Suppose \(f'(z)\) exists in a region \(\Omega\) except possibly at point \(w\) in \(\Omega\). Let \(T \subset \Omega\) be a triangle whose interior is also contained in \(\Omega\) and contains \(w\). Show that \[ \int_T f(z)\,dz = 0. \]
Erratum: the statement is false as written in the source, which puts no hypothesis on \(f\) at \(w\). For \(f(z) = 1/(z - w)\) on \(\Omega = \mathbb{C}\), \(f'\) exists except at \(w\), \(f\) is not even defined at \(w\), and for any positively oriented triangle \(T\) containing \(w\) in its interior \(\int_T f(z)\,dz = 2\pi i \ne 0\); setting \(f(w) = 0\) does not change the integral. The corrected hypothesis is that \(f\) is continuous on \(\Omega\), or more generally bounded near \(w\).