Math 411: Honours Complex Variables - University of Alberta

36 CHAPTER 5. CAUCHY’S INTEGRAL THEOREM AND FORMULA
Then g is continuous on D and holomorphic on D \ {z}. By Lemma 5.1, g has an
antiderivative on BR(z0). Since ∂Br(z0) is closed, this means that
�
0 =
∂Br(z0)
using the identity �
yields the claim.
�
g(ζ)dζ =
∂Br(z0)
∂Br(z0)
�
f(ζ) 1
dζ −f(z)
ζ −z ∂Br(z0) ζ −z dζ,
� �� �
=2πi
1 dζ = 2πi established by Lemma 5.2. Division by 2πi
ζ−z
Corollary 5.3.1 (Mean Value Equation). Let D ⊂ C be open, let f : D → C be
holomorphic, and let z0 ∈ D and r > 0 be such that Br[z0] ⊂ D. Then we have
Proof. Parametrize ∂Br(z0) as
The Cauchy Integral Formula then yields
f(z0) = 1
� 2π
f(z0 +re
2π 0
iθ )dθ.
γ: [0,2π] → C, θ ↦→ z0 +re iθ .
f(z0) = 1
�
2πi
= 1
2πi
= 1
2π
γ
� 2π
0
� 2π
0
f(ζ)
dζ
ζ −z0
f(z0 +re iθ )
re iθ ire iθ dθ
f(z0 +re iθ )dθ.
Lemma 5.3. Let D ⊂ RN be open, and let f: [a,b]×D → R be continuous such that
: [a,b]×D → R all exist and are continuous. Define
∂f ∂f
,..., ∂x1 ∂xN
� b
g: D → R, x ↦→ f(t,x)dt.
Then g is continuously partially differentiable with
for x ∈ D and j = 1,...,N.
� b
∂g ∂f
(x) = (t,x)dt
∂xj ∂xj
a
a