Goursat's integral lemma for complex-differentiable functions

From Companal

This article gives the statement, and possibly proof, of a basic fact in complex analysis.
View a complete list of basic facts in complex analysis

Statement

Suppose is an open subset, and is complex-differentiable at every point. Suppose is a triangle lying completely inside . Then, we have:

Proof

Define:

Note that can be split into four congruent triangles by joining the midpoints of the sides. Since is the sum of evaluated for each of the pieces, at least one of the pieces must satisfy:

Pick this triangle, and again divide it into four triangles. Repeating this process we get a descending sequence of triangles, each having side-lengths half of its predecessor:

and with the further property that:

On the other hand, we have that:

Now let be the unique point in the intersection of all the s. Since is differentiable at the point , we have:

Define now:

Then . Observe that:

Thus, we get:

Applying the length estimates and size estimates now gives the result. Fill this in later