Biholomorphic Riemann Surfaces can have different differential structure? I know that for real manifolds, being diffeomorphic does not imply having same differential structure.
For example $ \mathbb R $ with atlas $\{ (\mathbb R , x) \}$ and $ \mathbb R $ with atlas $\{ (\mathbb R , x^3) \}$ are diffeomorphic but two charts are not compatible.
Is it similar for biholomorphic riemann surfaces? 
i.e. does there exist biholomorphic Riemann surfaces having holomorphically  uncompatible charts?
 A: The same basic idea as $x^3$ vs $x$ can produce a huge number of examples - we can modify the holomorphic structure by composing it with any nasty homeomorphism. In detail:
Let $(\Sigma,A = \{\phi_i\})$ be any Riemann surface; i.e. a topological surface $\Sigma$ equipped with an atlas of homeomorphisms $\phi_i : \Sigma \supset U_i \to \phi_i(U_i) \subset \mathbb C$ such that $\phi_i \circ \phi_j^{-1}$ is holomorphic wherever defined. Let $\psi : \Sigma \to \Sigma$ be any homeomorphism that is not holomorphic as a map $(\Sigma,A) \to (\Sigma,A).$
I claim that we can replace each $\phi_i$ with $\tilde\phi_i = \phi_i \circ \psi$ to obtain a holomorphic atlas $\tilde A$ on $\Sigma$ which is biholomorphic to the original one, but not compatible. 
First, on the overlaps we have $$\tilde \phi_i \circ \tilde \phi_j^{-1} = \phi_i \circ \psi \circ \psi^{-1} \circ \phi_j^{-1} = \phi_i \circ \phi_j^{-1},$$ so the fact that $A$ is an atlas tells us that $\tilde A$ is an atlas.
Secondly, $\psi$ is a biholomorphism $(\Sigma, \tilde A) \to (\Sigma, A):$ for any charts $\tilde \phi_i, \phi_j$ we have $$ \phi_j \circ \psi \circ \tilde \phi_i^{-1} = \phi_j \circ \psi \circ \psi^{-1} \circ \phi_i^{-1} = \phi_j \circ \phi_i^{-1},$$ which again is holomorphic because $A$ is a holomorphic atlas.
Finally, the fact that $\psi : (\Sigma,A) \to (\Sigma,A)$ is not holomorphic tells us that there are some overlapping charts on which $\phi_i \circ \psi \circ \phi_j^{-1}=\tilde \phi_i \circ \phi_j^{-1}$ is not holomorphic, so $\tilde \phi_i$ is not compatible with $A$.
