A particular solution for $y''+a_1y' + a_2y = Ae^{i\omega x}$ I'm trying to show that for the equation $y''+a_1y' + a_2y = Ae^{i\omega x}$, there is a solution of $\phi(x)$ of the form $$\phi(x)=\frac{A}{\vert p(i\omega)\vert}e^{i(\omega x - \alpha)}$$
where $ p(i\omega)=\vert p(i\omega)\vert e^{ix}$, and $p(i\omega) \neq 0$. The last statement is confusing and I'm unsure how to interpret it. Why would $p(i\omega)=\vert p(i\omega)\vert e^{ix}$? I think that $$p(i \omega)=\omega^2+a_1i\omega+a_2$$
which is a complex number with $Re = \omega ^2+a_2$, and $Im = a_1\omega$. It does not seem like we will ever end up to such an identity, with the standard definition of complex norm. To be precise, we have that $$\vert p(i\omega)\vert=\sqrt{(\omega^2 +a_2)^2 + (a_1\omega)^2}$$
As for validating the solution, one would compute $L(\frac{A}{\vert p(i\omega)\vert}e^{i(\omega x - \alpha)})$, take the term with the norm as a common factor. I imagine the norm is cancelled at some point. What am I missing here?

I'm sure $p(i\omega)=\vert p(i\omega)\vert e^{ix}$ should be $p(i\omega)=\vert p(i\omega)\vert e^{i\alpha}$
The form $e^{i\omega x}$ is preserved by derivatives so for the particular solution try:
$$y_p = K e^{i\omega x} \tag{1}$$
into
$$ y''+a_1y' + a_2y = Ae^{i\omega x} \tag{2}$$
factor out $e^{i\omega x}$
$$-K {\omega}^2 + iKa_1\omega + Ka_2 = A \tag{3}$$
$$K = \frac{A}{-{\omega}^2 + ia_1\omega + a_2} \tag{4} $$
$$ y_p = \frac{Ae^{i\omega x}}{-{\omega}^2 + ia_1\omega + a_2} \tag{5} $$
The characteristic equation of this DE is:
$$p(z) = z^2 + a_1z + a_2 \tag{6}$$
So
$$p(i\omega) = -{\omega}^2 + ia_1\omega + a_2 \tag{7}$$
In $re^{i\theta}$form:
$$p(i\omega) = \sqrt{(a_2-{\omega}^2)^2 + (a_1\omega)^2 } \,
 e^{i \arctan{\left(\frac{a_1\omega}{a_2-{\omega}^2}\right)}} \tag{8}$$
Substitute $(8)$ into $(5)$
$$ y_p = \frac{Ae^{i\omega x}}{\sqrt{(a_2-{\omega}^2)^2 + (a_1\omega)^2 } \,
 e^{i \arctan{\left(\frac{a_1\omega}{a_2-{\omega}^2}\right)}}} \tag{9} $$
$$ y_p = \frac{Ae^{i(\omega x - \alpha)}}{|p(i\omega)|} \tag{10} $$
Where
$$\alpha = arg(p(i\omega)) = \arctan{\left(\frac{a_1\omega}{a_2-{\omega}^2}\right)}\tag{11}$$
