Definition $:$ Let $G \subseteq \mathbb C^n$ be a domain. The boundary $\partial G$ of $G$ is said to be smooth at $z_0 \in \partial G$ if there is an open neighbourhood $U = U(z_0) \subseteq \mathbb C^n$ and a function $\rho \in \mathcal C^{\infty} (U; \mathbb R)$ such that$:$
$1.$ $U \cap G = \{z \in U\ :\ \rho (z) \lt 0 \}.$
$2.$ $\left (d \rho \right )_z \neq 0$ for all $z \in U.$
The function $\rho$ is called a local defining function (or boundary function).
Proposition $:$ Let $\partial G$ be smooth at $z_0,$ and let $\rho_1, \rho_2$ be two local defining functions on $U = U(z_0).$ Then there is a $\mathcal C^{\infty}$- function $h$ on $U$ such that
$1.$ $h \gt 0$ on $U.$
$2.$ $\rho_1 = h \cdot \rho_2$ on $U.$
$3.$ $\left (d \rho_1 \right )_z = h(z) \cdot \left (d \rho_2 \right)_z$ for all $z \in U \cap \partial G.$
The proof of the above proposition starts with the fact that after a change of coordinates, we may assume that $z_0 = 0$ and $\rho_2 = y_n.$
But I don't get the fact that how do we make $\rho_2 = y_n.$ Any suggestion in this regard would be highly solicited.
Thanks in advance.
EDIT $:$ What I have observed is the following $:$
If possible, by translation, we may assume that $z_0 = 0.$ Then by the definition of local defining function we have $\rho (0) = 0$ and $(d \rho)_0 \neq 0.$ If possible, by rotation, we may assume that $(d \rho)_{0} = (0, 0, \cdots, 0, \lambda).$ Hence by implicit function theorem there exist neighbourhoods $U' = U' (\vec {0})$ in $\mathbb R^{2n - 1}$ and $U'' = U''(0)$ in $\mathbb R$ and a $\mathbb C^{\infty}$-function $h : U' \longrightarrow U''$ such that $$\partial G \cap \left (U' \times U'' \right ) = \left \{\left (z', x_n + ih(z',x_n) \right ) :\ (z', x_n) \in U' \right \}.$$ Hence if we make the coordinate change $\widetilde {y_n} = y_n - h(z', x_n)$ in the last real coordinate then locally $\partial G$ is of the form $\left \{\left (z', x_n + i \widetilde {y_n} \right ) :\ \widetilde {y_n} = 0 \right \}.$ But I still don't understand how do we make $\rho = y_n$ locally.