Goursat's integral lemma for complex-differentiable functions

Statement

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

${\int }_{\partial \mathrm{△}}f(z)dz=0$

Proof

Define:

$\alpha (\mathrm{△})={\int }_{\partial \mathrm{△}}f(z)dz$

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

$|\alpha ({\mathrm{△}}^1)|\le |\alpha (\mathrm{△})|$

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:

$\mathrm{△}\supset {\mathrm{△}}^1\supset {\mathrm{△}}^2\supset \dots$

and with the further property that:

$|\alpha (\mathrm{△})|\le 4^n|\alpha ({\mathrm{△}}^n)|$

On the other hand, we have that:

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

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

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

Define now:

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

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

${\int }_{\partial {\mathrm{△}}^n}f(c)dz={\int }_{\partial {\mathrm{△}}^n}{f}^{\prime }(c)(z-c)=0$

Thus, we get:

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

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