$\mu_5=\{\xi \in \mathbb{C}| f(\xi)=0\}$,is $\mu_5\subset\mathbb{C}/\{0\}$ a finite subgroup?

Question

Consider $f(x)\in \mathbb{Z[x]}$ such that $f(x)=x^5-1$. Now let $\mu_5=\{\xi \in \mathbb{C}| f(\xi)=0\}$,is $\mu_5\subset\mathbb{C}/\{0\}$ a finite subgroup? I considered the factorization of $f(x)$ whose elements are in $\mu_5$ but i have problems in showing that $\mu_5$ is closed for inverse.

Answer 1

$\xi^5-1=0\implies\frac{1}{\xi^5}(\xi^5-1)=0\implies(1-\frac{1}{\xi^5})=0\implies\left( \frac{1}{\xi}\right)^5-1=0\implies\frac{1}{\xi}\in \mu_{5}$

Answer 2

Hint: A polynomial of degree $n$ has $n$ zeros in its splitting field. Here $\Bbb C$ is algebraically closed and the polynomial $f(x)=x^5-1\in{\Bbb Z}[x]$ has all its 5 roots in $\Bbb C$. These roots are the fifth roots of unity, which also form a cyclic group of order 5.