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I know I can write the frequency response of a Generalized linear-phase system as: $$H(e^{j\omega}) = A(\omega) e^{-j\left(\alpha \omega - \beta\right)}$$ where $A(\omega)$ is real.
I need to prove that $A(\omega)$ is real and I can't find anywhere the demonstration.
Let's assume that $A(\omega)$ is complex valued:
$$A(\omega)=B(\omega)e^{j\phi_A(\omega)}\tag{1}$$
where $B(\omega)$ is a (possibly bipolar) real-valued function, and $\phi_A(\omega)$ is a (real-valued) phase function.
The phase of
$$H(e^{j\omega})=A(\omega)e^{-j(\alpha\omega-\beta)}=B(\omega)e^{j\phi_A(\omega)}e^{-j(\alpha\omega-\beta)}=B(\omega)e^{j\phi(\omega)}\tag{2}$$
is then
$$\phi(\omega)=\phi_A(\omega)-(\alpha\omega-\beta)\tag{3}$$
(where I define the phase such that jumps by $\pi$ are taken care of by the sign of $B(\omega)$, not by $\phi(\omega)$). For the phase $\phi(\omega)$ to be linear, $\phi_A(\omega)$ must be constant or linear:
$$\phi_A(\omega)=a\omega+b\tag{4}$$
With ($4$) the total phase becomes
$$\phi(\omega)=(a-\alpha)\omega+(b-\beta)=-\alpha'\omega+\beta'\tag{5}$$
Combining $(2)$ and $(5)$ we get
$$H(e^{j\omega})=B(\omega)e^{j\phi(\omega)}=B(\omega)e^{-j(\alpha'\omega-\beta')}\tag{6}$$
which has the same form as the equation in your question. Note that in $(1)$ we have defined $B(\omega)$ to be real-valued.
This shows that $(6)$ (or, equivalently, the equation in your question) is the most general form of the frequency response of a linear phase filter. If $A(\omega)$ were chosen to be complex-valued, then $A(\omega)$ must have a linear phase (otherwise $H(e^{j\omega})$ wouldn't have a linear phase), and that linear phase can be absorbed by the constants $\alpha$ and $\beta$, such that the expression for the frequency response can always be reduced to the form $(6)$ with a real-valued function $B(\omega)$.
