Another pre calc proof help please?? I don’t get how you’re supposed to prove. I am extremely uncomfortable in this topic.

Guest Dec 10, 2018

#1**+1 **

\(\text{The first statement is false}\\ |z| \in \mathbb{R},~\forall z \in \mathbb{C}\\ i |z| \not \in \mathbb{R}\)

\(|z \cdot w | =\\ |(z_r w_r -z_i w_i) + i(z_r w_i + z_i w_r)| = \\ \sqrt{(z_r w_r -z_i w_i)^2 + (z_r w_i + z_i w_r)^2} = \\ \sqrt{-2 w_i z_i w_r z_r+w_i^2 z_i^2+w_r^2 z_r^2 + z_i^2 w_r^2+2 w_i z_i w_r z_r+w_i^2 z_r^2}=\)

\(\sqrt{w_r^2 z_r^2 + z_r^2 w_i^2 + z_i^2 w_r^2 + w_r^2 w_i^2}=\\ \sqrt{(z_r^2+z_i^2)(w_r^2+w_i)^2} = \\ \sqrt{z_r^2 + z_i^2}\sqrt{w_r^2+w_i^2} =\\ |z||w|\)

.Rom Dec 10, 2018