# Field Norm of Complex Number is Positive Definite

## 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.

$\blacksquare$