Extension of $|\cdot|_\infty$ on $\mathbb R$ to $\mathbb C$ Let $|\cdot|$ be the usual absolute value on $\mathbb C$. My question is:


  Is the only extension of $|\cdot|$ on $\mathbb R$ to $\mathbb C$ $|\cdot|$ itself? 


I'm not sure about the uniqueness. I want to show any two extensions of $|\cdot|$ to $\mathbb C$ induce same topologies, which means that one is a positive power of the other (and so are equal).
Is it true any two absolute values on $\mathbb C$ induce the same topology (as $\mathbb C$ is a finite dimensional vector space over $\mathbb R$)?

Thank you!
Any absolute value on $\mathbf{C}$ restricting to the usual absolute value on $\mathbf{R}$ is in particular a norm on $\mathbf{C}$ as a two-dimensional real vector space. All norms on a finite-dimensional vector space induce the same topology. Thus your absolute value $|\cdot|$ on $\mathbf{C}$ restricts to a continuous homomorphism $S^1\to \mathbf{R}^+$, which must be trivial as $\mathbf{R}^+$ has no nontrivial compact subgroups, so you must have $|re^{i\theta}|=|r|$ for all $r$ and $\theta$.
Another elementary argument;

For roots of unity, it's clear they have norm 1. Now let $c$ be with irrational angle times pi, then take powers of it so that the angle is super close to $0$ mod $2 \cdot pi$, then use the triangle inequality in both directions its norm is close to norm $1$.

EDIT:
Since many people seem to be confused, I will add more details:

If $x$ is a root of unity of order $n$, $|x|^n=|x^n|=1$, proving $|x|=1$. In paricular this is true for $x=i$. Now let $x$ be on the unit circle, with angle $r \cdot 2pi$. Then it suffices to show that for arbitarily large $n$, $|x|^n$ is arbitarily close to $1$ (since if the norm was more than $1$, it'd explode, and if it was less, it would tend to $0$).

It is well known that if $a$ is irrational, then $na$ mod $1$ is gets arbitarily small. Applying this on our $a=r$, we find $x^n$ with $x^n-1=c+d\cdot i$ with $c$ close to $1$, and $d$ close to $0$. Then since we know $|i|=1$, subtracting $d\cdot i$ can only change the norm by $d$, so the norm is $c$ up to $+-d$, finishing the proof.