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(T_n) = 0

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

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 C_1.

 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 T_n. Thus w belongs to T = T₁ and since f(z) is analytic on T  

Now let be given. For there exists such that when  

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