Cauchys integral formula

24.10 CAUCHY’S INTEGRAL FORMULA
y
C
γ
C1
C2
x
Figure 24.11 The contour used to prove the result (24.43).
Consider two closed contours C and γ in the Argand diagram, γ being sufficiently small
that it lies completely within C. Show that if the function f(z) is analytic in the region
between the two contours then
0
0
f(z) dz = f(z) dz.
(24.43)
C
γ
To prove this result we consider a contour as shown in figure 24.11. The two close parallel
lines C1 and C2 join γ and C, which are ‘cut’ to accommodate them. The new contour Γ
so formed consists of C, C1 , γ and C2 .
Within the area bounded by Γ, the function f(z) is analytic, and therefore, by Cauchy’s
theorem (24.40),
0
f(z) dz = 0.
(24.44)
Γ
Now the parts C1 and C2 of Γ are traversed in opposite directions, and in the limit lie on
top of each other, and so their contributions to (24.44) cancel. Thus
0
0
f(z) dz + f(z) dz = 0.
(24.45)
C
γ
The sense of the integral round γ is opposite to the conventional (anticlockwise) one, and
so by traversing γ in the usual sense, we establish the result (24.43). A sort of converse of Cauchy’s theorem is known as Morera’s theorem, which
states that if f(z) is a continuous
function of z in a closed domain R bounded by
/
a curve C and, further, C f(z) dz = 0, then f(z) is analytic in R.
24.10 Cauchy’s integral formula
Another very important theorem in the theory of complex variables is Cauchy’s
integral formula, which states that if f(z) is analytic within and on a closed
851
COMPLEX VARIABLES
contour C and z0 is a point within C then
0
f(z)
1
dz.
f(z0 ) =
2πi C z − z0
(24.46)
This formula is saying that the value of an analytic function anywhere inside a
closed contour is uniquely determined by its values on the contour§ and that the
specific expression (24.46) can be given for the value at the interior point.
We may prove Cauchy’s integral formula by using (24.43) and taking γ to be a
circle centred on the point z = z0 , of small enough radius ρ that it all lies inside
C. Then, since f(z) is analytic inside C, the integrand f(z)/(z − z0 ) is analytic
in the space between C and γ. Thus, from (24.43), the integral around γ has the
same value as that around C.
We then use the fact that any point z on γ is given by z = z0 + ρ exp iθ (and
so dz = iρ exp iθ dθ). Thus the value of the integral around γ is given by
2π
0
f(z)
f(z0 + ρ exp iθ)
dz =
I=
iρ exp iθ dθ
ρ exp iθ
γ z − z0
0
2π
f(z0 + ρ exp iθ) dθ.
=i
0
If the radius of the circle γ is now shrunk to zero, i.e. ρ → 0, then I → 2πif(z0 ),
thus establishing the result (24.46).
An extension to Cauchy’s integral formula can be made, yielding an integral
expression for f (z0 ):
f(z)
1
f (z0 ) =
dz,
(24.47)
2πi C (z − z0 )2
under the same conditions as previously stated.
Prove Cauchy’s integral formula for f (z0 ) given in (24.47).
To show this, we use the definition of a derivative and (24.46) itself to evaluate
f(z0 + h) − f(z0 )
h
0
f(z)
1
1
1
dz
= lim
−
h→0 2πi C
h
z − z0 − h z − z0
0
f(z)
1
= lim
dz
h→0 2πi C (z − z0 − h)(z − z0 )
0
f(z)
1
dz,
=
2πi C (z − z0 )2
f (z0 ) = lim
h→0
which establishes result (24.47). §
The similarity between this and the uniqueness theorem for the Laplace equation with Dirichlet
boundary conditions (see chapter 20) is apparent.
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