Sean w , z ∈ C {\displaystyle w,z\in {C}} Demostrar que:
a) | | z | | = | | z ¯ | | {\displaystyle ||z||=||{\overline {z}}||}
Demostración
Sea z ¯ = a − i b y z = a + i b {\displaystyle {\overline {z}}=a-ib\qquad y\qquad z=a+ib} entonces
| | z ¯ | | = a 2 + ( − b ) 2 y | | z | | = a 2 + b 2 {\displaystyle ||{\overline {z}}||={\sqrt {a^{2}+(-b)^{2}}}\qquad y\qquad ||z||={\sqrt {a^{2}+b^{2}}}}
∴ | | z | | = | | z ¯ | | {\displaystyle \therefore \qquad ||z||=||{\overline {z}}||}
b) | | z w | | = | | z | | | | w | | {\displaystyle \qquad ||zw||=||z||||w||}
| | z w | | 2 = ( z w ) ( z w ¯ ) {\displaystyle ||zw||^{2}=(zw)({\overline {zw}})}
| | z w | | 2 = ( z w ¯ ) ( w z ¯ ) {\displaystyle ||zw||^{2}=(z{\overline {w}})(w{\overline {z}})}
| | z w | | 2 = | | z | | 2 | | w | | 2 {\displaystyle \qquad ||zw||^{2}=||z||^{2}||w||^{2}}
∴ | | z w | | = | | z | | | | w | | {\displaystyle \therefore \qquad ||zw||=||z||||w||}