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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 $z_0$, $$\int _\gamma f=0$$

please any one help me with this problem i did't have any idea for how to solve this problem please help me

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

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You need the following theorem:

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 $z \in D$.

It is known as Riemman's Removable Singularity Theorem.

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  • $\begingroup$ @7667...so the answer is 0 beacuse by cauchy theorem $\endgroup$
    – user495078
    Commented Oct 28, 2017 at 4:33
  • $\begingroup$ 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 $\endgroup$
    – reuns
    Commented Oct 28, 2017 at 5:46

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