# Tagged Questions

numbers $z$ such that $z^n=1$ for some natural number $n$; here usually $z$ is in $\mathbb C$ or some other field

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### A spiralling sequence based on integer divisors. Has anyone noticed this before?

Firstly, please excuse the informal style of my explanation, as I am not a mathematician, although I am aware that this can be explained in more formal terms. I have mapped integers to points on a ...
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### Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

Using $\text{n}^{\text{th}}$ root of unity $$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$ Prove that $$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
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### Drunkard's walk on the $n^{th}$ roots of unity.

Fix an integer $n\geq 2$. Suppose we start at the origin in the complex plane, and on each step we choose an $n^{th}$ root of unity at random, and go $1$ unit distance in that direction. Let $X_N$ ...
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### Is $\sqrt 7$ the sum of roots of unity?

Let $a_n$ and $b_n$ be 2 sequences of $n$ rationals. Is it possible that $\sqrt 7 = \sum_{m=1}^{n} a_m (-1)^{b_m}$ ? Is it possible that $\sqrt{17}$ = $\sum_{m=1}^{n} a_m (-1)^{b_m}$ ? How to ...
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### A little more on $\sqrt[3]{\cos\bigl(\tfrac{2\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{4\pi}7\bigr)}+\sqrt[3]{\cos\bigl(\tfrac{8\pi}7\bigr)}$

Using a special case of an identity by Ramanujan, we find that given the roots $x_i$ of $$x^3 + x^2 - (3 n^2 + n)x + n^3=0\tag1$$ which, since its discriminant is negative, always has three real ...
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### Roots of unity in non-Abelian groups: when do they form subgroups?

I haven't studied group theory in earnest beyond first courses, so my notation may be nonstandard and my question may be a 'standard fact', so bear with me: Consider a group $G$, and for each natural ...
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### Why is $\sqrt [n] 1$ not an expression “in radicals” of a root of unity?
$\sum (-1)^n b_n$ is representative of an alternating series. We look at whether $\sum b_n$ converges and if $b_{n+1}<b_n$ $\forall n\in \mathbb{Z}$. What if our alternating series is of the form \$...