Why do we want to consider the complexification of the tangent space in complex geometry?

Given a complex manifold $$M$$, its complexified tangent bundle is $$TM \otimes \mathbb C$$. It is quite confusing for me as to why we want to do this since at each point $$TM$$ can already be viewed as a complex vector space.

A complex manifold is given by two alternative definitions. First you have $$M$$ a "nice" topological space that has an atlas $$\phi_j \colon U_j \longrightarrow V_j\subset \mathbb{C}^n$$ such that $$\phi_j\circ\phi_i^{-1}$$ is a biholomorphism. This way you'll get a complex tangent bundle.
Another way is to consider a differentiable manifold of (real) dimension $$2n$$ together with an automorphism $$J$$ of the tangent bundle $$TM$$ such that $$J^2= -Id$$. Complexifying $$TM$$ gives a diagonalization for $$J$$ in the fibers. Locally, it decomposes $$TM\otimes \mathbb{C} = E_i \oplus E_{-i}$$ as the sum of eigenspaces. Note that $$TM\otimes \mathbb{C}$$ has real rank $$4n$$ and from $$J^2=-Id$$, $$E_i$$ and $$E_{-i}$$ both have real rank $$2n$$. This is called an almost complex structure. If $$[E_{-i},E_{-i}] \subset E_{-i}$$ then this decomposition is global and we call $$E_i$$ (which has complex rank $$n$$) the holomorphic tangent bundle of $$M$$ and $$J$$ a complex structure.
See Huybrecht's book. From the begining he starts this discussion exploring the identification of $$\mathbb{R}^2$$ and $$\mathbb{C}$$.
• I am confused. In your first definition, apparently, the manifold is also a smooth manifold of dimension $2n$, which means its tangent space at each point can be viewed as a complex vector space of dimension $n$. This is not consistent with the second definition, since the complexification will give you a vector space of complex dimension 2n. – Keith Sep 30 '18 at 15:35