I have this question, from several complex variables:
Start with the differential form: $$\omega(z)=\sum_{\nu=1}^{n} \frac{(-1)^{\nu-1}\bar{z}_{\nu}}{|z|^{2n}} d\bar{z}[\nu] \wedge dz, $$
where $dz= dz_1 \wedge ... \wedge dz_n$ and $d\bar{z}[\nu] = dz_1 \wedge ... dz_{\nu - 1} \wedge dz_{\nu +1} ... \wedge dz_n$.
Since $d(\bar{z}_{\nu} d\bar{z}[\nu]) = d\bar{z}_{\nu} \wedge d\bar{z}[\nu] = (-1)^{\nu -1} d\bar{z}$, we have
$$d\omega =\sum_{\nu=1}^{n} \frac{\partial}{\partial \bar{z}_{\nu}} \bigg( \frac{\bar{z}_{\nu}}{|z|^{2n}} \bigg) d\bar{z} \wedge dz.$$
It's this line, right at the beginning, which I can't seem to compute. I'm pretty new to differential forms and exterior derivatives, so I can't see exactly how this is arrived at. I dont understand the first computation either ( $d(\bar{z}_{\nu} d\bar{z}[\nu]) = d\bar{z}_{\nu} \wedge d\bar{z}[\nu]$ etc. ). Any help or hints would be great, thanks.
Recall the product rule $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{|\alpha|} \alpha \wedge d\beta$. Hence, we have
$$ d(\bar{z}_{\nu} d\bar{z}[\nu]) = d(\bar{z}_{\nu}) \wedge d\bar{z}[\nu] + \bar{z}_{\nu} \wedge d(d\bar{z}[\nu]). $$
Applying the product rule again, you see that $d(d\bar{z}[\nu])$ is a sum of a wedge of forms, each containing $d^2 z_i$ and since $d^2 = 0$, it vanishes.
Hence,
$$ d(\bar{z}_{\nu} d\bar{z}[\nu]) = d(\bar{z}_{\nu}) \wedge d\bar{z}[\nu] = d\bar{z}_{\nu} \wedge d\bar{z}_1 \wedge \dots \wedge d\bar{z}_{\nu - 1} \wedge d\bar{z}_{\nu + 1} \wedge \dots d\bar{z}_n \\ = (-1)^{\nu - 1} d\bar{z}_1 \wedge \dots \wedge d\bar{z}_n = (-1)^{\nu - 1} d\bar{z} $$
where we used the anti-symmetry of the wedge product to move the factor $d\bar{z}_{\nu}$ into the correct place, by applying $\nu - 1$ adjacent swaps.
See if this is enough to get you going for the second calculation.