Field Norm of Complex Number is Positive Definite
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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 positive definite on $\C$.
Proof
First it is shown that $\map N z = 0 \iff z = 0$.
| \(\ds z\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0 + 0 i\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map N z\) | \(=\) | \(\ds 0^2 + 0^2\) | Definition of $N$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
Let $z = x + i y$.
| \(\ds \map N z\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \map N {x + i y}\) | \(=\) | \(\ds 0\) | Definition of $z$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds x^2 + y^2\) | \(=\) | \(\ds 0\) | Definition of $N$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds 0\) | Square of Real Number is Non-Negative | ||||||||||
| \(\ds b\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds 0\) | Definition of $z$ |
Then we have:
| \(\ds \map N z\) | \(=\) | \(\ds \map N {x + i y}\) | Definition of $z$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds x^2 + y^2\) | Definition of $N$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | Square of Real Number is Non-Negative |
Hence the result by definition of positive definite.
$\blacksquare$