If $f(x)$, $\phi_n(x) \in \mathbb{C}$, $\;\;-\infty<a<x<b<\infty$ and
$$\langle f, \phi_n\rangle = \int_a^b f(x)\overline{\phi_n(x)}\;dx,$$
then why does:
$$\overline{\langle f, \phi_n\rangle }\langle f, \phi_n\rangle = |\langle f, \phi_n\rangle|^2,$$
that is:
$$\overline{\langle f, \phi_n\rangle }\langle f, \phi_n\rangle = \int_a^b \overline{f(x)}\phi_n(x)\;dx\int_a^b f(x)\overline{\phi_n(x)}\;dx $$
$$ = \left| \int_a^b f(x)\overline{\phi_n(x)}\;dx \right|^2 = |\langle f, \phi_n\rangle|^2.$$
The function $\phi_n(x)$ belongs to an orthonormal set in $L^2(a,b)$, which is the space of square-integrable functions on $[a,b]$.
Thank you for any help =)
For any $z\in \mathbb C$, you have that $$z\overline z = |z|^2$$ because $z=a+bi,\overline z = a-bi$ and $$z\overline z = (a+bi)(a-bi) = a^2 - (bi)^2 = a^2 + b^2 = |z|^2.$$