Proving that $\sum^n_{k=1} e^{ik\theta}=\sum^n_{i=1}\cos k\theta +i\sum^n_{k=1}\sin k\theta$. Prove:
$$\sum^n_{k=1} e^{ik\theta}=\sum^n_{i=1}\cos k\theta +i\sum^n_{k=1}\sin k\theta$$
Thanks a lot!!
I tried:
With Euler's identity I can get $\sin x= \dfrac{e^{ix} - e^{-ix}}{2i}$ and the cosine too. But i'm lost trying to development it.

Given that
$$ e^{ik\theta} = \cos(k\theta) + \textbf{i} \sin(k\theta) $$
Then we get
$$ \sum_{k=1}^n e^{ik\theta} = \sum_{k=1}^n \Big( \cos(k\theta) + \textbf{i} \sin(k\theta) \Big) $$
which gives
$$ \sum_{k=1}^n e^{ik\theta} = \sum_{k=1}^n \cos(k\theta) + \textbf{i} \sum_{k=1}^n \sin(k\theta) $$
