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.
 A: 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.
A: The general term is 
$$e^{i\theta+i2k\pi/n}=e^{i\theta}e^{i2k\pi/n}=e^{i\theta}(e^{i2\pi/n})^k.$$
You can factor out $e^{i\theta}$ and a geometric series of common ratio $e^{i2\pi/n}$ remains. As the index $k$ runs from $0$ to $n-1$, the numerator of the summation will be
$$e^{i2n\pi/n}-1$$ which is obviously $0$.
