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Towards Cauchy’s Theorem
Adrian Down
October 11, 2005
1 General theorem
1.1 Theorem
Today we work towards the general statement of Cauchy’s Theorem. The
complete theorem will not be presented until Thursday
Theorem (Cauchy’s). Let S ⊂ C be a domain, and f : S → C be differen-
tiable, and γ be a piecewise closed path in S.
w(γ, z) = 0, ∀z /∈ S
γ
f = 0
Proof. To be given on Thursday.
1.2 Hypothesis
The hypothesis is that γ does not wind around any point outside of S. We
show that this hypothesis is necessary.
Proposition. Suppose ∃ a closed path γ and a given S, and w(γ, z 0 ) 6 = 0 for
some z 0 ∈/ S. Then ∃ f : S → C differentiable such that
γ
f 6 = 0.
Proof. Consider the function,
f (z) =
z − z 0
Let S = C r { 0 }, and γ be the unit circle. Then,
γ
z − z 0
dz = 2πıw(γ, z 0 )
2 Cauchy’s theorem for triangles
2.1 Statement
Proposition. Let T be any path tracing a triangle. Let f be differentiable
in an open set S containing all points in the interior to the triangle. Then
T
f = 0
Note. The assumption of Cauchy’s theorem is automatically satisfied, since
w(T, z) = 0, ∀z /∈ S
This special case will help us work towards the statement of the theorem on
an arbitrarily-shaped path.
Definition (Diameter). The diameter of a set is the supremum of the dis-
tances between points in the set.
diameter of S = sup { | w − z| : w, z ∈ S }
Proof. Let
c =
γ
f
Our goal is to prove that c = 0.
2.3 There is a z 0 which is in all of the triangles
We want to show
∞ ⋂
n=
Tn 6 = ∅
For each n, choose some zn ∈ Tn. Since Tn+1 ⊂ Tn, zn+1 ∈ Tn. Using the
definition of diameter,
|zn − zn+1| ≤ 2
−n d = diameter of Tn
This statement is true regardless of how zn is chosen.
Thus the sequence of points zn is Cauchy. That the sequence is Cauchy
follows from the triangle inequality, which can be used to deal with |zm − zn|
in terms of |zn+1 − zn|. Since { zn } is Cauchy, it has a limit, call it z 0 , by
the completeness property of the real numbers.
Fix any k. Then
n ≥ k ⇒ zn ∈ Tk
Tk is a closed set. The limit of a convergent sequence in a closed set is in the
set, so the limit of the sequence { zn }, which is z 0 , is in Tk. This holds for
any choice of k, so z 0 ∈ Tk, ∀k.
2.4 Linear approximation of f
Recall that our goal is to show that c is 0; we do so by getting an upper
bound on the contour integrals around the component triangles.
Let z 0 be such that,
z 0 ∈
⋂^ ∞
n=
Tn
We have just proven that such a z 0 exists. We use the fact that f is assumed
to be differentiable on S and z 0 ∈ S to make a linear approximation of f
around z 0.
f (z) = f (z 0 ) + (z − z 0 )f
′ (z 0 ) + |z − z 0 |η(z)
where η(z) → 0 as z → z 0
We assumed that f was differentiable at every point, but we only use differ-
entiability at one point, z 0. However, the assumption that f is differentiable
at all points of S because we do not know a priori at which point z 0 we will
be taking the derivative.
Integrate this expression for f around Tn,
∫
Tn
f (z) =
Tn
f (z 0 ) + (z − z 0 )f
′ (z 0 ) + |z − z 0 |η(z)
Two of the terms are 0, because they have anti-derivatives and the path Tn
is closed. The constant function f (z 0 0) has an anti-derivative, as does the
linear function (z − z 0 ), thus
Tn
f (z 0 )dz = f
′ (z 0 )
Tn
(z − z 0 )dz = 0
The integral of f around Tn simplifies,
∣ ∣ ∣ ∣
Tn
f
Tn
|z − z 0 |η(z)dz
2.5 Estimation lemma
We use the estimation lemma. To do so, we need an upper bound on the
integrand. By the definition of the diameter,
|z − z 0 | ≤ δ ∀z, z 0 ∈ Tn
The diameter of Tn is related to the diameter of T 0 by δ = 2
−n d. Thus, using
the estimation lemma,
∣ ∣ ∣ ∣
Tn
f
Tn
|z − z 0 |η(z)dz
≤ L(Tn) · 2
−n d · max z∈Tn
|η(z)|
L(Tn) is related to L(T 0 ) by L(Tn) = 2
−n L(T 0 ).
∣ ∣ ∣ ∣
Tn
f
−n L(T 0 ) · 2
−n d · max z∈Tn
|η(z)|
The two factors of 2
−n combine to give us the 4
−n that will cancel.
3.2 Star-shaped sets
3.2.1 Definitions
Definition (Notation). Let [z, w] denote the path tracing the segment from
z to w,
t → z + t(w − z)
t ∈ [0, 1]
Definition (Star-shaped). Let E ⊂ C, and z? ∈ E. E is star-shaped with
center z? if ∀z ∈ E, the segment [z?, z] ⊂ E.
Note. • Any convex set is star-shaped, including sets that are shaped like
stars.
- Important examples of sets that are not star-shaped are annuli.
- A star-shaped set can have many centers, but our definition requires
only that there be 1.
- The most important example of a star-shaped set is the open disk. This
is where the Cauchy’s theorem on star-shaped sets is most useful.
Recall the following, which we used in a previous proof,
Fact. If f is differentiable at z, then
h
− 1
[z,z+h]
f → f (z) as h → 0
3.2.2 Cauchy’s theorem for star-shaped sets
Proposition.
Let
S ⊂ C be a star-shaped open set
f : S → C be differentiable
γ be a closed path in S
Then,
∫
γ
f = 0
Proof. It is enough to construct an anti-derivative F such that F
′ ≡ f ,
∀z ∈ S. The result then follows from the Fundamental Theorem of contour
integration. A good choice for the anti-derivative is
F (z) =
[zstar ,z]
f
We need to show that F
′ = f. To do so, write down the difference
quotient. We can take |h| sufficiently small so that the path is in S, since
z ∈ S and S is open.
F (z + h) − F (z)
h
= h
− 1
[z?,z+h]
f − h
− 1
[z?,z]
f
We want to make use of our triangle lemma. To do so, connect z?, z, and
z + h. Define the triangular path T
T = [z?, z] + [z 1 , z + h] − [z?, z + h]
We need to show that for sufficiently small triangles, the triangle is com-
pletely in S, i.e. given z ∈ S, ∃δ > 0 such that ∀|h| < δ, the solid triangle
with vertices z, z?, z + h ⊂ S. Consider z ∈ S. Since s is open, ∃δ > 0 such
that Nδ(z) ⊂ S. Then for h < δ, the segment [z, z + h] ⊂ Nδ(z) ⊂ S. We can
then evoke the star-shaped condition, so that each of the points in [z, z + h]
can be connected to the center z?.
We can then apply the Cauchy’s theorem for triangles to get
T
f =
[z?,z]
f +
[z,z+h]
f −
[z?,z+h]
f
Therefore,
F (z + h) − F (z)
h
= h
− 1
[z,z+h]
f
Using the fact preceding the proof,
F (z + h) − F (z)
h
= h
− 1
[z,z+h]
f → f (z) as h → 0
Thus
lim h→ 0
F (z + h) − F (z)
h
exists and −−→ h→ 0
f (z)
Proof. The proof involves a picture. Introduce four extra segments between
the large circle and the small circle. Take the total path Γ traversing the inner
and outer circle and the four additional segments. We use the same trick as
before, traversing the four paths between the inner and outer circle twice
each, once in each direction. The contributions from all of the additional
segments cancel, and we are left with,
Γ
g =
γ
g −
γr
g
We want the integral over Γ to be 0. We will prove this next time.
