= Solution
Let $f=u+iv$ be analytic on a neighbourhood of a bounded planar region $\Omega$ and its piecewise smooth boundary $C$, oriented with $\Omega$ on the left. The <Cauchy integral theorem> states
$$
\boxed{\oint_Cf(z)\,dz=0.}
$$
For a region with holes, $C$ includes the clockwise inner boundaries as well as the counterclockwise outer one. In particular, the integral around a closed contour bounding a region throughout which $f$ is analytic is zero.
To prove this classical form, expand the line integral and apply <Green's theorem> to its real and imaginary parts:
$$
\begin{aligned}
\oint_C f\,dz
&=\oint_C(u\,dx-v\,dy)+i\oint_C(v\,dx+u\,dy)\\
&=\iint_\Omega(-v_x-u_y)\,dx\,dy+i\iint_\Omega(u_x-v_y)\,dx\,dy.
\end{aligned}
$$
Both integrands vanish by the <Cauchy-Riemann equations>, $u_x=v_y$ and $u_y=-v_x$. Analyticity supplies the continuous first partial derivatives needed by this <Green's theorem> proof. The requirement that the enclosed region contain no singularities is essential.
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