d c operator
= d c operator
{title2=$d^c$}
With the convention $d^c=i(\bar\partial-\partial)$ and with $J$ acting on forms by applying the <complex structure> to every argument,
$$
d^c=J^{-1}dJ.
$$
Indeed, $J$ acts on a $(p,q)$-form by $i^{p-q}$; conjugating $d=\partial+\bar\partial$ therefore multiplies its two type components by $-i$ and $i$, respectively.