Field Norm of Complex Number is not Norm

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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 by Counterexample:

Let $z_1 = z_2 = 1$.


\(\ds \map N {z_1 + z_2}\) \(=\) \(\ds \cmod {z_1 + z_2}^2\) Definition of $N$
\(\ds \) \(=\) \(\ds 2^2\)
\(\ds \) \(=\) \(\ds 4\)


\(\ds \map N {z_1} + \map N {z_2}\) \(=\) \(\ds \cmod {z_1}^2 + \cmod {z_2}^2\) Definition of $N$
\(\ds \) \(=\) \(\ds 1^2 + 1^2\)
\(\ds \) \(=\) \(\ds 2\)

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 $\C$.