[step:Show that the CR functions give an immersion on the real slice]
Define $F:U\to \mathbb C^{m+1}$ by $F(x)=(f_1(x),\dots,f_{m+1}(x))$. We prove that $dF_x:T_xM\to T_{F(x)}\mathbb C^{m+1}$ is injective after shrinking $U$.
Fix $x\in U$ and let $v\in T_xM$ satisfy $dF_x(v)=0$. Complexify the real tangent space by setting $\mathbb C T_xM:=T_xM\otimes_{\mathbb R}\mathbb C$. The real-analytic inclusion $U\hookrightarrow U^{\mathbb C}$ induces a canonical complex-linear identification
$\mathbb C T_xM\cong T_{1,0,x}U^{\mathbb C}$
under which the real slice corresponds to the fixed-point set of complex conjugation and the subspace $T_{0,1,x}M\subset \mathbb C T_xM$ identifies with the holomorphic distribution $\mathcal L_x$. Let $v_{1,0}$ denote the image of $v$ under this identification. Since $f_\alpha=Z_\alpha|_U$, the real differential $df_{\alpha,x}(v)$ equals $dZ_{\alpha,x}(v_{1,0})$. The equality $dF_x(v)=0$ therefore gives $dZ_{\alpha,x}(v_{1,0})=0$ for every $1\le \alpha\le m+1$. Hence $v_{1,0}\in \bigcap_{\alpha=1}^{m+1}\ker dZ_{\alpha,x}=\mathcal L_x$, so $v_{1,0}$ lies in $T_{0,1,x}M$. Because $v$ is real, this forces $v\in T_{1,0,x}M\cap T_{0,1,x}M=\{0\}$ by the hypersurface-type hypothesis. Therefore $v=0$, and $dF_x$ is injective.
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