= Solution
The only possible singularity of $f(z)/(z-z_0)$ inside the contour is at $z_0$, with <residue> $f(z_0)$. It is at most a simple pole; when $f(z_0)=0$, it is removable. Thus the <residue theorem> gives the <Cauchy integral formula>,
$$
\boxed{\oint_C\frac{f(z)}{z-z_0}dz=2\pi i\,f(z_0),}
$$
for a counterclockwise simple contour containing $z_0$ once, with $f$ analytic on a neighbourhood of the contour and its interior. More generally the answer is $2\pi i\,\operatorname{Ind}(C,z_0)f(z_0)$, where the <winding number> accounts for multiple turns or reversed orientation. A clockwise simple contour has the opposite sign.
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