Question

(FUVEST) sendo i a unidade imaginária(
i^2=-1), pergunta-se: quantos números reais a existem para os quais (a+i)^4 é um número real ?

Answer 1

$\displaystyle (a+i)^4=w\\ \\ a+i=R_k\\ \\ R_k=\sqrt[4]w\left[\cos\left(\frac{2\pi k}{4}\right)+i\sin\left(\frac{2\pi k}{4}\right)\right]\\ \\ \text{Donde }k\in\{0,1,2,3\}\\ \\ R_0=\sqrt[4]w\\ R_1=\sqrt[4]w\,i\\ R_2=-\sqrt[4]w\\ R_3=-\sqrt[4]w\,i$

Lo de arriba cuando W>0 ===> a=0

Y si W <0

$\displaystyle R'_k=\sqrt[4]w\left[\cos\left(\frac{\pi+2\pi k}{4}\right)+i\sin \left(\frac{\pi+2\pi k}{4}\right)\right]\\ \\ R'_0=\sqrt[4]w\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)=\sqrt[4]w\cdot\frac{\sqrt2}{2}(1+i)\\ \\ R'_1=\sqrt[4]w\left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)=\sqrt[4]w\cdot\frac{\sqrt2}{2}(-1+i)\\ \\ R'_2=-\sqrt[4]w\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)=-\sqrt[4]w\cdot\frac{\sqrt2}{2}(1+i)$

$R'_3=-\sqrt[4]w\left(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)=-\sqrt[4]w\cdot\frac{\sqrt2}{2}(-1+i)$

entonces a = 1 ó a=-1