Let S=z C:z^2+√6iz-3=0.

Question

Let $S={z\in C:{z}^{2}+\sqrt{6}iz-3=0}$. Then $\sum _{z\in S}{z}^{8}$ is equal to :

Accepted Answer

$\begin{gathered} z=\frac{-\sqrt{6}i\pm \sqrt{-6+12}}{2} \ =\frac{-\sqrt{6}i\pm \sqrt{6}}{2} \ =\frac{\sqrt{6}}{2}(\pm 1-i) \end{gathered}$

$\begin{gathered} \alpha =\frac{-\sqrt{6}}{2}(1+i),\beta =\frac{\sqrt{6}}{2}(1-i) \ \alpha =\frac{-\sqrt{6}}{2}	imes \sqrt{2}{e}^{i\pi /4},\beta =+\frac{\sqrt{6}}{2}	imes \sqrt{2}-i\pi /4 \ \alpha =-\sqrt{3}{e}^{i\pi /4},\beta =+\sqrt{3}{e}^{-i\pi /4} \end{gathered}$

$\begin{gathered} {\alpha }^{8}+{\beta }^{8}=81\cdot {e}^{i2\pi }+81\cdot {e}^{i(-2\pi )} \ =81(1+1)=162 \end{gathered}$

Solutions · 1