14.8. THE CAUCHY FORMULA 357
Consider the following picture where you have a large circle of radius R and a smallcircle of radius r centered at z, a point on the inside of γR. The Cauchy integral formulagives f (z) in terms of the values of f on the large circle.
Theorem 14.8.1 If f is differentiable on Ω an open set, then f ′ is continuous andin fact, f (z) = 1
2πi∫
γr
f (w)w−z dw where B(z0,r)⊆Ω,γr the counter clockwise boundary of the
ball.
Proof: Let B(z0,r+ ε)⊆Ω. Let z ∈ B(z0,r) and let γr be the counter clockwise orien-tation of the circle of radius r which is the boundary of B(z0,r). Pick δ > 0 such that thefollowing picture is applicable, the small circle having radius δ .
−γδ
z
γr
Γ1Γ2
There are two closed curves Γ1,Γ2 which intersect in vertical lines in the above picture,these contours each contained in a star shaped region on which the derivative of f exists.Therefore, there exists a primitive of w→ f (w)− f (z)
w−z valid on those two star shaped regions.Then adding the contour integrals which are each 0,∫
γr
f (w)− f (z)w− z
dw =∫
γδ
f (w)− f (z)w− z
dw
Since f ′ (z) exists, the integral on the right converges as δ → 0 to 0. Then it follows that∫γr
f (w)w− z
dw =∫
γr
f (z)w− z
dw = n( f ,z)(2πi) f (z)
However, from the above argument about the winding number, n( f ,z) = 1. Therefore,f (z) = 1
2πi∫
γr
f (w)w−z dw. Then using the above representation
f (z+h)− f (z)h
=1
2πi
∫γr
f (w)(− 1(w− z)(h−w+ z)
)dw
Passing to a limit as h → 0, this yields f ′ (z) = 12πi∫
γr
f (w)(w−z)2 dw. If desired, you could
continue taking derivatives and so f ′ is continuous and in fact, f is infinitely differentiable.■
Corollary 14.8.2 The same Cauchy integral formula holds if f is only differentiable onthe inside of the large circle and continuous on its closure.
Proof: Letting z be on the inside of the large circle, you could shrink the large circle bydecreasing its radius to r̂ and be in the situation of the above theorem. Thus
f (z) =1
2πi
∫γ r̂
f (w)w− z
dw =1
2π
∫ 2π
0
f(z0 + r̂eit
)z0 + r̂eit − z
r̂eitdt