Show that $ \left|\sum _{S (\theta ) } z _k\right|= \left|\sum _{S (\theta ) } e ^{-i \theta } z _k \right|$. Let $z _k = |z _k |e ^{i \alpha _k}  $ and let $S(\theta ) $ be the set of all $k $ for which $\cos(\alpha _k - \theta) >0 $, $1 \le k \le n $.
Then $$ \left|\sum _{S (\theta ) } z _k\right|= \left|\sum _{S (\theta ) } e ^{-i \theta } z _k \right|$$
Suppose this is simple but I couldn't see it.
Thanks in advance!

Hint: $|e^{-i\theta}| = 1$. The identity has nothing to do with the definition of $S(\theta)$.
