[proofplan] We first put the point at the origin and straighten the real tangent hyperplane by a holomorphic affine change of coordinates. The [implicit function theorem](/theorems/52) then writes the hypersurface as a graph $v=\phi(z',u)$ with vanishing first-order jet. The weighted quadratic Taylor expansion separates the Levi, or Hermitian, quadratic part from the pluriharmonic quadratic part. Strict pseudoconvexity makes the Hermitian part positive definite after choosing the correct side, so a complex linear change normalizes it to $|z'|^2$, and a holomorphic quadratic change of the normal coordinate removes the pluriharmonic part. The remaining Taylor terms have exactly the stated weighted order, and the leading model is therefore the Heisenberg quadric. [/proofplan]

[step:Straighten the point and the real tangent hyperplane] Choose initial holomorphic coordinates \begin{align*} \zeta=(\zeta_1,\dots,\zeta_n,\tau)\in\mathbb C^n\times\mathbb C \end{align*} near $p$, with $\tau=s+it$, such that $p$ corresponds to $0$. Let $T_0M$ denote the real tangent hyperplane to $M$ at $0$. Since $M$ is a smooth real hypersurface, after a complex linear change of coordinates we may assume \begin{align*} T_0M=\{(\zeta',s+it)\in\mathbb C^n\times\mathbb C:t=0\}. \end{align*} Let $H_0M$ denote the complex tangent space at $0$, namely the maximal complex-linear subspace of $T_0M$; in these coordinates $H_0M=\mathbb C^n\times\{0\}$. Choose the sign of $t$ so that the Levi form of a local defining function for the pseudoconvex side is positive definite on $H_0M$. Since $T_0M=\{t=0\}$, the projection \begin{align*} \pi:M\to\mathbb C^n\times\mathbb R,\qquad (\zeta',s+it)\mapsto (\zeta',s) \end{align*} has invertible differential at $0$ when restricted to $M$. By the smooth implicit function theorem, after shrinking the coordinate neighbourhood there is a real-valued function \begin{align*} \phi:U\subset \mathbb C^n\times\mathbb R\to\mathbb R \end{align*} of class $C^\infty$, defined on a neighbourhood $U$ of $0$, such that \begin{align*} M=\{(\zeta',s+it):t=\phi(\zeta',s)\}. \end{align*} Moreover, \begin{align*} \phi(0,0)=0,\qquad d\phi_{(0,0)}=0, \end{align*} because $0\in M$ and $T_0M=\{t=0\}$. [/step]

[step:Separate the weighted quadratic Taylor terms]Let $z=(z_1,\dots,z_n)$ denote the $\mathbb C^n$ variable. Since $\phi$ is $C^\infty$ and $d\phi_{(0,0)}=0$, Taylor expansion to second order in the real variables $(x_1,y_1,\dots,x_n,y_n,s)$ gives \begin{align*} \phi(z,s)=Q(z,\bar z)+R_0(z,s), \end{align*} where $Q$ is a real-valued homogeneous quadratic polynomial in $(z,\bar z)$ and $R_0$ is $C^\infty$ with \begin{align*} R_0(z,s)=O\left(|z|^3+|s||z|+s^2\right). \end{align*} The absence of a term linear in $s$ follows from $d\phi_{(0,0)}=0$, and the terms $s z_j$, $s\bar z_j$, and $s^2$ are included in the displayed remainder because they have sizes bounded by $|s||z|$ and $s^2$. Every real-valued quadratic polynomial $Q$ in $(z,\bar z)$ has a unique decomposition \begin{align*} Q(z,\bar z)=H(z,\bar z)+\operatorname{Re} P(z), \end{align*} where \begin{align*} H(z,\bar z)=\sum_{j,k=1}^n h_{jk}z_j\bar z_k \end{align*} for a Hermitian matrix $H_0=(h_{jk})_{j,k=1}^n$, and \begin{align*} P:\mathbb C^n\to\mathbb C \end{align*} is a holomorphic homogeneous polynomial of degree $2$.[/step]

[guided]The point of this step is to isolate the terms that affect the Levi form from the terms that can be removed by a holomorphic change of the normal variable. We write $z_j=x_j+iy_j$ and regard $\phi$ as a smooth real-valued function of the real variables \begin{align*} (x_1,y_1,\dots,x_n,y_n,s)\in\mathbb R^{2n+1}. \end{align*} Because $\phi(0,0)=0$ and $d\phi_{(0,0)}=0$, the Taylor polynomial of ordinary degree at most $2$ has no constant or linear part. The terms of weighted degree at most $2$ are therefore precisely the quadratic terms involving only $z$ and $\bar z$; the terms involving $s$ are already part of the allowed error. More explicitly, terms of the form $s z_j$ and $s\bar z_j$ are bounded by a constant times $|s||z|$, while $s^2$ is already one of the allowed terms. The smooth Taylor remainder of order at least $3$ in the real variables is bounded by a constant times $|(z,s)|^3$, and for $(z,s)$ near $0$ this is bounded by a constant times \begin{align*} |z|^3+|s||z|+s^2. \end{align*} Thus there is a real-valued homogeneous quadratic polynomial $Q$ in $(z,\bar z)$ and a smooth remainder $R_0$ satisfying \begin{align*} \phi(z,s)=Q(z,\bar z)+R_0(z,s), \end{align*} with \begin{align*} R_0(z,s)=O\left(|z|^3+|s||z|+s^2\right). \end{align*} Now decompose $Q$. A real-valued quadratic expression in complex variables is a sum of terms $z_j\bar z_k$, terms $z_jz_k$, and the complex conjugates of the $z_jz_k$ terms. The $z_j\bar z_k$ part is Hermitian because $Q$ is real-valued; hence it has the form \begin{align*} H(z,\bar z)=\sum_{j,k=1}^n h_{jk}z_j\bar z_k \end{align*} with $h_{kj}=\overline{h_{jk}}$. The remaining pure holomorphic and anti-holomorphic quadratic terms are the real part of a holomorphic homogeneous polynomial \begin{align*} P:\mathbb C^n\to\mathbb C. \end{align*} Therefore \begin{align*} Q(z,\bar z)=H(z,\bar z)+\operatorname{Re}P(z). \end{align*} This split is exactly what is needed: $H$ is the Levi quadratic term, while $\operatorname{Re}P$ is pluriharmonic and can be absorbed into the imaginary part of the normal coordinate.[/guided]

[step:Normalize the Hermitian Levi quadratic form] For the defining function \begin{align*} \rho(\zeta',\tau)=t-\phi(\zeta',s), \end{align*} the Levi form at $0$ on $H_0M=\mathbb C^n\times\{0\}$ is represented, with the sign fixed by the chosen side, by the Hermitian matrix associated with $H(z,\bar z)$. Replacing $t$ by $-t$ replaces $\rho$ by $-\rho$ to second order in the normal direction and therefore multiplies the Levi form by $-1$ on $H_0M$. Strict pseudoconvexity lets us choose the sign for which this Hermitian form is positive definite. By the spectral theorem for positive definite Hermitian matrices, there is a complex linear isomorphism \begin{align*} A:\mathbb C^n\to\mathbb C^n \end{align*} such that \begin{align*} H(Az,\overline{Az})=|z|^2 \end{align*} for every $z\in\mathbb C^n$. Replacing $\zeta'$ by $Az$ and keeping the normal coordinate $\tau$ unchanged, the equation of $M$ becomes \begin{align*} t=|z|^2+\operatorname{Re}P_1(z)+R_1(z,s), \end{align*} where \begin{align*} P_1:\mathbb C^n\to\mathbb C \end{align*} is a holomorphic homogeneous polynomial of degree $2$, and \begin{align*} R_1(z,s)=O\left(|z|^3+|s||z|+s^2\right). \end{align*} [/step]

[step:Remove the pluriharmonic quadratic terms by changing the normal coordinate] Define a holomorphic coordinate change \begin{align*} (z,w):\mathbb C^n\times\mathbb C\to\mathbb C^n\times\mathbb C,\qquad (z,\tau)\mapsto (z,\tau-iP_1(z)). \end{align*} Write \begin{align*} w=u+iv. \end{align*} Since $P_1$ has degree $2$, this map has complex-linear differential equal to the identity at $0$, so it is a local biholomorphic coordinate change. On $M$, where $\tau=s+it$ and \begin{align*} t=|z|^2+\operatorname{Re}P_1(z)+R_1(z,s), \end{align*} we have \begin{align*} v=\operatorname{Im}(\tau-iP_1(z))=t-\operatorname{Re}P_1(z)=|z|^2+R_1(z,s). \end{align*} Also \begin{align*} u=\operatorname{Re}(\tau-iP_1(z))=s+\operatorname{Im}P_1(z), \end{align*} so \begin{align*} s=u-\operatorname{Im}P_1(z). \end{align*} Define \begin{align*} R:\mathbb C^n\times\mathbb R\to\mathbb R,\qquad R(z,u)=R_1(z,u-\operatorname{Im}P_1(z)) \end{align*} on a sufficiently small neighbourhood of $0$. Since $P_1(z)=O(|z|^2)$, the estimate for $R_1$ gives \begin{align*} R(z,u)=O\left(|z|^3+|u||z|+u^2\right). \end{align*} Thus, in the new holomorphic coordinates, \begin{align*} M=\{(z,w):v=|z|^2+R(z,u)\}. \end{align*} [/step]

[step:Identify the weighted order and the Heisenberg leading model] Rename $z$ as $z'=(z_1,\dots,z_n)$. The estimate \begin{align*} R(z',u)=O\left(|z'|^3+|u||z'|+u^2\right) \end{align*} means that there are a neighbourhood $V$ of $0$ in $\mathbb C^n\times\mathbb R$ and a constant $C>0$ such that \begin{align*} |R(z',u)|\le C\left(|z'|^3+|u||z'|+u^2\right) \end{align*} for all $(z',u)\in V$. Under the anisotropic scaling \begin{align*} (z',u)\mapsto (rz',r^2u), \end{align*} the three displayed error terms have weights $3$, $3$, and $4$, respectively. Hence the remainder has weighted order at least $3$. Discarding this weighted-order at least $3$ remainder leaves the model equation \begin{align*} v=|z'|^2, \end{align*} which is the Heisenberg quadric. This proves the stated normal form. [/step]