708 CHAPTER 35. ANALYTIC FUNCTIONS

TT 11

T 12

T 13 T 1

4z1 z2

z3

Thus ∫∂T

f (z)dz =4

∑k=1

∫∂T 1

k

f (z)dz. (35.5)

On the “inside lines” the integrals cancel because there are two integrals going in oppositedirections for each of these inside lines. Recall the method for evaluating a line integralwith a C1 parametrization.

Theorem 35.5.3 (Cauchy Goursat) Let f : Ω→ X , where Ω is an open subset of C and Xis a complex complete normed linear space, have the property that f ′ (z) exists for all z∈Ω

and let T be a triangle contained in Ω. Then∫∂T

f (w)dw = 0.

Proof: Suppose not. Then ∣∣∣∣∫∂T

f (w)dw∣∣∣∣= α ̸= 0.

From 35.5 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 .