let's prove that
$$
\arg \frac{1-zf(s-u)}{1-zf(s+u)}< \pi/2- \arg(1+z\bar{z}f(2u)).
$$
Then summing this up with an analogous inequality 
$$
\arg \frac{1-zf(t+u)}{1-zf(t-u)}< \pi/2- \arg(1+z\bar{z}f(-2u))
$$
we get what we need. 

Denote $zf(s-u)=A$, $zf(s+u)=B$, $z\bar{z}f(2u)=C$. Then for functions $\varphi_1(x)=e^{isx}$, $\varphi_2(x)=\bar{z}e^{iux}$, $\varphi_3(x)=\bar{z}e^{-iux}$ in $(L^2,\mu)$ we have $A=\langle\varphi_1,\varphi_2\rangle$, $B=\langle\varphi_1,\varphi_3\rangle$, $C=\langle\varphi_2,\varphi_3\rangle$. Thus the following matrix is non-negative definite as the Gram matrix of our functions
$$
\pmatrix{\|\varphi_1\|^2&A&B\\\bar{A}&\|\varphi_2\|^2&C\\\bar{B}&\bar{C}&\|\varphi_3\|^2}.
$$
We may increase diagonal elements upto 1. Of course, matrix remains non-negative definite and, moreover, becomes positive definite (since $|z|<1$, we strictly increase two diagonal elements of our matrix, and its determinant becomes strictly positive):
$$
M=\pmatrix{1&A&B\\\bar{A}&1&C\\\bar{B}&\bar{C}&1},\,\det M>0.
$$
What we have to prove is that $\arg (1-A)(1-\bar{B})=\arg \frac{1-A}{1-B}<\pi/2-\arg(1+C)$, or, in other words, $\Re (1-A)(1-\bar{B})(1+C)>0$ [we have $\arg(1-A),\arg(1-\bar{B}),\arg(1-C)\in (-\pi/2,\pi/2)$, thus if $\arg(1-A)+\arg(1-\bar{B})\geq \pi/2-\arg(1-C)$ we have $\arg (1-A)+\arg (1-\bar{B})+\arg (1-C)\in [\pi/2,3\pi/2]$, i.e. modulo $2\pi$ we have $\arg T\in [\pi/2,3\pi/2]$, where $T=(1-A)(1-\bar{B})(1-C)$, i.e. $\Re T\leq 0$.] 
Well, if I am not mistaken we have
$$
2\Re (1-A)(1-\bar{B})(1+C)=|-C+\bar{A}+B-1|^2+\det M>0
$$
as desired.