Let \(\Omega \subseteq \mathbb{C}\) be an open domain, and let \(T \subset \Omega\) be a closed triangle whose interior \(T^\circ \subset \Omega\). 1. Assuming \(f \in C^1(\Omega)\) (meaning \(f(z) = u(x, y) + i v(x, y)\) has continuous real partial derivatives), apply Green’s Theorem and the Cauchy-Riemann equations to prove Cauchy’s Theorem on triangles: \[\oint_{\partial T} f(z) \, dz = 0.\]
- State Goursat’s Theorem (which removes the assumption that \(f'\) is continuous) and outline Goursat’s proof via triangular subdivision.
Solution
Let \(f=u+iv\) and \(dz=dx+i\,dy\). Then \[ \oint_{\partial T}f(z)\,dz = \oint_{\partial T}(u\,dx-v\,dy) +i\oint_{\partial T}(v\,dx+u\,dy). \] Green’s theorem gives \[ \oint_{\partial T}(u\,dx-v\,dy) =-\iint_T(v_x+u_y)\,dA, \] and \[ \oint_{\partial T}(v\,dx+u\,dy) =\iint_T(u_x-v_y)\,dA. \] The Cauchy–Riemann equations make both integrands zero, so \[ \oint_{\partial T}f(z)\,dz=0. \]
Goursat’s theorem removes the \(C^1\) assumption: if \(f\) is complex differentiable at every point of an open set, then the integral around every triangle contained in the open set is zero.
For the standard proof, repeatedly subdivide a triangle into four congruent triangles and choose at each stage one whose boundary integral has modulus at least one quarter of its parent’s. This produces nested triangles \(T_n\) with diameter and perimeter scaled by \(2^{-n}\) and a unique common point \(z_0\). Differentiability at \(z_0\) gives \[ f(z)=f(z_0)+f'(z_0)(z-z_0)+\eta(z)(z-z_0), \qquad \eta(z)\to0. \] The constant and linear terms integrate to zero, while \[ \left|\oint_{\partial T_n}\eta(z)(z-z_0)\,dz\right| \le \sup_{T_n}|\eta|\,\operatorname{diam}(T_n)\,\operatorname{length}(\partial T_n). \] Multiplying by the factor \(4^n\) accumulated from the subdivisions cancels the \(4^{-n}\) geometric factor. Since \(\sup_{T_n}|\eta|\to0\), the original triangle integral is zero.