Assuming that there is a complex structure on $S^{2n}$ and it becomes a complex manifold, also assuming there are complex coordinate $z, w$ on $U, V$ respectively, where $U, V$ are open cover of $S^{2n}$. Let $\rho_U, \rho_V=1-\rho_U$ be partition of unity. Then we have a connection on the bundle $T^{(1,0)}S^{2n}$ defined by $$\nabla= \rho_U \nabla^U+\rho_V \nabla^V, $$ where $\nabla^U, \nabla^V$ are defined by $$\nabla^U\frac {\partial }{\partial z^i}=0; \ \ \ \nabla^V\frac {\partial }{\partial w^i}=0, \ \ i=1,\cdots, n.$$ Denote $A= (\frac {\partial w^j}{\partial z^i}), \ \omega_0=dA\cdot A^{-1}= \partial A\cdot A^{-1}$ and $(\frac {\partial }{\partial z^i})=(\frac {\partial }{\partial z^1},\cdots, \frac {\partial }{\partial z^n})^t=A(\frac {\partial }{\partial w^j})$, then we have $$\nabla(\frac {\partial }{\partial z^i})= \rho_U \nabla^U (\frac {\partial }{\partial z^i}) +\rho_V \nabla^V(\frac {\partial }{\partial z^i})=\rho_VdA\cdot A^{-1}(\frac {\partial }{\partial z^i}), $$ $$\nabla^2(\frac {\partial }{\partial z^i})=(d\rho_V\wedge\omega_0 +\rho_V(1-\rho_V)\omega_0\wedge\omega_0)(\frac {\partial }{\partial z^i})=\Omega|_U (\frac {\partial }{\partial z^i}).$$

Then it is easy to see the top Chern class of the the bundle $T^{(1,0)}S^{2n}$ is trivial if $n>1$, that is, $$c_n(T^{(1,0)}S^{2n})=\det (\frac {\sqrt{-1}}{2\pi}\Omega)=0.$$ On the other hand, we have $c_n(T^{(1,0)}S^{2n}) = e(TS^{2n})$ the Euler class of tangent bundle of $S^{2n}$, a contradiction.

My question is:

If the open ball $B^{2n}\subset R^{2n}$ is a complex manifold, is there a holomorphic isomorphism from $B^{2n}$ to a open subset of $C^n$? Or equivalently, does complex manifold $B^{2n}$ can be covered by one complex coordinates card?

In 1977, R Hamilton (J. D. G., vol.12, no.1, 1-45) showed that such isomorphism exists if the complex structure on $B^{2n}$ is sufficiently close to the structure on $C^n \supset B^{2n}$.