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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: $$∮_\gamma \frac{f(w)}{w-z}dw=0$$

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Yes. By Cauchy's theorem, the integral for points outside the contour is zero, since the integrand is analytic.

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