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.
Def: The diameter of T, is denoted as d(T) and is defined as the maximum length of the three sides of T.
Next we give two lemmas, the second is taken from Real Analysis and thus its proof will not be given in this note.
H: x is a fixed real number
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.
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
(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.
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
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
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