Pullback of a complex one-form

Question

Let f be a holomorphic function from $\mathbb{C}^n$ to $\mathbb{C}^n$. Let $g_j$ be the jth coordinate function from $\mathbb{C}^n$ to $\mathbb{C}$.

Show that $f^*dz_j$= $\partial g_j$.

I understand that this $\partial g_j$ means the Dolbeaut operator of first taking exterior derivative and then projecting on the (1,0) component.

I tried writing $dz_j=dx_i+idy_j$ and taking $f^*$ of that. Then I distributed $d$ linearly but I cannot see how to get to $\partial g_j$.

Help?

Answer 1

Note that because $f$ is holomorphic, each component $f_j$ is holomorphic, and so $\bar\partial f_j = 0$. On the other hand, because $$df_j = d(f^*z_j) = f^*dz_j,$$ we conclude that $f^*dz_j = df_j = \partial f_j + \bar\partial f_j = \partial f_j$.