Sum of series $\sin \theta+\sin 2 \theta+\sin 3\theta+\dots$ I need to prove sum of series: $$\sin \theta+\sin 2 \theta+\sin 3\theta+\dots=\sum_{n=1}^\infty \sin n\theta$$
by using in the first place the complex numbers.

De Moivre's Theorem states that $e^{ix}=\cos{x} + i \sin{x}$, so $\sin{x} = \Im{(e^{ix})}$ (i.e. the imaginary part of the expression).
So,
$\sin{\theta} + \sin{2\theta} + \sin{3\theta} + \ldots = \Im{(e^{i\theta})} + \Im{(e^{2i\theta})} + \Im{(e^{3i\theta})} + \ldots = \Im{(e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \ldots)}$
Using the fact that taking the imaginary part is a linear operation. And the thing at the end is a geometric series with first term and ratio both equal to $e^{i\theta}$, which should be enough to get you the answer.
