Fie [tex]z_1=a+ib[/tex] si [tex]z_2=c+id[/tex].
Atunci:
[tex]|z_1|=|a+ib|=\sqrt{a^2+b^2}\\[/tex]
[tex]|z_1|=1 \implies \sqrt{a^2+b^2}=1\implies a^2+b^2=1[/tex]
Analog obtinem [tex]c^2+d^2=1[/tex]
[tex]|z_1+z_2|=|a+c+i(b+d)|=\sqrt{(a+c)^2+(b+d)^2}=[/tex]
=[tex]\sqrt{a^2+2ac+c^2+b^2+2bd+d^2}=\sqrt{a^2+b^2+c^2+d^2+2ac+2bd}=[/tex]
[tex]=\sqrt{1+1+2ac+2bd}=\sqrt{2+2ac+2bd[/tex]
Dar [tex]|z_1+z_2|=\sqrt3[/tex]
[tex]\implies \sqrt{2+2ac+2bd}=\sqrt3\implies 2+2ac+2bd=3[/tex]
[tex]\implies2ac+2bd=1[/tex]
[tex]|z_1-z_2|=|a+ib-(c+id)|=|a+ib-c-id|=|a-c+i(b-d)|=[/tex]
[tex]=\sqrt{(a-c)^2+(b-d)^2}=\sqrt{a^2-2ac+c^2+b^2-2bd+d^2}=[/tex]
[tex]=\sqrt{a^2+b^2+c^2+d^2-2ac-2bd}=\sqrt{2-(2ac+2bd)}=[/tex]
[tex]=\sqrt{2-1}=1[/tex]
Deci raspunsul corect este B.
