Let $K$ be an algebraic number field such that $[K:\mathbb{Q}]\geq 2$. Let $\sigma:K\rightarrow \mathbb{C}$ be a complex embedding of $K$ ( $\sigma(K)\nsubseteq\mathbb{R}$). Let $\tau:K\rightarrow \mathbb{C}$ be a real embedding of $K$ ( $\tau(K)\subseteq\mathbb{R}$). For each $x\in K$, we can define two absolute values $${\|x\|}_1=|\sigma(x)|\text{ and }{\|x\|}_2=|{\tau}(x)|$$
How to prove that ${\|\text{ }\|}_1\text{ and }{\|\text{ }\|}_2$ are not equivalent ?
My idea:
${\|\text{ }\|}_1\text{ and }{\|\text{ }\|}_2$ will be equivalent iff $\exists$ $a>0$ such that ${\|x\|}_{1}^{a}={\|x\|}_{2}$ for all $x\in K$. In particular, if we take $x\in\mathbb{Q}$, we get $a=1$. So, ${\|x\|}_{1}={\|x\|}_{2}$ for all $x\in K$, or equivalently, $|\sigma(x)|=|\tau(x)|$ for all $x\in K$. I am trying to find a suitable $x$ such that $|\sigma(x)|\neq|\tau(x)|$.
Any ideas ?
Pick an element $ x \in K $ which is mapped to a non-real by the embedding $ \sigma $, and let $ \sigma(x) = a + bi $, $ \tau(x) = c $. By considering $ -x $ if necessary, we may assume $ c > 0 $. If $ |\sigma(x)| \neq |\tau(x)| $ then we are done, otherwise we have $ a^2 + b^2 = c^2 $. We then have $ \sigma(x+1) = (a+1) + bi $ and $ \tau(x+1) = c+1 $, and
$$ |\sigma(x+1)|^2 = (a+1)^2 + b^2 = c^2 + 2a + 1 < c^2 + 2c + 1 = |\tau(x+1)|^2 $$
where the inequality is strict since $ b \neq 0 $.