A holomorphic function on an annulus can have a nonzero closed-contour integral

The point $1$ lies in $A$. For $z\in A$, let $\delta =\frac{1}{2}\min \{\mathrm{\mid }z\mathrm{\mid }-1\mathrm{/}2,2-\mathrm{\mid }z\mathrm{\mid }\}>0$; if $\mathrm{\mid }w-z\mathrm{\mid }<\delta$, [L1] gives $1\mathrm{/}2<\mathrm{\mid }w\mathrm{\mid }<2$, so $A$ is open.

Given $z=r\exp (i\theta )$ and $w=s\exp (i\varphi )$ in $A$ as in [L3], the radial paths $t\mapsto ((1-t)r+t)\exp (i\theta )$ and $t\mapsto ((1-t)s+t)\exp (i\varphi )$ stay in $A$, and the unit-circle arc $t\mapsto \exp (i((1-t)\theta +t\varphi ))$ joins their unit endpoints. By [L7] these paths are continuous; the first, the arc, and the reversal of the second concatenate to join $z$ to $w$, so [L4] makes $A$ path-connected and connected.

Steps 1.1 and 1.2 show that $A$ is nonempty, open, and connected, hence a complex domain by [L5]; since $0\mathrm{\notin }A$, [L2] makes $f(z)=1\mathrm{/}z$ holomorphic on $A$.

The unit circle is a closed rectifiable contour in $A$, while [L6] gives its integral as $2\pi i\mathrm{\ne }0$. Thus the displayed domain, function, and contour refute the claim.