Let $a, b, c, d$ be complex numbers, but such that
$b = \displaystyle \frac{a}{k}, d = \displaystyle \frac{c}{k}$ with $k$ real.
Moreover,
$$ab^* = \frac{|a|^2}{k} = cd^* = \frac{|c|^2}{k} = r$$
Where $b^*$ is the complex conjugate of $b$ and so on.
Is it possible to prove that
$$ad^* = cb^* = r$$
?
Thank you anyway!
No. Counterexample:
$$b=e^{i\phi_1}, d=e^{i\phi_2}\quad\Rightarrow\quad a=ke^{i\phi_1},c=ke^{i\phi_2}$$
gives
$$ad^*=ke^{i(\phi_1-\phi_2)}\quad\textrm{and}\quad cb^*=ke^{i(\phi_2-\phi_1)}$$
both of which are not real if $\phi_1-\phi_2\neq \pi n,\quad n\in\mathbb{Z}$.