ita complexos

por Gravidade10 Seg 03 Fev 2014, 16:15

Considere a equação: ita complexos <a href=

" /> sendo x um numero real a soma dos quadrados das soluções desta equação é
R:6

por PedroCunha Seg 03 Fev 2014, 16:46

Olá, Gravidade10.

$16\left( \frac{1-ix}{1+ix} \right)^3 = \left( \frac{1+i}{1-i} - \frac{1-i}{1+i} \right)^4 \\\\ \circ \frac{1+i}{1-i} = \frac{(1+i) \cdot (1+i)}{(1+i) \cdot (1-i)} \therefore \frac{1^2 + 2i + i^2}{1^2 - i^2} \therefore \frac{2i}{2} \therefore i \\\\\circ \frac{1-i}{1+i} \therefore \frac{(1-i) \cdot (1-i)}{(1+i) \cdot (1-i)} \therefore \frac{1^2 - 2i + i^2}{1^2 - i^2} \therefore \frac{-2i}{2} \therefore - i \\\\16 \cdot \left( \frac{1-ix}{1+ix} \right)^3 = \left( \frac{1+i}{1-i} - \frac{1-i}{1+i} \right)^4 \therefore 16 \cdot \left( \frac{1-ix}{1+ix} \right)^3 = (2i)^4 \therefore 16 \cdot \left( \frac{1-ix}{1+ix} \right)^3 = 16 \therefore \\\\ \frac{(1-ix)^3}{(1+ix)^3} = 1 \therefore (1-ix)^3 = (1+ix)^3 \therefore \\\\ 1^3 -3\cdot 1^2 \cdot ix + 3 \cdot 1 \cdot (ix)^2 - (ix)^3 = 1^3 + 3 \cdot 1^2 \cdot ix + 3 \cdot 1 \cdot (ix)^2 + (ix)^3 \\\\ 1 - 3ix - 3x^2 + ix^3 = 1 + 3ix - 3x^2 - ix^3 \therefore \\\\ 2ix^3 - 6ix \therefore 2ix \cdot (x^2 - 3) = 0 \therefore x = 0 \text{ ou } x = \pm \sqrt3 \\\\\circ 0^2 + (\sqrt3)^2 + (-\sqrt3)^2 = 6$