# show that for any closed curve $\gamma$ containing $z_0$, $\int _\gamma f=0$

Let $f(z)$ be holomorphic on a simple connected region A, except possibly not holomorphic at $z_0\in A$. Suppose, however, that f is bounded in absolute value near $z_0$. show that for any closed curve $\gamma$ containing $z_0$, $$\int _\gamma f=0$$

• $f$ will have a removable singularity at $z_0$. Commented Oct 28, 2017 at 4:16
• Perhaps Morera's theorem would be al alternative approach? Commented Oct 28, 2017 at 17:03

If $f:D \to A$ is holomorphic function, except for $z_{0} \in D$ and f is bounded near $z_{0}$, then f can be extended to holomorphic function for all $z \in D$.
• The answer is that$\int_\gamma f(z)dz = \lim_{\epsilon \to 0} \int_{|z-z_0| = \epsilon} f(z)dz = 0$ using the Cauchy integral theorem, or if you prefer the Green theorem (assuming $f$ is regularized such that $f'$ is continuous). This is one of the steps in proving the Cauchy integral formula and hence Riemann's theorem on removable singularities. @nareshnayer Commented Oct 28, 2017 at 5:46