= Solution
Use the complex-bilinear extension of the <Nijenhuis tensor> expression to complex <vector fields>. If $U,V\in T^{1,0}X$, then $JU=iU$ and $JV=iV$, so
$$
N(U,V)=-4[U,V]-4iJ[U,V]=-8[U,V]^{0,1}.
$$
For $U,V\in T^{0,1}X$, the same substitution with $-i$ gives $N(U,V)=-8[U,V]^{1,0}$. For inputs of opposite types, the two leading <Lie brackets> cancel, as do the two terms involving $J$, so $N(U,V)=0$. Applying the <type decomposition of the complexified tangent bundle> to both inputs consequently gives
$$
\boxed{N(\alpha,\beta)=-8[\alpha',\beta']''-8[\alpha'',\beta'']',\qquad
N(\alpha,\beta)''=-8[\alpha',\beta']''.}
$$
Thus $N=0$ precisely when the <Lie bracket> of any two sections of either <eigenbundle> remains in that same <eigenbundle>. In other words, \b[$N=0$ if and only if both tangent-type distributions are <involutive distributions>]. This proves the equivalence directly and does not assume that an arbitrary <almost complex structure> already has holomorphic coordinates.
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