A Proof of the Cauchy Integral Theorem

On this page you will find a rigorous proof of the Cauchy Integral Theorem. We begin with some background notation.

Let T be a closed triangular region. Notice that since T is closed it contains its boundary.

T

d(T)

Def: The diameter of T, is denoted as d(T) and is defined as the maximum length of the three sides of T.

Rem: If

Next we give two lemmas, the second is taken from Real Analysis and thus its proof will not be given in this note.

Lemma 1:

H: x is a fixed real number

H:

C: x = 0

Proof: Let us deny the conclusion which says that Since we are given that x is nonnegative, it follows that x > 0. Let and thus . Since x > 0, this is a contradiction. Therefore x = 0 and the proof is complete.

Lemma 2: (from Real Analysis)

H: T1, T2, T3, … is a sequence of closed triangular regions

H: T1 T2 T3

H: d(Tn) = 0

C: There is one and only one point in common to every Tn.

The Cauchy Integral Theorem for triangles:

H: C is a triangle and T is the closed region in and on C.

H: f(z) is analytic on T.

C:

Proof:

Let T be named T1 and let C be named C1.

T1 = T

Divide T into 4 triangular regions as shown.

C14 = C2

Let C11, C12, C13, and C14 be the 4 generated triangles interior to C1.

Consider , , , and .

Let C2 be the triangle corresponding to the largest of these 4 numbers. Clearly

=

and

(the length of C) = 2 . Next let T2 be the closed triangular region with boundary C2 so that clearly d(T) = 2 d(T2). (Remember that d(T) is defined as the maximum length of the three sides of T.)

Next divide T2 into 4 triangular regions in the same way as before, see the diagram below.

……………………………C3

Let C3 be the triangle for which is the maximum among the 4 moduli and let T3 be the closed triangular region with boundary C3. Then

and therefore .

In addition and d(T) = 2d(T2) = 22d(T3). Therefore and d(T) = 22d(T3).

.

.

.

Continue in this way. That is divide T3 into 4 triangular regions etc.. We thus generate

the triangles C1, C2, C3, C4, C5, … and the corresponding closed triangular regions T1, T2, T3, T4, T5, … such that

1)

2)

3)

Note that and thus by Lemma 2, there exists w in common to every Tn. Thus w belongs to T = T1 and since f(z) is analytic on T

Now let be given. For there exists such that when