Question

Determine z E C tal que i.z²=-z (conjugado)

Answer 1

Seja  Z = a + bi,  temos:

$i\,.\,(a+bi)^2~=~-(\,\overline{a+bi}\,)\\\\\\i\,.\,(a^2+2.a.bi+b^2i^2)~=~-(a-bi)\\\\\\a^2i+2abi^2-b^2i~=~-a+bi\\\\\\\boxed{-2ab+(a^2-b^2).i~=~-a+bi}\\\\\\Igualando~as~partes~Reais~e~as~partes~Imaginarias\\\\\\\left\{\begin{array}{ccc}-2ab&=&-a\\a^2-b^2&=&b\end{array}\right\\\\\\Pela~1^a~equacao,~temos:\\\\\\-2ab~=~-a\\\\\\b~=~\frac{-a}{-2a}\\\\\\\boxed{b~=~\frac{1}{2}}$

$Pela~2^a~equacao,~temos:\\\\\\a^2-b^2~=~b\\\\\\a^2~-~\left(\frac{1}{2}\right)^2~=~\frac{1}{2}\\\\\\a^2~=~\frac{1}{2}+\frac{1}{4}\\\\\\a~=~\sqrt{\frac{3}{4}}\\\\\\\boxed{a~=~\frac{\sqrt{3}}{2}}$

Obs.: Podemos mostrar que a raiz negativa de "a" não verifica a equação substituindo a = -(√(3))/2 na equação.