Goursat's integral lemma for complex-differentiable functions

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 $U\subset \mathbb {C}$ is an open subset, and $f:U\to \mathbb {C}$ is complex-differentiable at every point. Suppose $\triangle$ is a triangle lying completely inside $U$. Then, we have:

$\int _{\partial \triangle }f(z)\,dz=0$

Proof

Define:

$\alpha (\triangle )=\int _{\partial \triangle }f(z)\,dz$

Note that $\triangle$ can be split into four congruent triangles by joining the midpoints of the sides. Since $\alpha (\triangle )$ is the sum of $\alpha$ evaluated for each of the pieces, at least one of the pieces $\triangle ^{1}$ must satisfy:

$|\alpha (\triangle ^{1})|\leq |\alpha (\triangle )|$

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:

$\triangle \supset \triangle ^{1}\supset \triangle ^{2}\supset \ldots$

and with the further property that:

$|\alpha (\triangle )|\leq 4^{n}|\alpha (\triangle ^{n})|$

On the other hand, we have that:

$L(\partial \triangle ^{n})={\frac {1}{2^{n}}}L(\partial \triangle )$

Now let $c\in \triangle$ be the unique point in the intersection of all the $\triangle ^{n}$s. Since $f$ is differentiable at the point $c$, we have:

$\lim _{z\to c}{\frac {f(z)-f(c)}{z-c}}=0$

Define now:

$g(z):={\frac {f(z)-f(c)}{z-c}}-f'(c)$

Then $g(c)=0$. Observe that:

$\int _{\partial \triangle ^{n}}f(c)\,dz=\int _{\partial \triangle ^{n}}f'(c)(z-c)=0$

Thus, we get:

$\alpha (\triangle ^{n})=\int _{\partial \triangle ^{n}}g(z)(z-c)\,dz$

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