smooth ( complex valued) functions "divisible" by $\mathbb{z}^{2}$

Question

Let $U \subset \mathbb{C}$ be an open set containing $0$. Is it possible to construct a smooth function $f : U \rightarrow \mathbb{C}$ satisfying:

1. $f(0 ) = 0$
2. The function $g:U \rightarrow \mathbb{C}$, defined by $g(z) = z^2 f(z)$, is a diffeomorphism onto its image.

(Hope: no).

Answer 1

Any function $h$ satisfying $|h(z)|\le C|z|^2$ in a neighborhood of $0$ satisfies $Dh(0)=0$ in the real sense. This falls right out of the definition of "differentiable". And yes, $Dh(0)=0$ iff $J_h(0)$ is the zero matrix.