I want to show that the usual norm on $\mathbb{C}$
$$|z| = \sqrt{z \bar z}$$
is the unique multiplicative norm including the euclidian norm on $\mathbb{R}$:
$$|x| = \sqrt{x^2}$$
My Idea: Let $|\cdot|$ be any norm on $\mathbb{C}$ admitting the conditions above. Let $z \in \mathbb{C}$. Then we have:
$$|z| = |\sqrt{z \bar z}e^{i \arg{z}}| = \underbrace{|\sqrt{z \bar z}|}_{\in \mathbb{R}}\cdot|e^{i \arg z}|$$
At this point the proof would be complete if I'd be able to show that any element on the unit circle has to have norm 1. But I don't know how to proceed, or whether there is a more elegant way to proof the statement.
Suppose the norm of some $e^{i\theta}$ is strictly smaller than $1$. Then, since $e^{i\theta}\neq 1$ and the norm is multiplicative, the norms of the sequence $e^{in\theta}$ tend to zero. But the unit circle is compact, so we get a limit point on the circle with norm zero, contrary to the definition of a norm. Therefore no point on the circle can have norm smaller than $1$. Suppose now some point $e^{i\alpha}$ has norm larger than $1$. Then since $e^{-i\alpha}e^{i\alpha}=1$, and the norm of $1$ is $1$, we get a point on the unit circle whose norm is strictly smaller than $1$, which as we have seen, is a contradiction. Therefore all points on the unit circle must have norm equal to $1$, and therefore, since every complex number is of the form $re^{i\theta}$ for some $r\geq 0$, we recover the usual absolute value of complex numbers.