Complex Analysis Geometric Series The question is:
Let $n$ be a positive integer, let $h$ be a positive integer not divisible by $n$, and let 
$$ w = \cos\left(\frac{2\pi}{n}\right) + i \sin\left(\frac{2\pi}{n}\right) $$
Show that
$$ 1 + w^h + w^{2h} + w^{3h} + \dots + w^{(n-1)h} = 0$$
I believe I do something with a geometric series, but I am not sure where to begin.

Note that
$$w = \cos\left(\frac{2\pi}{n}\right) + i \sin\left(\frac{2\pi}{n}\right) = e^{\frac{2\pi i}{n}}.$$
Use the geometric series
$$\sum_{k=0}^{n-1} z^k = \frac{z^n-1}{z-1}.$$
Your case delivers
$$\sum_{k=0}^{n-1} \left( e^{\frac{2\pi i}{n}} \right)^{kh}=\sum_{k=0}^{n-1} \left( \left( e^{\frac{2\pi i}{n}} \right)^{h} \right)^k =\sum_{k=0}^{n-1} \left( e^{\frac{2\pi i h}{n}} \right)^k = \frac{\left( e^{\frac{2\pi i h}{n}} \right)^n -1}{e^{\frac{2\pi ih}{n}} -1} = \frac{e^{2\pi i h}-1}{e^{\frac{2\pi i h}{n}}-1} = \frac{1- 1}{e^{\frac{2\pi i}{n}}-1} = 0$$
as $h \in \mathbb{N}$.
