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Let $X$ be a compact Riemann surface with boundary. Let us shrink each connected component of the boundary into a point. We get a closed topological surface $Z$ with several marked points (which came from the components of the boundary of $X$) such that the complement $Z_0$ of these points carries a complex structure.

Does there exist a complex structure on $Z$ such that it induces on $Z_0$ its original complex structure?

Rmk. I think that if such a complex structure exists, it is unique.

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    $\begingroup$ If there is, it would mean that there is homomorphic diffeomorphism $\mathbb{C}\to \mathbb{D}$ which is not the case. $\endgroup$
    – Anton Petrunin
    Commented Oct 15, 2017 at 16:28
  • $\begingroup$ @AntonPetrunin: Could you please add more details? $\endgroup$
    – asv
    Commented Oct 15, 2017 at 16:37

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No. Complex structure on $X$ "knows" whether an end is a hole or a puncture. The conformal invariant responsible for this is called the extremal length of the family of closed curves surrounding the hole/puncture. See Ahlfors, Lectures on quasiconformal mappings.

The simplest case is a sphere with a puncture (conformally equivalent to the plane) vs sphere with a hole (conformally equivalent to a disk). These two things are conformally not equivalent. This is essentially Anton Petrunin's remark.

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