Does there exist a non-open connected set where the derivative of $f$ vanishes without $f$ being constant?

Question

Let $D \subsetneq \Bbb C$ be a non-open connected set and let $f : D \longrightarrow \Bbb C$ be an analytic function with $f'(z) = 0,$ for all $z \in D.$ Does it necessarily imply that $f$ is constant on $D?$

Any help in this regard will be highly appreciated. Thank you very much.

Answer 1

By definition analyticity of $f$ on $D$ means it is analytic in some open set containing $D$. If $D$ has more than one point then $D$ is uncountable. Hence it has a limit point. Since the zeros of $f'$ have a limit point we get $f'=0$ in the connected component of the domain that contains $D$. Hence $f$ is constant on $D$.