4.1 complex nubers — ગણિત ધોરણ 11 સાયન્સ — Question

$x=\frac{\sqrt{2} \pm \sqrt{2-8}}{2}=\frac{\sqrt{2} \pm \sqrt{6} i}{2}$

$\alpha=\frac{\sqrt{2}+\sqrt{6} i}{2}=\sqrt{2} e^{\frac{i \pi}{3}}$ $\beta=\sqrt{2} e^{\frac{-i \pi}{3}}$

$\alpha^{14}=2^7 e^{\frac{i 14 \pi}{3}}=128\left[e^{\frac{i 2 \pi}{3}}\right]$

$\beta^{14}=128\left[e^{\frac{-i 2 \pi}{3}}\right]$

$\alpha^{14}+\beta^{14}=128(2) \cos \left(\frac{2 \pi}{3}\right)=-128$