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(B) Given the equation $x^2+2x+2=0$. Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4-8}}{2} = -1 \pm i$. Let $\alpha = -1+i$ and $\beta = -1-i$.

Vedclass · Accepted Answer

$-256$

Vedclass · Answer

(B) Given the equation $x^2+2x+2=0$. Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4-8}}{2} = -1 \pm i$. Let $\alpha = -1+i$ and $\beta = -1-i$. In polar form,$\alpha = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4})$ and $\beta = \sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4})$. Then $\alpha^{15} = (\sqrt{2})^{15} (\cos \frac{45\pi}{4} + i \sin \frac{45\pi}{4}) = 2^{7.5} (\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) = 2^{7.5} (-\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}})$. Similarly,$\beta^{15} = (\sqrt{2})^{15} (\cos \frac{75\pi}{4} + i \sin \frac{75\pi}{4}) = 2^{7.5} (\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) = 2^{7.5} (-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}})$. Adding these,$\alpha^{15} + \beta^{15} = 2^{7.5} (-\frac{2}{\sqrt{2}}) = 2^{7.5} (-\sqrt{2}) = -2^8 = -256$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2+2x+2=0$,then $\alpha^{15}+\beta^{15}=$

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