Question

((4-3i)(12-5i))/√2i Ache o módulo

Answer 1

$\sf z = a+b\cdot i\\\\ M{\'o}dulo \ de\ z : \\\\ |z| = \sqrt{a^2+b^2 }$

Temos :

$\displaystyle \sf z = \frac{(4-3\cdot i)\cdot (12-5\cdot i)}{\sqrt{2}\cdot i} \\\\\\ |z| = \left| \frac{(4-3\cdot i)\cdot (12-5\cdot i)}{\sqrt{2}\cdot i} \right | \\\\\\ |z| = \frac{|(4-3\cdot i)\cdot (12-5\cdot i)|}{|\sqrt{2}\cdot i|} \\\\\\ |z| = \frac{|(4-3i)|\cdot |(12-5\cdot i)| }{|\sqrt{2}\cdot i|} \\\\\\ |z| = \frac{\sqrt{4^2+(-3)^2}\cdot \sqrt{12^2+(-5)^2}}{\sqrt{0^2+\left(\sqrt{2}\right)^2}} \to |z|=\frac{\sqrt{25}\cdot \sqrt{144+25}}{\sqrt{2}}$

$\displaystyle \sf |z| = \frac{5\cdot 13}{\sqrt{2}} = \frac{65}{\sqrt{2}} \\\\\\ \huge\boxed{\sf |z| = \frac{65\sqrt{2}}{2} \ }\checkmark$