- Let \(\Omega\) be an open subset of C and let \(T \subset \Omega\) be a triangle whose interior is also contained in Ω. Suppose that f is a function holomorphic in Ω except possibly at a point w inside \(T\) . Prove that if \(f\) is bounded near w, then
\[ \int_ {T} f (z) d z = 0. \]
Solution
Choose \(r>0\) so small that the closed square \[ Q_r=\{w+x+iy:|x|\le r,\ |y|\le r\} \] lies in the interior of \(T\) and in a neighborhood on which \(|f|\le M\) for some \(M\).
The compact polygonal region \(T\setminus\operatorname{int}(Q_r)\) can be triangulated into finitely many triangles whose closures avoid \(w\). On each such triangle \(f\) is holomorphic in a neighborhood of the triangle and its interior, so Goursat’s theorem gives zero integral around its boundary. Summing over the triangulation cancels all interior edges and yields \[ \int_T f(z)\,dz=\int_{\partial Q_r}f(z)\,dz, \] where \(\partial Q_r\) is positively oriented.
The perimeter of \(Q_r\) is \(8r\), so \[ \left|\int_{\partial Q_r}f(z)\,dz\right| \le 8Mr. \] Letting \(r\downarrow0\) gives \[ \int_T f(z)\,dz=0. \]