Let assume that $\Sigma_n$ is a sequence of compact surfaces in $\mathbb{R}^3$ of fixed genus. We assume that the surfaces are Alexandrov embedded, that is to say there exits an immersion $i_n$ from $\Omega$ a $3$-manifold into $\mathbb{R}^3$ such that $\partial \Omega$ is topologically like $\Sigma_n$ and such that $i_n(\partial \Omega) = \Sigma_n$.
Then we assume that $\Sigma_n$ converges to $\Sigma$ in the sense that around each point there exists local conformal parametrization $u_n :\mathbb{D} \rightarrow \mathbb{R}^3$ such that $u_n$ converge in $C^2_{loc}(\mathbb{D})$ to $u :\mathbb{D} \rightarrow \mathbb{R}^3$.
Is that possible that $\Sigma$ get branch point? i.e. locally $u$ looks like $z \mapsto z^k$.
This question come from constant mean curvature surface theory in some pertubative setting. Hence I can assume that the branch point are isolated.
For instance: does there exists a sequence of smooth Alexandrov embedded surfaces which converge in the sens defined above to $\omega\circ z\mapsto z^3$ where $\omega$ is the inverse of the stereographic projection?