Show that $ \left|\sum _{S (\theta ) } z _k\right|= \left|\sum _{S (\theta ) } e ^{-i \theta } z _k \right|$.

Question

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!

Answer 1

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

Answer 2

In the RHS just take out the factor $e^{-i \theta}$ and simplify so that it's magnitude is 1 and it is equal to LHS.

Answer 3

It seems that $\theta$ is not depending on the choice of the summation index set $S(\theta)$, it is constant regarding to the summation process and thus can be pulled out: $$ \left| \sum_{S(\theta)} e^{-i\theta} z_k \right| = \left| e^{-i\theta} \sum_{S(\theta)} z_k \right| = \left| e^{-i\theta}\right| \left| \sum_{S(\theta)} z_k \right| = \left| \sum_{S(\theta)} z_k \right| $$