How to show that the only absolute value on the field of complex numbers is the standard absolute value How do I show that the only absolute value on the field of complex numbers $\mathbb{C}$, whose restriction to $\mathbb{R}$ is the standard absolute value, i.e., $|z|=\sqrt{z\bar{z}}$?
I know that in finite-dimensional vector spaces all norms are equivalent but how can I use it here? Can someone provide me a proof using the standard definition of equivalent absolute values on a field (number field)?
 A: Let $|-|_1$ be an absolute value on $\mathbb C$ such that $|z|_1 = |z|$ holds for all $z$ real.  You want to show that this remains true for $z$ complex.
By the equivalence of norms on finite dimensional vector spaces, there exist positive constants $C$ and $D$ such that $C|z| \leq |z|_1 \leq D |z|$ for all $z \in \mathbb C$.  It follows from here that the topology on $\mathbb C$ coming from the norm metric with respect to $|-|_1$ is the ordinary Euclidean topology, and in particular, the map $\mathbb C \rightarrow [0,\infty), z \mapsto |z|_1$ is continuous when $\mathbb C$ is given the Euclidean topology.
If $w$ is any root of unity (that is, if there exists a positive integer $n$ such that $w^n = 1$), the equation $w^n = 1$ implies immediately that $|w|_1^n = |w^n|_1 = |1|_1 = 1$, and hence $|w|_1 = 1$.  Since the roots of unity are dense in the unit circle, the continuity of the function $z \mapsto |z|_1$ implies that $|w|_1 = 1$ for all complex numbers $w$ on the unit circle.
Now, let $z$ be any complex number.  Since there exists a complex number $w$ on the unit circle such that $wz$ is real, we get
$$|z|_1 = |wz|_1 = |wz| = |z|.$$
