[proofplan] We prove that $\int_{\partial T} f(z) \, dz = 0$ by a bisection argument due to Goursat. Repeatedly bisect the triangle, choosing at each stage the sub-triangle whose boundary integral carries at least $1/4$ of the total. After $n$ steps the nested triangles shrink to a point $z_0$ where holomorphicity provides a linear approximation with controllable error. The ML inequality then forces the original integral to vanish. [/proofplan] [step:Bisect the triangle and select the sub-triangle with the largest integral] Let $I = \int_{\partial T} f(z) \, dz$. Connect the midpoints of the three edges of $T$ to form four congruent sub-triangles $T_1^{(1)}, T_2^{(1)}, T_3^{(1)}, T_4^{(1)}$. Each interior edge is traversed once in each direction by adjacent sub-triangles, so these contributions cancel and \begin{align*} \int_{\partial T} f(z) \, dz = \sum_{j=1}^{4} \int_{\partial T_j^{(1)}} f(z) \, dz. \end{align*} By the triangle inequality, at least one sub-triangle $T^{(1)}$ satisfies \begin{align*} \left| \int_{\partial T^{(1)}} f(z) \, dz \right| \geq \frac{|I|}{4}. \end{align*} [guided] Why do interior edges cancel? When two sub-triangles share an edge, their boundary orientations traverse that edge in opposite directions. For example, if $T_1^{(1)}$ and $T_2^{(1)}$ share the segment from midpoint $M_1$ to midpoint $M_2$, then $\partial T_1^{(1)}$ traverses this segment from $M_1$ to $M_2$ while $\partial T_2^{(1)}$ traverses it from $M_2$ to $M_1$. The line integral over a path traversed in opposite directions produces opposite signs: $\int_{M_1}^{M_2} f \, dz + \int_{M_2}^{M_1} f \, dz = 0$. After cancellation, the only surviving edges are those on $\partial T$ itself, giving $\sum_{j=1}^4 \int_{\partial T_j^{(1)}} f \, dz = \int_{\partial T} f \, dz$. We then choose the sub-triangle whose integral has the largest absolute value. By the triangle inequality applied to the sum $I = \sum_{j=1}^4 \int_{\partial T_j^{(1)}} f \, dz$: \begin{align*} |I| \leq \sum_{j=1}^4 \left|\int_{\partial T_j^{(1)}} f \, dz\right|. \end{align*} Since four non-negative real numbers summing to at least $|I|$ cannot all be strictly less than $|I|/4$ (otherwise their sum would be strictly less than $|I|$), at least one satisfies $|\int_{\partial T^{(1)}} f \, dz| \geq |I|/4$. This is the sub-triangle we select for the next iteration. [/guided] [/step] [step:Iterate the bisection to produce a nested sequence shrinking to a point $z_0$] Repeat the bisection: at the $n$-th stage, select $T^{(n)} \subseteq T^{(n-1)}$ satisfying \begin{align*} \left| \int_{\partial T^{(n)}} f(z) \, dz \right| \geq \frac{|I|}{4^n}, \qquad \ell(\partial T^{(n)}) = \frac{\ell(\partial T)}{2^n}, \qquad \operatorname{diam}(T^{(n)}) = \frac{\operatorname{diam}(T)}{2^n}. \end{align*} The sequence $(T^{(n)})_{n \geq 0}$ is nested and each $T^{(n)}$ is compact with $\operatorname{diam}(T^{(n)}) \to 0$. By the nested compact sets theorem, $\bigcap_{n=0}^\infty T^{(n)} = \{z_0\}$ for a unique point $z_0 \in T \subseteq U$. [guided] Each bisection halves every edge, so the perimeter scales as $\ell(\partial T^{(n)}) = \ell(\partial T)/2^n$ and the diameter as $\operatorname{diam}(T^{(n)}) = \operatorname{diam}(T)/2^n$. The integral lower bound $|\int_{\partial T^{(n)}} f \, dz| \geq |I|/4^n$ follows by induction: at each step we select the sub-triangle carrying at least $1/4$ of the parent's integral. The nested compact sets theorem (Cantor's intersection theorem) applies because: - Each $T^{(n)}$ is a non-empty compact subset of $\mathbb{C}$ (a closed, bounded triangle). - The sequence is nested: $T^{(n+1)} \subseteq T^{(n)}$ for all $n$. - The diameters shrink to zero: $\operatorname{diam}(T^{(n)}) = \operatorname{diam}(T)/2^n \to 0$. Under these conditions, the intersection $\bigcap_{n=0}^\infty T^{(n)}$ is a single point $\{z_0\}$. Since every $T^{(n)} \subseteq T \subseteq U$ and $U$ is open, the limit point $z_0$ lies in the open set $U$, so $f$ is holomorphic at $z_0$. This is where we will exploit the local linear approximation of $f$. [/guided] [/step] [step:Use holomorphicity at $z_0$ to decompose $f$ into a linear part and a controlled error] Since $f$ is holomorphic at $z_0$, write \begin{align*} f(z) = f(z_0) + f'(z_0)(z - z_0) + \varphi(z)(z - z_0), \end{align*} where $\varphi: U \to \mathbb{C}$ satisfies $\varphi(z) \to 0$ as $z \to z_0$. The functions $z \mapsto f(z_0)$ and $z \mapsto f'(z_0)(z - z_0)$ possess antiderivatives on $\mathbb{C}$ (namely $f(z_0) z$ and $f'(z_0)(z - z_0)^2 / 2$), so the [Fundamental Theorem of Contour Integration](/theorems/339) gives \begin{align*} \int_{\partial T^{(n)}} f(z_0) \, dz = 0, \qquad \int_{\partial T^{(n)}} f'(z_0)(z - z_0) \, dz = 0. \end{align*} Therefore \begin{align*} \int_{\partial T^{(n)}} f(z) \, dz = \int_{\partial T^{(n)}} \varphi(z)(z - z_0) \, dz. \end{align*} [guided] The idea is to exploit the complex differentiability of $f$ at $z_0$. Writing $f(z) = f(z_0) + f'(z_0)(z - z_0) + \varphi(z)(z - z_0)$ is simply a restatement of the definition of the derivative. The error function is defined by $\varphi(z) = \frac{f(z) - f(z_0)}{z - z_0} - f'(z_0)$ for $z \neq z_0$, and $\varphi(z_0) = 0$. The key property is $\varphi(z) \to 0$ as $z \to z_0$. The constant term $f(z_0)$ has antiderivative $f(z_0) \cdot z$, and the linear term $f'(z_0)(z - z_0)$ has antiderivative $f'(z_0)(z - z_0)^2/2$. By the Fundamental Theorem of Contour Integration, both integrate to zero over any closed contour. So the entire integral $\int_{\partial T^{(n)}} f(z) \, dz$ is carried by the error term $\varphi(z)(z - z_0)$, which we can make arbitrarily small as $T^{(n)}$ shrinks to $z_0$. [/guided] [/step] [step:Apply the ML inequality to force $I = 0$] Let $\varepsilon > 0$. Since $\varphi(z) \to 0$ as $z \to z_0$, there exists $N$ such that for all $n \geq N$, every $z \in T^{(n)}$ satisfies $|\varphi(z)| < \varepsilon$. For $z \in \partial T^{(n)}$, we have $|z - z_0| \leq \operatorname{diam}(T^{(n)}) = \operatorname{diam}(T)/2^n$. The ML inequality applied to $\int_{\partial T^{(n)}} \varphi(z)(z - z_0) \, dz$ with \begin{align*} \sup_{z \in \partial T^{(n)}} |\varphi(z)(z - z_0)| \leq \varepsilon \cdot \frac{\operatorname{diam}(T)}{2^n} \end{align*} gives \begin{align*} \frac{|I|}{4^n} \leq \left| \int_{\partial T^{(n)}} \varphi(z)(z - z_0) \, dz \right| \leq \varepsilon \cdot \frac{\operatorname{diam}(T)}{2^n} \cdot \frac{\ell(\partial T)}{2^n} = \varepsilon \cdot \frac{\operatorname{diam}(T) \cdot \ell(\partial T)}{4^n}. \end{align*} Cancelling $1/4^n$ from both sides: $|I| \leq \varepsilon \cdot \operatorname{diam}(T) \cdot \ell(\partial T)$. Since $\varepsilon > 0$ was arbitrary, $I = 0$. [guided] The ML inequality states that $\left|\int_\gamma g(z) \, dz\right| \leq \sup_{z \in \gamma^*} |g(z)| \cdot \ell(\gamma)$ for any piecewise $C^1$ path $\gamma$. We apply it to $g(z) = \varphi(z)(z - z_0)$ on $\gamma = \partial T^{(n)}$. The bound $|\varphi(z)| < \varepsilon$ holds on $T^{(n)}$ for $n$ large enough because $T^{(n)}$ shrinks to $\{z_0\}$ and $\varphi$ is continuous with $\varphi(z_0) = 0$. The factor $|z - z_0|$ is at most $\operatorname{diam}(T^{(n)})$, and the path length is $\ell(\partial T^{(n)})$. Both quantities carry a factor of $1/2^n$, which together produce $1/4^n$ — exactly cancelling the $1/4^n$ lower bound on the integral. This leaves $|I| \leq \varepsilon \cdot C$ where $C = \operatorname{diam}(T) \cdot \ell(\partial T)$ is a fixed constant. Since $\varepsilon$ was arbitrary, $I = 0$. [/guided] [/step]