TITLE: What is the image of $\zeta_3$ under the non-identity embedding of $\mathbb{Q}(\zeta_3)$ in $\mathbb{C}$?
QUESTION [5 upvotes]: What is the image of $\zeta_3$ under the non-identity embedding of $\mathbb{Q}(\zeta_3)$ in $\mathbb{C}$?

REPLY [3 votes]: If you define an algebraic extension of $\Bbb Q$ as $K={\Bbb Q}[x]/(f(x))$ where $f(x)$ is an irreducible polynomial of degree $d$, you obtain the $d$ embeddings of $K$ by sending $x\mapsto\alpha$ where $\alpha$ is a complex root of $f(x)$ (they always exist, thanks to Gauss).
In the case of ${\Bbb Q}(\zeta_3)$, this is just ${\Bbb Q}[x]/(x^2-x+1)$ because $x^2-x+1$ is the quadratic irreducible factor of $x^3-1$. Thus you obtain two complex embeddings sending
$$
x=\zeta_3\mapsto\alpha^\pm=\frac12(1\pm\sqrt{-3}).
$$
If you decide to identify $\zeta_3=\alpha^{+}=(1+\sqrt{-3})/2$, a moment of thought (in the form of a short computation) will convince you that the "other" root, i.e. the other embedding, $\alpha^{-}$ is just $\zeta_3^2$. Mind that the two roots (and so the two embeddings) are switched by complex conjugation.
The identities $\zeta_3^2=\overline{\zeta}_3=\zeta_3^{-1}$ are also clear observing that $1$, $\alpha^{+}$ and $\alpha^{-}$ are the three complex roots of $x^3-1$, so that their product must be $1$.
Adrian's answer gives the generalization to the $p$-th cyclotomic field