Consider the set of complex numbers with absolute value 1. Some points on this set are n-th roots of unity. But there are points that aren't. Is there a closed-form expression for even one of those points which are not n-th roots of unity?
A point on the unit circle that is not an n-th root of unity.
complex numbersroots-of-unity
Related Solutions
From Vieta's formula, $\omega_1+\omega_2+1 = 0$, where $\omega_1,\omega_2$ are two distinct complex roots of unity.
You already knew $\omega_1 = \omega_2^2$ and $\omega_2=\omega_1^2$. Either way you get $\omega+\omega^2+1 = 0$, and multiply both sides by $\omega^4$ to get the result.
Since a good proof of this on MathOverflow has already been linked in the comments, I will provide an alternate and original proof.
Firstly, note that we may write $f(x)=x^r g(x)$ for some monic $g(x)\in\mathbb{Z}[x]$ with no roots at $x=0$.
Let $n=\deg(g)$, and let $\alpha_1,\cdots,\alpha_n$ be the roots of $g(x)$. Define the sequence
$$A_k=\alpha_1^k+\alpha_2^k+\cdots+\alpha_n^k$$
It is a fact that the fixed field of the Galois group of a splitting field of a polynomial in $\mathbb{Q}[x]$ is $\mathbb{Q}$, so since $A_k$ is fixed by every automorphism in $\text{Gal}(\mathbb{Q}(\alpha_1,\cdots,\alpha_n)/\mathbb{Q})$, then $A_k\in\mathbb{Q}$. Furthermore, since $\alpha_i$ are algebraic integers, then so is $A_k$, and since $A_k$ is also rational, then it must be an integer. (note that this fact can also be proven by combining of Newton's identities and Vieta's formulas)
Now, note that since $|\alpha_i|\leq 1$, then
$$|A_k|\leq |\alpha_1|^k+|\alpha_2|^k+\cdots+|\alpha_n|^k\leq n$$
Combined with the fact that $A_k$ is an integer, this means that $A_k$ only takes on finitely many values. Writing, $g(x)=x^n-b_{n-1}x^{n-1}-\cdots-b_1x-b_0$, we know that $A_k$ satisfies the linear recurrence
$$A_k=b_{n-1}A_{k-1}+\cdots+b_1A_{k-n+1}+b_0A_{k-n}$$
Since $A_k$ satisfies a linear recurrance and takes on only finitely many values, then $A_k$ is periodic with some period $m$. This means that $A_{mk}=A_0=n$ for all $k\geq 0$. Therefore, for $|x|<1$, we have that \begin{equation} \begin{split} \frac{n}{1-x}&=\sum_{k=0}^\infty nx^k\\ &=\sum_{k=0}^\infty A_{mk}x^k\\ &=\sum_{k=0}^\infty \sum_{i=1}^n \alpha_i^{mk}x^k\\ &=\sum_{i=1}^n\sum_{k=0}^\infty\alpha_i^{mk}x^k\\ &=\sum_{i=1}^n\frac{1}{1-\alpha_i^mx}\\ \end{split} \end{equation} and therefore, multiplying both sides by $1-x$, and taking the limit as $x\rightarrow 1^-$, we have that \begin{equation} n=\lim_{x\rightarrow 1^-}\sum_{i=1}^n\frac{1-x}{1-\alpha_i^mx}\\ \end{equation} Notice that if $\alpha_i$ is an $m$-th root of unity, then $\lim\limits_{x\rightarrow 1^-}\frac{1-x}{1-\alpha_i^mx}=1$ , and otherwise $\lim\limits_{x\rightarrow 1^-}\frac{1-x}{1-\alpha_i^mx}=0$. Combining this with the formula above tells us that $g(x)$ has precisely $n$ roots $\alpha_i$ which are $m$-th roots of unity (this is every root of $g(x)$).
Equivalently, every root of $f(x)=x^r g(x)$ is either $0$ or a root of unity, as desired.
Best Answer
Since $n^\textrm{th}$ roots of unity are the values of the form $\zeta_{n,m} = e^{2\pi i m/ n}$ for integers $m,n$ where (if considering primitive roots) $\gcd(m,n) = 1$. So there are only countably many $n^\textrm{th}$ roots of unity, while the unit circle is an uncountable set.
So the question becomes can we represent any of those others in a closed form easily. And we can, we can represent any point on the unit circle as $e^{i \theta}$ so for any $\theta \neq 2\pi \frac{m}{n}$, then $e^{i\theta}$ will be a point on the unit circle that isn't a root of unity. So for example, we have $e^{ni}$ is on the unit circle but isn't a root of unity for any integer $n$.