Is every connected Lie subgroup of the additive group $\mathbb{C}^n$ already a complex vector subspace? I wonder if every connected Lie subgroup of the additive group $\mathbb{C}^n$ already is a complex vector subspace. This seems correct to me. In general, I would be happy to find a good book on complex Lie groups.
Thanks for your help!
Edit: What if we moreover assume that the subgroup is a global analytic set in $\mathbb{C}^n$, i.e. the vanishing set of some holomorphic functions  in several variables?
 A: The connected Lie subgroups of $\mathbb{C}^n$ are exactly the real vector subspaces. We can see this as follows: the connected abelian Lie groups are all products of copies of $\mathbb{R}$ and $S^1$ (e.g. by considering the exponential map from their Lie algebras). Since $\mathbb{C}^n$, as a group, is torsion-free, all its subgroups must be torsion-free, which means $S^1$ can't occur as a subgroup. So the connected Lie subgroups must have the form $\mathbb{R}^k$ for some $k$, and smooth homomorphisms $\mathbb{R}^k \to \mathbb{C}^n$ must be $\mathbb{R}$-linear (in fact much weaker conditions would suffice, continuity or even measurability; this is related to the Cauchy functional equation).
If by Lie group you meant "complex Lie group" then Dietrich Burde's argument in the comments finishes it; the same exponential map argument as above shows that the connected abelian complex Lie groups are all quotients of $\mathbb{C}^k$ by a lattice, and then torsion-freeness implies the lattice must be trivial, so we get $\mathbb{C}^k$ on the nose, and similarly holomorphic homomorphisms $\mathbb{C}^k \to \mathbb{C}^n$ must be $\mathbb{C}$-linear.
