Sum of a finite complex series

Let
$$C=\cos\theta+\cos(\theta+ \frac{2\pi}{n})+ \cos(\theta+ \frac{4\pi}{n})+...+\cos(\theta+ \frac{(2n-2)\pi}{n})$$
and
$$S=\sin\theta+\sin(\theta+ \frac{2\pi}{n})+ \sin(\theta+ \frac{4\pi}{n})+...+\sin(\theta+ \frac{(2n-2)\pi}{n})$$
Show that $C+iS$ forms a geometric series and hence show that $C=0$ and $S=0$

Then $C+iS$ = $(\cos\theta+\cos(\theta+ \frac{2\pi}{n})+ \cos(\theta+ \frac{4\pi}{n}))+i(\sin\theta+\sin(\theta+ \frac{2\pi}{n})+ \sin(\theta+ \frac{4\pi}{n}))$
$$e^{i\theta}+exp (i(\theta+\frac{2\pi}{n}))+...+exp (i(\theta+\frac{(2n-2)\pi}{n}))$$

This is a far as I've got as I'm a little stuck where to go from here. Feel like I'm quite close to the answer though.

Answer

Yes, you are close indeed:
\begin{align}
&\exp(i\theta)+ \exp(i(\theta+2\pi/n)) +\ldots+ \exp(i(\theta+2\pi(n-1)/n))\\
&= \exp(i\theta) \sum_{k=0}^{n-1} \exp(2\pi ik/n) = \exp(i\theta) \sum_{k=0}^{n-1} \exp(2\pi i/n)^k\\

&= \exp(i\theta) \frac{1-\exp(2\pi i/n)^n}{1-\exp(2\pi i/n)} =
\exp(i\theta) \frac{1-\exp(2\pi i)}{1-\exp(2\pi i/n)} = 0,
\end{align}
and hence $C+iS = 0,$ implying that $C=S=0$.

EDIT: By request, we can avoid using the sigma notation as follows. Put $q=\exp(2\pi i/n)$ for clarity. Then
\begin{align}
&\exp(i\theta)+ \exp(i(\theta+2\pi/n)) +\ldots+ \exp(i(\theta+2\pi(n-1)/n))\\
&= \exp(i\theta) \big( 1+\exp(2\pi i/n) + \ldots + \exp(2\pi i(n-1)/n) \big)\\
&= \exp(i\theta) \big( 1+q + \ldots +q^{n-1} \big)

= \exp(i\theta) \frac{1-q^n}{1-q},
\end{align}
etc.