z0). The first of two remarkable
formulas is known simply as the Cauchy integral formula.
Theorem 5.9 Cauchy’s Integral Formula
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(z0) = 1
�
f(z)
dz. (1)
2πi C z − z0
Proof Let D be a simply connected domain, C a simple closed contour in
D, and z0 an interior point of C. In addition, let C1 be a circle centered at
z0 with radius small enough so that C1 lies within the interior of C. By the
principle of deformation of contours, (5) of Section 5.3, we can write
�
�
f(z)
dz =
z − z0
f(z)
dz.
z − z0
(2)
C
We wish to show that the value of the integral on the right is 2πif(z0). To this
end we add and subtract the constant f(z0) in the numerator of the integrand,
�
�
f(z) f(z0) − f(z0)+f(z)
dz =
dz.
C1 z − z0 C1 z − z0
�
�
1
f(z) − f(z0)
= f(z0) dz +
dz. (3)
z − z0
z − z0
C1
From(6) of Section 5.3 we know that
�
1
dz =2πi
z − z0
(4)
C1
C1
C1