Positive semi definiteness of complex matrices [CSIR 2015] Consider the following subsets of the complex plane:
$$
\Omega_1=\left\{ C\in \mathbb{C} :  \begin{bmatrix}
        1  & C\\
        \overline C &1
        \end{bmatrix}\text{ is positive semi-definite} \right\}
$$
$$
\Omega_2=\left\{ C\in \mathbb{C} :  \begin{bmatrix}
        1  & C &C\\
        \overline C &1 &C\\
        \overline C & \overline C & 1
        \end{bmatrix}\text{ is positive semi-definite} \right\}
$$
Let $\overline {\mathbb{D}}= \{z \in \mathbb{C} :\vert z \vert \le 1\}$.Then


*

*$\Omega_1=\overline {\mathbb{D}},\Omega_2=\overline {\mathbb{D}}$.

*$\Omega_1\neq\overline {\mathbb{D}},\Omega_2=\overline {\mathbb{D}}$.

*$\Omega_1=\overline {\mathbb{D}},\Omega_2\neq\overline {\mathbb{D}}$.

*$\Omega_1\neq\overline {\mathbb{D}},\Omega_2\neq\overline {\mathbb{D}}$.
$\Omega_1$ is positive semidefinite implies that its determinant $1-\vert z \vert^2 \ge 0$ impliying that $\vert z \vert \le 1$.Thus $\Omega_1=\overline {\mathbb{D}}$.
As far as $\Omega_2$  is concerned it being positive semi definite would suggest that the determinant of every principal submatrix of $\Omega_2$ is non-negative.
The two diagonal ones which are basically same as $\Omega_1$ give 
$z \in \mathbb{C} :\vert z \vert \le 1$.But the other submatrices result in the relation that $z^2-z$  is real and non-negative.This definitely means:
$\Omega_2\neq \overline {\mathbb{D}}$. Right?
That would mean 3 is the right answer? Am I correct?

*

*Indeed, $\Omega_1=\overline{\mathbb D}$ because the application of Sylvetser's criterion to this case yields that $A(z)=\begin{bmatrix}1&z\\\overline z&1\end{bmatrix}$ is positive semi-definite if and only if $\det A(z)=1-|z|^2\ge 0$.

*It holds $\Omega_2\neq\overline{\mathbb D}$, and, though one could use Sylvester's criterion in a more subtle way, I think the easiest way to show it is the following.

Let $z\in\mathbb C$. $\det \begin{bmatrix}1&z&z\\ \overline z &1&z\\ \overline z &\overline z &1\end{bmatrix}=1-3|z|^2+2|z|^2\mathfrak{Re}\,z \implies-1\in \overline{\mathbb{D}}\setminus\Omega_2$

