51.7. THE GENERAL CAUCHY INTEGRAL FORMULA 1633

Denote by∫

∂T f (z)dz, the expression,∫

γ(z1,z2,z3,z1)f (z)dz. Consider the following pic-

ture.

TT 11

T 12

T 13 T 1

4z1 z2

z3

By Lemma 50.0.11 ∫∂T

f (z)dz =4

∑k=1

∫∂T 1

k

f (z)dz. (51.7.20)

On the “inside lines” the integrals cancel as claimed in Lemma 50.0.11 because there aretwo integrals going in opposite directions for each of these inside lines.

Theorem 51.7.1 (Cauchy Goursat) Let f : Ω→ X have the property that f ′ (z) exists forall z ∈Ω and let T be a triangle contained in Ω. Then∫

∂Tf (w)dw = 0.

Proof: Suppose not. Then ∣∣∣∣∣∣∣∣∫∂T

f (w)dw∣∣∣∣∣∣∣∣= α ̸= 0.

From 51.7.20 it follows

α ≤4

∑k=1

∣∣∣∣∣∣∣∣∫∂T 1

k

f (w)dw∣∣∣∣∣∣∣∣

and so for at least one of these T 1k , denoted from now on as T1,∣∣∣∣∣∣∣∣∫

∂T1

f (w)dw∣∣∣∣∣∣∣∣≥ α

4.

Now let T1 play the same role as T , subdivide as in the above picture, and obtain T2 suchthat ∣∣∣∣∣∣∣∣∫

∂T2

f (w)dw∣∣∣∣∣∣∣∣≥ α

42 .

Continue in this way, obtaining a sequence of triangles,

Tk ⊇ Tk+1,diam(Tk)≤ diam(T )2−k,

and ∣∣∣∣∣∣∣∣∫∂Tk

f (w)dw∣∣∣∣∣∣∣∣≥ α

4k .