Pull-backs of $\sum dx_i \wedge dy_i$ under radial diffeomorphisms of $\mathbb C^n$ Consider $\mathbb C^n$ with coordinates $(z_1,\dots,z_n)$, $z_j=x_j+iy_j$. Let $\omega=\sum dx_i\wedge dy_i$. Let us call by a radial diffeomorhpism $\varphi$ of $\mathbb C^n$ a diffemorphism of $\varphi:\mathbb C^n\to \mathbb C^n$ that sends any $z\in \mathbb C^n$ to $f(|z|)\cdot z$, where $f$ is a smooth positive function on $\mathbb R^+$.
Question. Let $\varphi$ be a radial diffeomorphism of $\mathbb C^n$. Is it true that $\varphi^*(\omega)$ is a $(1,1)$-form on $\mathbb C^n$? 
I think I can prove the above statement but find it a bit surprising, since the diffeo $\varphi$ is of course not a bi-holomorphism unless $f$ is constant. I would be grateful for a short proof /reference/ or a counterexample for this statement.  
 A: Yes, it's true, and it follows immediately from a simple calculation:  Since $f$ is assumed to be a function of $|z|$ such that $\phi(z) = f(|z|)z$ is a smooth diffeomorphism, it follows that $\mathrm{d}f$ is a multiple of $\mathrm{d}\bigl(|z|^2\bigr)$, so $\partial f$ is a multiple of $\partial\bigl(|z|^2\bigr)$.  Now
$$
\phi^*(\omega) = \tfrac{i}{2} \,\phi^*\left(\mathrm{d}z_j\wedge\mathrm{d}\overline{z_j}\right)
= \tfrac{i}{2} \mathrm{d}(fz_j)\wedge\mathrm{d}(f\overline{z_j})
=  \tfrac{i}{2}\left(f^2 \mathrm{d}z_j\wedge\mathrm{d}\overline{z_j} + f\,\mathrm{d}f\wedge(z_j\,\mathrm{d}\overline{z_j}-\overline{z_j}\,\mathrm{d}z_j) \right).
$$
The $(2,0)$ and $(0,2)$ parts of the right hand side vanish if and only if 
$$
\partial f \wedge (\overline{z_j}\,\mathrm{d}z_j) 
= \overline\partial f \wedge (z_j\,\mathrm{d}\overline{z_j}) = 0.\tag1
$$
In particular, these equations will hold since they are the same as 
$$
\partial f \wedge \partial (|z|^2) =\overline \partial f \wedge \overline\partial (|z|^2) = 0,\tag2
$$
which holds, by hypothesis.
Addendum:  In fact, a stronger result holds:  If $n\ge 2$ and $f$ is any smooth (real-valued) function on $\mathbb{C}^n$, then the mapping $\phi:\mathbb{C}^n\to\mathbb{C}^n$ defined by $\phi(z) = f(z)z$ pulls back the form $\omega$ to be of type $(1,1)$ if and only if $f$ is constant on the level sets of $|z|$.  The proof is a little more subtle in the $n=2$ case, but, basically what you have to observe is that (1) implies that there is a (complex) function $g$ such that
$$\mathrm{d}f = g\,(\overline{z_j}\,\mathrm{d}z_j)
+\overline{g}\, \overline{z_j}\,\mathrm{d}z_j\,.\tag3
$$ 
Taking the exterior derivative of (3) yields a relation
$$
0 = \mathrm{d}g\wedge(\overline{z_j}\,\mathrm{d}z_j)
+\mathrm{d}\overline{g}\wedge\overline{z_j}\,\mathrm{d}z_j
+(\overline{g}-g)\,\omega,\tag 4
$$
which, when $n>2$, implies by elementary exterior algebra that $g = \overline{g}$, so that $\mathrm{d}f = g\mathrm{d}(|z|^2)$, implying that $f$ is constant on the level sets of $|z|$.  When $n=2$, one has to examine (4) more carefully, but it still implies that $g = \overline{g}$.
