Field Norm of Complex Number is not Norm

Theorem
Let $\C$ denote the set of complex numbers.

Let $N: \C \to \R_{\ge 0}$ denote the field norm on complex numbers:
 * $\forall z \in \C: \map N z = \cmod z^2$

where $\cmod z$ denotes the complex modulus of $z$.

Then $N$ is not a norm on $\C$.

Proof
Proof by Counterexample:

Let $z_1 = z_2 = 1$.

Then:

But:

So we have that for these instances of $z_1$ and $z_2$:
 * $\map N {z_1 + z_2} > \map N {z_1} + \map N {z_2}$

and so the triangle inequality is not satisfied.

Hence by definition $N$ is not a norm on $\C$.