# Show that $\int_{\gamma} f(z)\ dz =0$.

Suppose that $f$ is analytic on a disk $B(a;r)$ except for finitely many points $a_1 , a_2 , \cdots , a_n$ such that $\lim_{z \rightarrow a_i } (z - a_i ) f(z) = 0$ for $i = 1,2, \cdots, n$. Then show that $\int_{\gamma} f(\zeta)\ d\zeta = 0$ for any simple closed contour $\gamma$ on $B(a;r) \setminus \{a_1 , a_2 , \cdots , a_n \}$.

I have proved this result for $n=1$ which is easy to prove using Cauchy's deformation theorem and estimation theorem. But I have failed to generalize it for any finite case. Please help me.

• The case $n > 1$ is exactly the same. $\int_{|z| = r} f(z) = \sum_{|a_m| < r} \int_{|z-a_m| = \epsilon} f(z)dz$. Note $\lim_{\epsilon \to 0} \int_{|z-a| = \epsilon} f(z)dz= Res(f(z),a)$ Oct 28, 2017 at 6:57
What about the points, either they are distinct or repeated.If they are distinct then we can consider them as a simple pole and by given condition, $lim_{z\to a_i}(z-a_i)f(z)=0$ implies $Res(f,a_i)=0$ then simply using the residue theorem your result follows.
By Riemann's theorem on removable singularities, your function $f$ is the restriction to $B(a;r)\setminus\{a_1,\ldots,a_n\}$ of a holomorphic function $F$. Therefore$$\int_\gamma f(\zeta)\,\mathrm d\zeta=\int_\gamma F(\zeta)\,\mathrm d\zeta=0,$$by Cauchy's integral theorem.