Given an arbitrary matrix $A=\left[a_{i,j}\right]\in\mathbb{C}^{2\times2} $
Now define the set $$ \mathcal K = \left\lbrace z\in\mathbb C : \left|z-a_{1,1}\right|\cdot\left|z-a_{2,2}\right|=\left|a_{1,2}\right|\cdot\left|a_{2,1}\right| \right\rbrace$$
And the set of so-called equiradial matrices
$$\omega(A) = \left\lbrace B\in\mathbb C^{n\times n} : b_{1,1}=a_{1,1}, b_{2,2}=a_{2,2}, \left|b_{2,1}\right|=\left|a_{2,1}\right|, \left|b_{1,2}\right|=\left|a_{1,2}\right| \right\rbrace$$
Now I want to show, that the set of all spektra of those matrices $B$ is equal to $\mathcal K$
Necessarily $B$ is of the form $$B = \begin{pmatrix} a_{1,1} & \left|a_{1,2}\right|\exp{i\psi_1} \\ \left|a_{2,1}\right|\exp{i\psi_2} & a_{2,2} \end{pmatrix}$$ where $\psi_1$, $\psi_2$ are some real numbers.
Now fix one of those matrices $B$ and calculate the eigenvalues gives, putting them into the defining equality in $\mathcal K$ demonstrates, that they really belong to $\mathcal K$
But I am stuck on the other direction. Now I want to fix any $z \in \mathcal K$ and show that there is a corresponding matrix $B$ in $\omega(A)$ that has this $z$ as an eigenvalue. It's quite obvious that there must exist such an $B$ but how to chose $\psi_1$ and $\psi_2$ depending on $z$?