sum of the complex e-function i have to show that, for $y \in \mathbb{R}$ with $ e(y)=e^{2\pi iy}$ that
$\left | \sum_{n=0}^{N-1} e(\mathbf{h \cdot t_n}) \right| = \sum_{n=0}^{N-1}e(\mathbf{h \cdot t_n}- \theta) \Longrightarrow \left | \sum_{n=0}^{N-1} e(\mathbf{h \cdot t_n}) \right| = \sum_{n=0}^{N-1}cos(2\pi (\mathbf{h}\cdot t_n -\theta)) $ 
$\theta \in \mathbb{R} , \mathbf{t_n} \in [0,1)^s,\mathbf{h} \in \mathbb{Z^s}$ and $ N,s \in \mathbb{N} $
$ \mathbf{h \cdot t_n}= \langle\mathbf{h} \vert\mathbf{t_n} \rangle$ is the inner product.
I have tried to split the e function and also tried to replace the e function with $e(\alpha i) = cos(\alpha)+i \sin(\alpha)$,  but it didn't work.
please help me 

The sum $\sum\limits_{n = 0}^{N-1}e(\mathbf{h}\cdot \mathbf{t}_n - \theta)$ is a real number (in fact, nonnegative) since it is the modulus of $\sum\limits_{n = 0}^{N-1} e(\mathbf{h}\cdot \mathbf{t}_n)$. So the sum equals
$$\operatorname{Re}\left\{\sum_{n = 0}^{N-1} e(\mathbf{h}\cdot \mathbf{t}_n - \theta)\right\} = \sum_{n = 0}^{N-1} \operatorname{Re} e(\mathbf{h}\cdot \mathbf{t}_n - \theta) = \sum_{n = 0}^{N-1}\cos[2\pi(\mathbf{h}\cdot \mathbf{t}_n - \theta)]$$
The result now follows.
