$f(z) = 1/z$ has a singularity at $z=0$. On the punctured disk $0<|z|<1$, the function is continuous but not uniformly continuous: as $z\to 0$, $f(z)\to\infty$ and the modulus of continuity cannot be bounded uniformly. On $0<|z|\leq 1$ the function is bounded below away from the singularity... Per key: the region is $0<|z|<1$ (open punctured disk).