Suppose we have an almost complex manifold $(M,J)$ and $f:M\rightarrow \mathbb{C}$ a smooth function such that $0$ is a regular value, and that $(\bar \partial f)_p=0 $ for any $p\in M$. Then I would like to see that $f^{-1}(0)=N$ is a complex submanifold ,i.e., that $J_p(T_pN)=T_pN$. Now I think I was able to do this but if we asked the condition that $(\partial f)_p=0$ I think my proof would also work , so I wanted to make sure things added up.
Recall that $T_p N= \ker d_p f$, and that since $f$ is a $0-$form $df=\partial f +\bar \partial f$. Now suppose that $x\in J_p(T_p N)$ and there exists an $y\in T_p N$ such that $x=J_p(y)$. We will have that $d_pf(x)=(\partial f)_p(x)+\bar (\partial f)_p(x)=(\partial f)_p(J_p(y))=i\partial_p(y)= i d_pf(y)=0$
And so we obtain that $J_p(T_pN)\subset T_pN$, since both subspaces have the same dimension we obtain that $T_pN=J_p(T_pN)$, and since our choice of $p$ was arbritary we get the desired result.
What do you think ? Thanks in advance.