Fix $z \in B(z_0, r)$ and define $g: B(z_0, r) \to \mathbb{C}$ by: \begin{align*}

g(w) = \begin{cases} \dfrac{f(w) - f(z)}{w - z} & w \neq z, \\[6pt] f'(z) & w = z. \end{cases} \end{align*} Since $f$ is holomorphic, $g$ is holomorphic on $B(z_0, r) \setminus \{z\}$ and continuous at $z$ (the limit of the difference quotient is $f'(z)$ by definition). **Claim:** $\int_{\partial T} g \, dw = 0$ for every triangle $T \subseteq B(z_0, r)$. If $z \notin T$, this follows from [Cauchy's theorem for triangles](/theorems/341) applied to $g$ (which is holomorphic on a neighbourhood of $T$). If $z \in T$ (including on the boundary), use Goursat's bisection argument directly: bisect $T$ repeatedly, selecting the sub-triangle carrying at least $1/4$ of $|\int_{\partial T} g|$. The nested sub-triangles shrink to a point $w_0$. If $w_0 \neq z$, the standard Goursat argument applies (using holomorphicity of $g$ at $w_0$). If $w_0 = z$, then *continuity* of $g$ at $z$ provides the estimate: for any $\varepsilon > 0$, eventually $|g(w) - g(z)| < \varepsilon$ on the $n$-th sub-triangle $T_n$. The constant and linear parts of $g(z) + (g(w) - g(z))$ integrate to zero (they have antiderivatives), and the ML inequality bounds the remainder by $\varepsilon \cdot \ell(\partial T_n) \cdot \operatorname{diam}(T_n) \leq \varepsilon \cdot (\ell(\partial T)/2^n)^2$. Since $|\int_{\partial T} g| \leq 4^n \cdot \varepsilon \cdot \ell(\partial T)^2/4^n = \varepsilon \cdot \ell(\partial T)^2$, and $\varepsilon$ is arbitrary, $\int_{\partial T} g = 0$. Since $B(z_0, r)$ is convex (hence star-shaped) and $g$ is continuous with all triangle integrals vanishing, the [star-shaped Cauchy theorem](/theorems/342) construction gives $\int_{\partial B(z_0, r)} g(w) \, dw = 0$. Expanding: \begin{align*} 0 = \int_{\partial B(z_0, r)} \frac{f(w) - f(z)}{w - z} \, dw = \int_{\partial B(z_0, r)} \frac{f(w)}{w - z} \, dw - f(z) \int_{\partial B(z_0, r)} \frac{1}{w - z} \, dw. \end{align*} The second integral equals $2\pi i$ (since $z$ lies inside $B(z_0, r)$ and the winding number of $\partial B(z_0, r)$ around $z$ is $1$, computed directly by parametrising $w = z_0 + re^{i\theta}$). Therefore: \begin{align*} \int_{\partial B(z_0, r)} \frac{f(w)}{w - z} \, dw = 2\pi i \cdot f(z), \end{align*} which gives $f(z) = \frac{1}{2\pi i} \int_{\partial B(z_0, r)} \frac{f(w)}{w - z} \, dw$. $\blacksquare$