Let \(f(z)\) be analytic in an open set \(\Omega\) except possibly at a point \(z_0\) inside \(\Omega\).
Show that if \(f(z)\) is bounded in near \(z_0\), then \(\displaystyle \int_\Delta f(z) dz = 0\) for all triangles \(\Delta\) in \(\Omega\).
Solution
Goal: If \(f\) is analytic on \(\Omega \setminus \theset{z_0}\) and bounded near \(z_0\), show \(\int_\Delta f = 0\) for every triangle \(\Delta \subseteq \Omega\).
1. \(z_0\) is a removable singularity of \(f\).
\(f\) is bounded in a punctured neighborhood of \(z_0\), so by Riemann’s theorem on removable singularities \(f\) extends holomorphically across \(z_0\); call the extension \(\tilde f\), which is holomorphic on all of \(\Omega\).
2. For every triangle \(\Delta \subseteq \Omega\), \(\int_\Delta \tilde f = 0\).
\(\tilde f\) is holomorphic on the open set \(\Omega\), so the Cauchy–Goursat theorem gives vanishing integrals over all closed triangles.
3. \(\int_\Delta f = 0\) for every triangle \(\Delta \subseteq \Omega\).
By 1, \(f\) and \(\tilde f\) agree on \(\Omega \setminus \theset{z_0}\); the boundary of a triangle is a finite union of segments not containing \(z_0\) (if \(z_0\) is a vertex or lies on an edge, perturb the triangle slightly or note the integral over the triangle equals the integral over a nearby one avoiding \(z_0\), by continuity of the holomorphic extension — standard reduction: subdivide \(\Delta\) so no sub-triangle has \(z_0\) in its interior and let the ones containing \(z_0\) shrink to zero measure), so \(\int_\Delta f = \int_\Delta \tilde f = 0\) by 2.
4. Q.E.D.
3 is the claim.