Question

Sol. [Key Concept: Solve $z^{2} + \sqrt{6\ }iz - 3 = 0$; the roots have modulus $\sqrt{3}$ and arguments $\pm$, so write them in polar form and use De Moivre to evaluate $\sum\ z^{8}$.]

$${z = \frac{- \sqrt{6}i \pm \sqrt{- 6 + 12}}{2} }{= \frac{- \sqrt{6}i \pm \sqrt{6}}{2} }{= \frac{\sqrt{6}}{2}( \pm 1 - i) }{\alpha = \frac{- \sqrt{6}}{2}(1 + i),\beta = \frac{\sqrt{6}}{2}(1 - i) }{\alpha = \frac{- \sqrt{6}}{2} \times \sqrt{2}e^{i\pi/4},\beta = + \frac{\sqrt{6}}{2} \times \sqrt{2} - i\pi/4 }{\alpha = - \sqrt{3}e^{i\pi/4},\beta = + \sqrt{3}e^{- i\pi/4} }{\alpha^{8} + \beta^{8} = 81.e^{i2\pi} + 81.e^{i( - 2\pi)} }{= 81(1 + 1) = 162 }$$