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.
 A: *

*Note that the matrices are Hermitian, so it is enough to check if the eigenvalues $\lambda\geq 0$ are non-negative, or equivalently, 
$$\mu~:=~1-\lambda~\leq~ 1.$$

*The characteristic polynomials read
$$ p_1(\lambda) ~=~\mu^2-|c|^2, $$
and
$$  p_2(\lambda) ~=~\mu^3+|c|^2(2{\rm Re}(c) -3\mu), $$
respectively.

*Define polar decomposition $c~=~re^{i\theta}~\in~\mathbb{C}$. 

*The roots are 
$$ \mu~=~\pm |c|,$$
and
$$ \mu = 2 r \cos\frac{\theta+2\pi p}{3},\qquad p\in\mathbb{Z},$$
respectively.

*Hence,
$$ \Omega_1~=~\{c\in \mathbb{C} \mid |c|\leq 1\}~=~\bar{D}, $$
while
$$   \Omega_2~=~\{re^{i\theta}\in \mathbb{C} \mid \forall p\in \mathbb{Z}:~ 2 r \cos\frac{\theta+2\pi p}{3}\leq 1\}~\neq~\bar{D}. $$
It is straightforward to check that
$$  \{c\in \mathbb{C} \mid |c|\leq \frac{1}{2}\}~\subsetneq~\Omega_2~\subsetneq~\bar{D}. $$

A: 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.
