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
Write \(\DD_\eps(z_0)\) for a disc in which \(f\) is bounded, say by \(\abs{f}\leq M\) here. Note that if \(z_0\) is not in the region enclosed by \(\Delta\), then \(\int_\Delta f = 0\) since \(f\) is holomorphic throughout \(\Delta\), so suppose \(z_0\) is in this region.
Since \(f\) is holomorphic in \(\Omega\smts{z_0}\), \(\Delta\) can be deformed to \(\bd \DD_\eps(z_0)\) without changing the integral. So \[ \abs{ \oint_\Delta f } &= \abs{\oint_{\bd\DD_\eps(z_0)} f } \\ &\leq \oint_{\bd \DD_\eps(z_0)} \abs{f} \\ &\leq \oint_{\bd \DD_\eps(z_0)} M \\ &= M\cdot 2\pi \eps \\ &\convergesto{\eps\to 0} 0 ,\] noting that the bound \(M\) is constant and remains an upper bound on smaller discs by the maximum modulus principle.