# Help with problem: Let $w$ a 15th-primitive-root of unity. Find all $n \in \Bbb N_{<0}$ such that $\sum_{i=0}^{n-1} w^{5i}=0$

we are starting to see complex numbers in my algebra class. So I have the following problem:

Let $w$ a 15th-primitive-root of unity. Find all $n \in \Bbb N_{<0}$ such that $\sum_{i=0}^{n-1} w^{5i}=0$

I thougth about seeing it as a geometric sum:

$\sum_{i=0}^{n-1} w^{5i}= \frac {(w^5)^n - 1}{w^5-1}$

But I don't know what should come next. As I'm just getting started with this subject I'm missing which properties from primitive roots should I use. Any suggestions?

• Note that $w^5$ is a primitive third root of unity, so the problem statement may be simplified a bit. Also, I advise you to thinking geometrically. – Arthur Nov 10 '16 at 21:37
• We can always write $n = 3m + k$ where $m,k$ are integers and $k \in \{0,1,2\}$. Write $w^{5n}$ in terms of $m$ and $k$ and use the definition of a primitive root of unity. – Winther Nov 10 '16 at 21:38

$$\sum_{i=0}^{n-1} w^{5i}=0\iff w^{5n}=1\stackrel{\text{Because primitive}}\iff 5n=15k\;,\;\;\text{for some}\;\;k\in\Bbb Z\iff$$
$$\iff n=3k\;,\;\;k\in\Bbb Z$$
Now choose the element as to belong to $\;\Bbb N_{<0}\;$ , which I presume means the negative integers, and that's all.