Suppose that \(f\) is holomorphic on a punctured open set \(\Omega\setminus\theset{w_0}\) and let \(T\subset \Omega\) be a triangle containing \(w_0\). Prove that if \(f\) is bounded near \(w_0\), then \(\int_T f(z) ~dz = 0\).
Solution
Because \(f\) is holomorphic on a punctured neighborhood of \(w_0\) and bounded there, Riemann’s removable-singularity theorem gives a holomorphic extension \[ \widetilde f:\Omega\to\mathbb C. \] On the boundary of \(T\), the functions \(f\) and \(\widetilde f\) agree. Therefore, by Cauchy’s theorem on the triangle, \[ \int_T f(z)\,dz =\int_{\partial T}\widetilde f(z)\,dz =0. \] Thus the puncture contributes no contour integral when the function is bounded near it.