I know the formula to find the order of an element in the form $m^k$ in a group with order $n$ is $n/\gcd(k,n)$. However I don't know how to apply that formula when it comes to complex elements.
For example, if they ask me to find the order of $e^{2\pi i \frac{8}{12}}$ in $\mathbb{C}^*$, how would I proceed? How would I be able to find the generators? Also, I thought $\mathbb{C}^*$ was a cyclic group of order infinite, so how can an element have a finite order?
Solution :
Best Answer
Our question is, given an element $z\in \mathbb{C}^*$, what is the lowest natural number $n$ satisfying $z^n=1$?
Since we require that $z^n=1$, it only makes sense that $z$ is a root of unity; i.e. $z=e^{q2\pi i}$ for some $q\in\mathbb{Q}$. However, if $q=\frac{a}{b}$, we can't say that the order of $z$ is $\frac{b}{a}$ unless $\frac{b}{a}$ is an integer (that contradicts the definition of "order"). Thus, if $\frac{a}{b}$ is a reduced fraction, then the order must be $b$, to allow $z^b=(e^{\frac{a}{b}2\pi i})^b=e^{a2\pi i} = 1$.