[example: Sphere and Hyperquadric]
For the sphere $M_S=\{|z|^2=1\}\subset\mathbb C^n$, take
\begin{align*}
\rho_S(z)=|z|^2-1=\sum_{m=1}^n z_m\bar z_m-1.
\end{align*}
We compute its Levi form by differentiating $\rho_S$ in the holomorphic and antiholomorphic coordinate directions. Treating $z_m$ and $\bar z_m$ as independent coordinates, for each pair $j,m$,
\begin{align*}
\partial_{z_j}(z_m\bar z_m)=(\partial_{z_j}z_m)\bar z_m+z_m(\partial_{z_j}\bar z_m)=\delta_{jm}\bar z_m+z_m\cdot 0=\delta_{jm}\bar z_m.
\end{align*}
Therefore
\begin{align*}
\frac{\partial\rho_S}{\partial z_j}=\partial_{z_j}\left(\sum_{m=1}^n z_m\bar z_m-1\right)=\sum_{m=1}^n\delta_{jm}\bar z_m-0=\bar z_j.
\end{align*}
Differentiating with respect to $\bar z_k$ gives
\begin{align*}
\frac{\partial^2\rho_S}{\partial z_j\partial\bar z_k}=\frac{\partial}{\partial\bar z_k}(\bar z_j)=\delta_{jk}.
\end{align*}
Let $p\in M_S$ and let
\begin{align*}
L=\sum_{j=1}^n a_j\partial_{z_j}\in T_p^{1,0}M_S.
\end{align*}
By the definition of the Levi form,
\begin{align*}
\mathcal L_{\rho_S,p}(L,\overline L)=\sum_{j,k=1}^n \frac{\partial^2\rho_S}{\partial z_j\partial\bar z_k}(p)a_j\overline{a_k}.
\end{align*}
Substituting $\frac{\partial^2\rho_S}{\partial z_j\partial\bar z_k}(p)=\delta_{jk}$ yields
\begin{align*}
\mathcal L_{\rho_S,p}(L,\overline L)=\sum_{j,k=1}^n \delta_{jk}a_j\overline{a_k}.
\end{align*}
The summand is $0$ when $j\ne k$, and when $j=k$ it is $a_j\overline{a_j}=|a_j|^2$, so
\begin{align*}
\mathcal L_{\rho_S,p}(L,\overline L)=\sum_{j=1}^n |a_j|^2.
\end{align*}
If $L\ne 0$, then some coefficient $a_j$ is nonzero, so the corresponding term $|a_j|^2$ is positive and all other terms are nonnegative. Hence the Levi form of the sphere is positive definite on every complex tangent space.
For the hyperquadric
\begin{align*}
M_Q=\{(z_1,z_2,w)\in\mathbb C^3:\operatorname{Im}w=|z_1|^2-|z_2|^2\},
\end{align*}
use the defining function
\begin{align*}
\rho_Q(z_1,z_2,w)=|z_1|^2-|z_2|^2-\operatorname{Im}w=z_1\bar z_1-z_2\bar z_2-\frac{w-\bar w}{2i}.
\end{align*}
The holomorphic first derivatives are
\begin{align*}
\frac{\partial\rho_Q}{\partial z_1}=\partial_{z_1}(z_1\bar z_1)-\partial_{z_1}(z_2\bar z_2)-\partial_{z_1}\left(\frac{w-\bar w}{2i}\right)=\bar z_1-0-0=\bar z_1.
\end{align*}
\begin{align*}
\frac{\partial\rho_Q}{\partial z_2}=\partial_{z_2}(z_1\bar z_1)-\partial_{z_2}(z_2\bar z_2)-\partial_{z_2}\left(\frac{w-\bar w}{2i}\right)=0-\bar z_2-0=-\bar z_2.
\end{align*}
\begin{align*}
\frac{\partial\rho_Q}{\partial w}=\partial_w(z_1\bar z_1)-\partial_w(z_2\bar z_2)-\partial_w\left(\frac{w-\bar w}{2i}\right)=0-0-\frac{1}{2i}=\frac{i}{2}.
\end{align*}
Differentiating these with respect to the antiholomorphic variables gives
\begin{align*}
\frac{\partial^2\rho_Q}{\partial z_1\partial\bar z_1}=\frac{\partial}{\partial\bar z_1}(\bar z_1)=1.
\end{align*}
\begin{align*}
\frac{\partial^2\rho_Q}{\partial z_2\partial\bar z_2}=\frac{\partial}{\partial\bar z_2}(-\bar z_2)=-1.
\end{align*}
\begin{align*}
\frac{\partial^2\rho_Q}{\partial w\partial\bar w}=\frac{\partial}{\partial\bar w}\left(\frac{i}{2}\right)=0.
\end{align*}
The remaining mixed second derivatives with distinct variables are $0$: $\bar z_1$ depends only on $\bar z_1$, $-\bar z_2$ depends only on $\bar z_2$, and $i/2$ is constant. Thus for
\begin{align*}
L=a_1\partial_{z_1}+a_2\partial_{z_2}+b\partial_w\in T_p^{1,0}M_Q,
\end{align*}
the Levi form is
\begin{align*}
\mathcal L_{\rho_Q,p}(L,\overline L)=1\cdot a_1\overline{a_1}+(-1)\cdot a_2\overline{a_2}+0\cdot b\overline b=|a_1|^2-|a_2|^2.
\end{align*}
The positive and negative values occur on actual complex tangent vectors. At $p=(p_1,p_2,p_w)\in M_Q$, the first derivatives above give
\begin{align*}
\frac{\partial\rho_Q}{\partial z_1}(p)=\bar p_1.
\end{align*}
\begin{align*}
\frac{\partial\rho_Q}{\partial z_2}(p)=-\bar p_2.
\end{align*}
\begin{align*}
\frac{\partial\rho_Q}{\partial w}(p)=\frac{i}{2}.
\end{align*}
Set
\begin{align*}
L_+=\partial_{z_1}+2i\bar p_1\partial_w.
\end{align*}
Then
\begin{align*}
L_+\rho_Q(p)=1\cdot\bar p_1+(2i\bar p_1)\cdot\frac{i}{2}.
\end{align*}
Since $(2i)(i/2)=i^2=-1$,
\begin{align*}
L_+\rho_Q(p)=\bar p_1-\bar p_1=0.
\end{align*}
Hence $L_+\in T_p^{1,0}M_Q$. In the formula $\mathcal L_{\rho_Q,p}(L,\overline L)=|a_1|^2-|a_2|^2$, this vector has $a_1=1$ and $a_2=0$, so
\begin{align*}
\mathcal L_{\rho_Q,p}(L_+,\overline{L_+})=|1|^2-|0|^2=1.
\end{align*}
Similarly, set
\begin{align*}
L_-=\partial_{z_2}-2i\bar p_2\partial_w.
\end{align*}
Then
\begin{align*}
L_-\rho_Q(p)=1\cdot(-\bar p_2)+(-2i\bar p_2)\cdot\frac{i}{2}.
\end{align*}
Since $(-2i)(i/2)=-i^2=1$,
\begin{align*}
L_-\rho_Q(p)=-\bar p_2+\bar p_2=0.
\end{align*}
Hence $L_-\in T_p^{1,0}M_Q$. In the same Levi-form formula this vector has $a_1=0$ and $a_2=1$, so
\begin{align*}
\mathcal L_{\rho_Q,p}(L_-,\overline{L_-})=|0|^2-|1|^2=-1.
\end{align*}
The sphere is the model with strictly positive Levi geometry, while this hyperquadric is the basic model where the Levi form has both positive and negative complex tangential directions.
[/example]