The triangle inequality and the non degeneracy properties are valid for the Euclidean norm on \(\RNrSpc{n}\).

iii.   \(\CAbs{ \CCnjgt{z} } = \CAbs{ z }\)

This identity follows by computation: if \(z=x+yi\), then

\(\CAbs{ \CCnjgt{z} }\)	\(=\)	\(\CAbs{ x-yi }\)
\(\)	\(=\)	\(\sqrt{ x^2 + (-y)^2}\)
\(\)	\(=\)	\(\sqrt{ x^2 + (y)^2}\)
\(\)	\(=\)	\(\CAbs{z}\)

iv.   \(\CAbs{ z }^2 = z\CCnjgt{z}\)

This identity follows by conjugation: if \(z=x+yi\), then

\(z\CCnjgt{z}\)	\(=\)	\((x+yi)(x-yi)\)
\(\)	\(=\)	\(x^2 + y^2\)
\(\)	\(=\)	\(\CAbs{z}^2\)

v.   \(\CAbs{wz} = \CAbs{w} \CAbs{z}\)

We offer two independent arguments: The first is by direct computation; the second uses properties of complex conjugation.

Proof 1   by direct computation. Suppose \(w=u+vi\), and \(z=x+yi\). Then we find

\(\CAbs{ w\cdot z }\)	\(=\)	\(\CAbs{ (u+vi)(x+yi) }\)
\(\)	\(=\)	\(\CAbs{ ux-vy + (uy+vx)i }\)
\(\)	\(=\)	\(\sqrt{ (ux-vy)^2 + (uy+vx)^2 }\)
\(\)	\(=\)	\(\sqrt{ u^2x^2 - 2uxvy +v^2y^2 + u^2y^2 +2uyvx + v^2x^2 }\)
\(\)	\(=\)	\(\sqrt{ (u^2 + v^2)(x^2 + y^2)}\)
\(\)	\(=\)	\(\CAbs{w} \CAbs{z}\)

Proof 2   using properties of complex conjugation. We also invoke the fact that the squaring function is injective (1 to 1) when restricted to the nonnegative numbers:

\(\CAbs{ w\cdot z }^2\)	\(= \)	\(w\cdot z\cdot \CCnjgt{w}{z}\)
\(\)	\(= \)	\(w\cdot z\cdot \CCnjgt{w}\cdot \CCnjgt{z}\)
\(\)	\(= \)	\(w\cdot \CCnjgt{w}\cdot z\cdot \CCnjgt{z}\)
\(\)	\(=\)	\(\CAbs{w}^2 \cdot \CAbs{z}^2\)
\(\)	\(=\)	\(\left( \CAbs{w}\cdot \CAbs{z} \right)^2\)

Now, each of these norms is a nonnegative real number. Therefore \(\CAbs{w\cdot z} = \CAbs{w}\cdot \CAbs{z}\) as claimed.