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! 
 A: 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 $\theta$.
A: You don't need all of Ostrowski's theorem to give that any extension of the absolute value of $\mathbb{R}$ is indeed the one used to. Here is an elementary argument. Let $||_1$ and $||_2$ be extensions of $||_{\infty}$ to $\mathbb{C}$ , where $||_1$ is the familiar extension (i.e the norm). Consider then the function $$f:\mathbb{C} \rightarrow \mathbb{R}$$
given by $f(z) = |z|_2/|z|_1.$ Then we note that since for $z= a+bi$ we have that $$|a+bi|_2 \leq |a|+|b| \leq \sqrt{2} \sqrt{a^2+b^2} = \sqrt{2} |a+bi|_1$$ we have that $$|a+bi|_2/|a+bi|_1 \leq \sqrt{2}.$$ But actually, $f(\alpha) \leq 1$ for all $\alpha$ as well, since if not, since the norm is multiplicative, we'd have that for $n$ large enough that $f(\alpha^n)=f(\alpha)^n > \sqrt{2}$ which is a contradiction. Thus, we see that $|\alpha|_2 \leq |\alpha|_1.$ This shows that the two valuations give the same topology, so the restriction of them to $\mathbb{R}$ shows that they're equal.
A: Yes, it is true that the usual absolute value $|\cdot|_\infty$ on $\Bbb{R}$ extends uniquely to $\Bbb{C}$, but no, it is false that all absolute values on $\Bbb{C}$ (or even on $\Bbb{R}$, for that matter) induce the same topology.
Here's a counterexample: with a bit of work one can prove that every $p$-adic absolute value $|\cdot|_p$ on $\Bbb{Q}$ extends (albeit non-uniquely) to an absolute value $|\cdot|_p$ on $\Bbb{C}$. Then observe that $|\cdot|_p$ cannot be equivalent to $|\cdot|_\infty$ because $|n|_p < 1$ for every $n \in \Bbb{Z}_{>0}$: in particular, every $|\cdot|_p$-ball of radius $>1$ contains the positive integers, but you can always find a positive integer outside of a $|\cdot|$-ball.
On the other hand, by one of Ostrowski's theorems we know that up to equivalence there is exactly one Archimedean valuation $|\cdot|_\infty$ on $\Bbb{Q}$ (you can find the proof on PlanetMath). Now, recall that $\Bbb{R}$ can be constructed as the completion of $\Bbb{Q}$ with respect to this absolute value. By the uniqueness of completions, this means that $|\cdot|_\infty$ is the unique Archimedean absolute value on $\Bbb{R}$. Finally, note that $\Bbb{C} \supset \Bbb{R}$ is an algebraic extension of degree $2$ and apply the following

Theorem. If $K$ is a complete field with respect to an absolute value $|\cdot|$, then $|\cdot|$ can be extended in a unique way to any given algebraic extension $L \supset K$. Furthermore, if $L$ is finite then $L$ is complete with respect to $|\cdot|$.
Proof. See Neukirch's Algebraic Number Theory, Theorem II.4.8.

A: Here is another elementary argument that the standard absolute value on $\mathbb{C}$ is the only extension of the standard absolute value on $\mathbb{R}$:
Denote our absolute value by $| \cdot |$. Note that for any $z=a+bi\in \mathbb{C}$, we have $$|z| \cdot |\overline{z}| = |z \overline{z}| = |a^2 + b^2| = a^2 + b^2$$ Thus, we're done if we can prove $|z| = |\overline{z}|$. It suffices to prove this for $z\in S^1$ since any other complex number can be written as a real multiple of such $z$.
Note that $|i|^2 = |i^2| = |-1| = 1 \implies |i| = 1$. So, if $z_n = a_n + i b_n$ is a sequence in $\mathbb{C}$ with $a_n \to a$ and $b_n \to b$ in $\mathbb{R}$, we have $$|z_n -a-bi| \le |a_n - a| + |i| |b_n - b| = |a_n - a| + |b_n - b|$$ where the right hand side converges to $0$ as $n \to \infty$. Thus, $z_n \to a+bi$ in the metric induced by our absolute value.
Note that if $\omega$ is a root of unity, then $|\omega| = 1$ using the relation $\omega^k = 1$ for some $k$. Given $z\in S^1$, we can choose a sequence of roots of unity $\omega_n$ such that $\omega_n \to z$ in the standard absolute value on $\mathbb{C}$. But, by the above argument, this implies $\omega_n \to z$ in our absolute value. Hence, by continuity of the absolute value, it follows $|z| = 1$.
Finally, we have for $z\in S^1$ that $|\overline{z}| = |1/z| = 1/|z| = 1 = |z|$ so we are done.
A: 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.
