My complex analysis textbook formulated the Cauchy's integral formula as the following:
let $f:U→C$ be holomorphic in an convex open set $U$ and let $\gamma:[a,b]→U$ be any closed curve on $U$. Then:
$$f(z)\text{Ind}_{\gamma}(z)=\frac{1}{2\pi i}∮_\gamma \frac{f(w)}{w-z}dw$$
for any $z \text{ in } U\ \backslash\ \gamma([a,b])$
Thus:
$$∮_\gamma \frac{f(w)}{w-z}dw=f(z)\text{Ind}_{\gamma}(z)2\pi i $$
Does that mean that the integral is always 0 if $z$ is not inside the region with border $\gamma$? Because if that happens $\text{Ind}_{\gamma}(z)$ would be 0 giving us:
$$∮_\gamma \frac{f(w)}{w-z}dw=0$$
Best Answer
Yes. By Cauchy's theorem, the integral for points outside the contour is zero, since the integrand is analytic.