# Cauchy Theorem for a triangle
## Theorem

Let $D\subseteq \mathbb {C}$ be a domain, $f:D\to \mathbb {C}$ a differentiable function. Let $T$ be a triangle such that ${\bar {T}}\subset D$. Then
$\left|\int \limits _{\partial T}f\right|=0$

## Proof

Assume
$\left|\int \limits _{\partial T}f\right|=c\geq 0$.
It will be shown that $c=0$.
First, subdivide $T$ into four triangles, marked $T^{1}$, $T^{2}$, $T^{3}$, $T^{4}$ by joining the midpoints on the sides. Then it is true that
$\int \limits _{\partial T}f=\sum \limits _{r=1}^{4}\left(\int \limits _{T^{r}}f\right)$.
Giving that
$c=\left|\int \limits _{\partial T}f\right|\leq \sum \limits _{r=1}^{4}\left|\int \limits _{T^{r}}f\right|$
Choose $r$ such that
$\left|\int \limits _{\partial T^{r}}f\right|\geq {\frac {1}{4}}c$
Defining $T^{r}$ as $T_{1}$, then
$\left|\int \limits _{\partial T_{1}}f\right|\geq {\frac {1}{4}}c$ and $L\left(\partial T_{1}\right)={\frac {1}{2}}L\left(\partial T\right)$
(where $L\left(\gamma \right)$ describes length of curve).
Repeat this process of subdivision to get a sequence of triangles
$T\supset T_{1}\supset T_{2}\supset \ldots \supset \ldots T_{n}\supset \ldots$
satisfying that
$\left|\int \limits _{\partial T_{n}}f\right|\geq \left({\frac {1}{4}}\right)^{n}c$ and $L\left(\partial T_{n}\right)=\left({\frac {1}{2}}\right)^{n}L\left(\partial T\right)$.
Claim: The nested sequence ${\bar {T}}\supset {\bar {T_{1}}}\supset {\bar {T_{2}}}\supset \ldots \supset {\bar {T_{n}}}\supset \ldots$ contains a point $z_{0}\in \bigcap \limits _{n=1}^{\infty }{\bar {T_{n}}}$. On each step choose a point $z_{n}\in T_{n}$. Then it is easy to show that $\left(z_{n}\right)$ is a Cauchy sequence. Then $\left(z_{n}\right)$ converges to a point $z_{0}\in \bigcap \limits _{n=1}^{\infty }{\bar {T_{n}}}$ since each of the ${\bar {T_{n}}}$s are closed, hence, proving the claim.
We can generate another estimate of $c$ using the fact that $f$ is differentiable. Since $f$ is differentiable at $z_{0}$, for a given $\varepsilon >0$ there exists $\delta >0$ such that
$0<\left|z-z_{0}\right|<\delta$ implies $\left|{\frac {f\left(z\right)-f\left(z_{0}\right)}{z-z_{0}}}-f'\left(z_{0}\right)\right|<\varepsilon$
which can be rewritten as
$0<\left|z-z_{0}\right|<\delta$ implies $\left|f\left(z\right)-f\left(z_{0}\right)-f'\left(z_{0}\right)\left(z-z_{0}\right)\right|<\varepsilon \left|z-z_{0}\right|$
For $z\in T_{n}$ we have $\left|z-z_{0}\right|<L\left(\partial \right)$, and so, by the Estimation Lemma we have that
$\left|\int \limits _{\partial T_{n}}\left\{f\left(z\right)-\left(f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)\right)\right\}dz\right|\leq \varepsilon L^{2}\left(\partial \right)$
As $f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)$ is of the form $\alpha z+\beta$ it has an antiderivative in D, and so $\int \limits _{\partial T_{n}}f\left(z_{0}\right)+f'\left(z_{0}\right)\left(z-z_{0}\right)=0$, and the above is then just
$\left|\int \limits _{\partial T_{n}}f\left(z\right)dz\right|\leq \varepsilon L^{2}\left(\partial \right)$
Notice that
$\left({\frac {1}{4}}\right)^{n}c\leq \left|\int \limits _{\partial T_{n}}f\left(z\right)dz\right|\leq \varepsilon L^{2}\left(\partial \right)=\left({\frac {1}{4}}\right)^{n}\varepsilon L^{2}\left(\partial \right)$
Giving
$c\leq \varepsilon L^{2}\left(\partial \right)$
Since $\varepsilon >0$ can be chosen arbitrary small, then $c=0$.