27 Oct 2006 ... Suppose that f is analytic on an inside C, except at a point z0 which ... C f(z)dz = 0
. We can use this to prove the Cauchy integral formula.
33
CAUCHY INTEGRAL FORMULA
P ROOF
October 27, 2006
Use the theorem to write Z C1
Z C1
f ( z)dz + f ( z)dz +
Z −C2
Z −C2
f ( z)dz = 0 f ( z)dz =
Z C1
f ( z)dz −
Z C2
f ( z)dz.
This is the deformation principle; if you can continuously deform C1 to C2 , without crossing points R where f is not analytic, then the value of C f ( z)dz is preserved. E XAMPLE
Suppose C is a simple, closed contour which contains the origin. Then Z C
dz = 2πi. z
Indeed, we know this is true for the unit circle, thus for any circle, and in fact for any simple closed contour.
33
Cauchy Integral Formula
Lecture 27 We start with a slight extension of Cauchy’s theorem. L EMMA Let C be a simple closed contour. Suppose that f is analytic on an inside C, except at a point z0 which satisfies lim |( z − z0 ) f ( z)| = 0. z→ z0
Then
R
C
f ( z)dz = 0.
P ROOF Let Cr be the contour which wraps around the circle of radius r centered at z0 exactly once in the clockwise direction. By theR deformation principle, if r is sufficiently small (i.e., if Cr is R in the interior of C), then C f ( z)dz = Cr f ( z)dz. Given any > 0, there exists a δ > 0 so that if | z − z0 | < δ, then |( z − z0 ) f ( z)| < , and | f ( z)| < | z− z0 | . If necessary, shrink δ so that Nδ ( z0 ) is in the interior of C. If 0 < r < δ, then Z Z f ( z)dz = f ( z)dz C
Cr
≤
Z Cr
m and |w − ( z + ∆z)| > m, so Z Z f ( z + ∆z) − f ( z) 1 f ( w ) M ≤ |∆z| − dw |dw| 2 ∆z 2πi C (w − z) |2π | C m3 ∆zM = . 2π m3 Professor Jeff Achter Colorado State University
52
M419: Introduction to Complex Variables Fall 2006
33
CAUCHY INTEGRAL FORMULA
October 27, 2006
Lecture 28
Now, start taking ∆z ever-smaller... In fact, we can use this to prove that derivatives of all orders exist, as follows. T HEOREM function
Suppose that f is continuous on and inside the simple closed contour C. Then each Fn ( z) =
is analytic in D, with derivative
Z C
f (w) dw (w − z)n
Fn0 ( z) = nFn+1 ( z).
P ROOF We omit the proof that F1 is continuous. At least if f is analytic, we have already shown that F1 is analytic, and its derivative is F2 . We now proceed by induction, and suppose that the theorem is true for all functions f and all orders less than n. Fix a value z, and construct the function g(w) = g z (w) =
f (w) . (w − z)
For any complex number α, let g(w) dw ( w − α )n C Z f (w) = . n C (w − α ) (w − z)
Gn (α ) =
Z
Suppose that Nr ( z) is any neighborhood of z whose closure is contained inside C. If w ∈ C, and if z + ∆z is inside Nr ( z), then |w − z| is bounded away from zero. Since f is continuous, this shows that Gn ( z + ∆z) is bounded on |∆z| < r. Then we have identities Fn ( z) = Gn−1 ( z)
1 1 Fn ( z + ∆z) − Gn−1 ( z + ∆z) = f (w) − n (w − ( z + ∆z)) (w − ( z + ∆z))n−1 (w − z) C Z f (w)(∆z) dw = n C ( w − z )( w − ( z + ∆z )) = ∆zGn ( z + ∆z). Z
dw
Moreover, since g(w) is continuous on and inside the contour C (away from z), we may assume that Gn−1 is differentiable, with derivative (n − 1) Gn ( z). We now show that Fn is continuous at z, and then differentiable. For the former, we have Fn ( z + ∆z) − Fn ( z) = Gn−1 ( z + ∆z) − Gn−1 ( z) + ∆zGn ( z + ∆z) Professor Jeff Achter Colorado State University
53
M419: Introduction to Complex Variables Fall 2006
34
BOUNDEDNESS AND THE MAXIMUM MODULUS PRINCIPLE
October 27, 2006
By the continuity of Gn−1 , lim ( Fn ( z + ∆z) − Fn ( z)) = lim ( Gn−1 ( z + ∆z) − Gn−1 ( z)) + lim ∆zGn ( z + ∆z)
∆z→0
∆z→0
∆z→0
But Gn is bounded, so
= 0, as desired. To show differentiability, use: lim
∆z→0
Fn ( z + ∆z) − Fn ( z) G ( z + ∆z) − Gn−1 ( z) = lim n−1 + Gn ( z + ∆z) ∆z→0 ∆z ∆z = (n − 1) Gn ( z) + Gn ( z)
since Gn is now continuous!
= nGn ( z) = nFn+1 ( z).
Lecture 29 In summary, we have the following representation of all derivatives of an analytic function: f
(n)
n! ( z) = 2πi
Z C
f (w) dw. ( w − z )n+1
T HEOREM [Morera] Suppose f is continuous on D. If for every closed contour C in D we have C f ( z ) dz = 0, then f is analytic in D.
R
P ROOF By hypothesis and the Cauchy-Goursat theorem, f admits an antiderivative, F. But if F has one derivative (namely, f ), then it has derivatives of all orders. In particular f = F 0 must have a derivative.
34
Boundedness and the Maximum Modulus Principle
Can we ever fit the entire complex plane into a finite circle? If we’re willing to use a function which is merely continuous, this is no problem; use f ( x) =
z . 1 + | z|
In this section we’ll prove Liouville’s theorem, which says that this is impossible if f is analytic. Professor Jeff Achter Colorado State University
54
M419: Introduction to Complex Variables Fall 2006
