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 $zf(s-u)=A$, $zf(s+u)=\bar{B}$, $z\bar{z}f(2u)=C$. 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 $A=\langle\varphi_1,\varphi_2\rangle$, $\bar{B}=\langle\varphi_1,\varphi_3\rangle$, $C=\langle\varphi_2,\varphi_3\rangle$. Thus the following matrix is non-negative definite as the Gram matrix of our functions
$$
\pmatrix{\|\varphi_1\|^2&A&\bar{B}\\\bar{A}&\|\varphi_2\|^2&C\\B&\bar{C}&\|\varphi_3\|^2}.
$$
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):
$$
M=\pmatrix{1&A&\bar{B}\\\bar{A}&1&C\\B&\bar{C}&1},\,\det M>0.
$$
What we have to prove is that $\arg (1-A)(1-B)=\arg \frac{1-A}{1-\bar{B}}<\pi/2-\arg(1-C)$, or, in other words, $\Re (1-A)(1-B)(1-C)>0$ 

[we have $\arg(1-A),\arg(1-B),\arg(1-C)\in (-\pi/2,\pi/2)$, thus if $\arg(1-A)+\arg(1-B)\geq \pi/2-\arg(1-C)$ then $\theta:=\arg (1-A)+\arg (1-B)+\arg (1-C)\in [\pi/2,3\pi/2]$, i.e. $(1-A)(1-B)(1-C)=e^{i\theta}R$, $R>0$, $\Re (1-A)(1-B)(1-C)=R\cos \theta\leq 0$.] 

This statement must have some elegant proof, but I do not see it, so let me provide at least some proof. 

First of all, 
$$
\det M=1-|A|^2-|B|^2-|C|^2+2\Re(ABC)=(1-|A|^2)(1-|B|^2)-|C-\bar{A}\cdot \bar{B}|^2> 0.
$$

Thus $C=\bar{A}\cdot \bar{B}+w$, $|w|< R:=\sqrt{(1-|A|^2)(1-|B|^2)}$. Thus what may be said about $\Re (1-A)(1-B)(1-C)$ is that
$$
\Re (1-A)(1-B)(1-C)>\Re (1-A)(1-B)(1-\bar{A}\cdot \bar{B})-R\cdot |1-A|\cdot |1-B|.
$$
Now we have
\begin{align*}
X:=\Re (1-A)(1-B)(1-\bar{A}\cdot \bar{B})=\Re (1-A)(1-B)\left((1-\bar{A})+\bar{A}\cdot (1-\bar{B})\right)=\\
=|1-A|^2(1-\Re B)+|1-B|^2(\Re A-|A|^2)
\end{align*}
Analogously $X=|1-A|^2(\Re B-|B|^2)+|1-B|^2(1-\Re A)$. Taking half sum of two expressions for $X$ and applying AM-GM for two summands we get
$$
X=\frac{|1-A|^2(1-|B|^2)+|1-B|^2(1-|A|^2)}2\geq \sqrt{(1-|B|^2)(1-|A|^2)}\cdot |1-A|\cdot |1-B|
$$
as desired.