Question

Daca puteti va rog sa ma ajutați la exercitiul 162

Answer 1

Fie  $z_1=a+ib$  si  $z_2=c+id$.
Atunci:
$|z_1|=|a+ib|=\sqrt{a^2+b^2}$
$|z_1|=1 \implies \sqrt{a^2+b^2}=1\implies a^2+b^2=1$
Analog obtinem $c^2+d^2=1$

$|z_1+z_2|=|a+c+i(b+d)|=\sqrt{(a+c)^2+(b+d)^2}=$
=$\sqrt{a^2+2ac+c^2+b^2+2bd+d^2}=\sqrt{a^2+b^2+c^2+d^2+2ac+2bd}=$
$=\sqrt{1+1+2ac+2bd}=\sqrt{2+2ac+2bd$

Dar $|z_1+z_2|=\sqrt3$
$\implies \sqrt{2+2ac+2bd}=\sqrt3\implies 2+2ac+2bd=3$
$\implies2ac+2bd=1$

$|z_1-z_2|=|a+ib-(c+id)|=|a+ib-c-id|=|a-c+i(b-d)|=$
$=\sqrt{(a-c)^2+(b-d)^2}=\sqrt{a^2-2ac+c^2+b^2-2bd+d^2}=$
$=\sqrt{a^2+b^2+c^2+d^2-2ac-2bd}=\sqrt{2-(2ac+2bd)}=$
$=\sqrt{2-1}=1$

Deci raspunsul corect este B.