alpha, beta are the roots of the equation x^2 | vedclass

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(C) Given the equation $x^2+2x+4=0$,we can write it as $(x+1)^2+3=0$.Solving for $x$,we get $x = -1 \pm \sqrt{3}i$.Since $\alpha$ lies in the $2^{nd}$

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$-2^{2024} \sqrt{3}$

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(C) Given the equation $x^2+2x+4=0$,we can write it as $(x+1)^2+3=0$.Solving for $x$,we get $x = -1 \pm \sqrt{3}i$.Since $\alpha$ lies in the $2^{nd}$ quadrant,$\alpha = -1 + \sqrt{3}i = 2(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})) = 2	ext{cis}(\frac{2\pi}{3})$.Then $\beta = -1 - \sqrt{3}i = 2	ext{cis}(-\frac{2\pi}{3})$.Using De Moivre's Theorem,$\alpha^{2024} = 2^{2024}	ext{cis}(\frac{2024 	imes 2\pi}{3}) = 2^{2024}	ext{cis}(\frac{4048\pi}{3}) = 2^{2024}	ext{cis}(\frac{4\pi}{3})$.Similarly,$\beta^{2024} = 2^{2024}	ext{cis}(-\frac{4\pi}{3})$.Now,$\alpha^{2024}-\beta^{2024} = 2^{2024}(	ext{cis}(\frac{4\pi}{3}) - 	ext{cis}(-\frac{4\pi}{3}))$.$= 2^{2024}((\cos(\frac{4\pi}{3}) + i\sin(\frac{4\pi}{3})) - (\cos(-\frac{4\pi}{3}) + i\sin(-\frac{4\pi}{3})))$.$= 2^{2024}(i\sin(\frac{4\pi}{3}) - i\sin(-\frac{4\pi}{3})) = 2^{2024}(i(-\frac{\sqrt{3}}{2}) - i(\frac{\sqrt{3}}{2}))$.$= 2^{2024}(-i\sqrt{3}) = -2^{2024}\sqrt{3}i$.Comparing with $ik$,we get $k = -2^{2024}\sqrt{3}$.

$\alpha, \beta$ are the roots of the equation $x^2+2x+4=0$. If the point representing $\alpha$ in the Argand diagram lies in the $2^{nd}$ quadrant and $\alpha^{2024}-\beta^{2024}=ik, (i=\sqrt{-1})$,then $k=$

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