= Solution
<Cauchy's integral theorem> says that if $f$ is <holomorphic> on an <open set> $U$ and a closed, <piecewise smooth> <contour> $\gamma$ is <null-homotopic> in $U$, then
$$
\boxed{\int_\gamma f(z)\,dz=0.}
$$
In particular, this holds for every closed <contour> when $U$ is <simply connected>, and for a simple closed <contour> whose boundary and enclosed region lie in $U$. For the later arguments it suffices to use the triangular version: the <integral> around a <triangle> is zero if the <triangle> and its interior lie in $U$ and $f$ is <holomorphic> there.
Both the holomorphicity condition and a suitable condition excluding holes inside the <contour> matter. <Continuity> alone, which is the starting hypothesis for this question, does not imply the theorem's conclusion; nor does holomorphicity on an arbitrary punctured <open set> imply zero <integral> around every loop.
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