Field Norm of Complex Number is not Norm
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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
Let $z_1 = z_2 = 1$.
Then:
\(\ds \map N {z_1 + z_2}\) | \(=\) | \(\ds \cmod {z_1 + z_2}^2\) | Definition of $N$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4\) |
But:
\(\ds \map N {z_1} + \map N {z_2}\) | \(=\) | \(\ds \cmod {z_1}^2 + \cmod {z_2}^2\) | 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 $\C$.
$\blacksquare$