I'm trying to find an example of a function which is holomorphic and bounded on $B(0,2)\backslash [0,1]$ which can not be extended to a holomorphic function on $B(0,2)$.

I was trying to use $e^{\text{something}}$ at first but someone recommended that I switch to trigonometric functions (since they are more poorly behavied) so I've been trying $$\sin{\frac{\pi}{z(z-3)}}.$$ The idea is to use the fact that sine will have an essential singularity at some point on $[0,1]$ so that defining a holomorphic extension at $z=0$ will fail.

Also, would be a graph of this function.

I can show/know that sine is continuous on $B(0,2)\backslash [0,1]$. I also know how to show that $\sin{\frac{\pi}{z}}$ has an essential singularity, but I don't think I can show that the function is bounded on $B(0,2)\backslash [0,1]$.

So I'm not sure what to do now. I'm having trouble describing a map from a bounded open subset of $\mathbb{C}$ whose image is bounded while also being discontinuous on the boundary of the closure of it's initial domain .I can try a non-isolated singularity maybe? Something like $\tan(1/z)$?