Suppose $G$ is a complex Lie group and $M$ is a complex manifold. Suppose we have an action of $G$ on $M$ which is a holomorphic map $G \times M \rightarrow M$. I have seen the claim that it is easy to prove that $Stab_G(M) := \{ g \in G: g m = m \quad \forall m \in M \}$ is a complex Lie subgroup. I know that it is a real Lie subgroup by the closed subgroup theorem; in particular it is also a differentiable submanifold. But what would be an argument that $Stab_G(M)$ is also a complex submanifold?
Also by the closed subgroup theorem, the stabilizer subgroups of individual elements $m \in M$, $Stab_G(m):= \{ g \in G: gm = m \}$ are also real Lie subgroups. Are they complex Lie subgroups, though?