Question

determinati modulul numarul
$z = \sqrt{2 } + 6i$
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Answer 1

Avem o formula :

Fie z = a+bi

|z| = √(a²+b²)

_________

|z| = √(√2²+6²) = √(2+36) = √38

Answer 2

$Forma \ generala \ a \ unui \ nr \ complex \ z \ este \ a+b\cdot i\\ \\ OBS: \ \boxed{|z|=\sqrt{a^2+b^2}}\\ \\ In \ cazul \ tau \ z=\sqrt{2}+6i \\ \\ a=\sqrt{2};~b=6\\ \\ \\ \Rightarrow |z|=\sqrt{(\sqrt{2})^2+6^2}=\sqrt{2+36}=\sqrt{38}\\ \\ \boxed{|z|=\sqrt{38}}$