consider the following subsets of complex plane $$\Omega_1=\left\{c\in\Bbb C:\begin{bmatrix}1&c\\\bar c&1\\ \end{bmatrix}\text{ is non-negative definite } \right\} $$
$$\Omega_2= \left\{c\in\Bbb C:
        \begin{bmatrix}
        1 & c & c \\
        \bar c & 1 & c \\
        \bar c & \bar c & 1 \\
        \end{bmatrix} \text{ is non-negative definite } \right\}$$
Let $$\bar D=\{z\in\Bbb C:|z|\le1\}$$
Then 1) $\Omega_1=\bar D,\Omega_2=\bar D$
2). $\Omega_1\neq\bar D, \Omega_2=\bar D$
3). $\Omega_1=\bar D, \Omega_2\neq\bar D$
4). $\Omega_1\neq\bar D, \Omega_2\neq\bar D$
How to proceed? Thank you.

The $2\times 2$ matrix has eigenvalues $1 \pm |c|$, so it is positive semidefinite if and only if $|c|\leq 1$.
For the $3\times 3$ matrix, observe that the determinant is $-4$ when $c=-1$, so the matrix is not positive semidefinite in this case.
