IME / ITA ⇒ (Escola Naval - 1987) Complexos Tópico resolvido

[ALDRINMOD]

[tex3]\sqrt{i}=[/tex3]

(A) [tex3]\pm \frac{\sqrt2}{2}(1+i)[/tex3]
(B) [tex3]\pm \frac{\sqrt2}{2}(1-i)[/tex3]
(C) [tex3]\pm (1+i)[/tex3]
(D) [tex3]\pm (1-i)[/tex3]
(E) [tex3]\pm i[/tex3]

[FilipeCaceres]

Olá Aldrin,

Seja
[tex3]z=\sqrt{i}[/tex3]

Reescrevendo,
[tex3]z=\left(cis\frac{\pi}{2}\right)^{\frac{1}{2}}=cis\frac{\pi}{4}[/tex3]
[tex3]z=\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}[/tex3]
[tex3]\boxed{Z=\frac{\sqrt{2}}{2}(1+i)}[/tex3]. Letra A

Abraço.