If (sqrt{3}+i)^8-(sqrt{3}-i)^8=alpha+i beta,t | vedclass

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(C) Let $z = \sqrt{3}+i$. Then $\bar{z} = \sqrt{3}-i$. We have $z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2e^{i\pi/6}$. Then $z^8 = 2^8 e^{i8

Vedclass · Accepted Answer

$384$

Vedclass · Answer

(C) Let $z = \sqrt{3}+i$. Then $\bar{z} = \sqrt{3}-i$. We have $z = 2(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) = 2e^{i\pi/6}$. Then $z^8 = 2^8 e^{i8\pi/6} = 256 e^{i4\pi/3} = 256(\cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3}) = 256(-\frac{1}{2} - i \frac{\sqrt{3}}{2}) = -128 - 128\sqrt{3}i$. Similarly,$\bar{z}^8 = \overline{z^8} = -128 + 128\sqrt{3}i$. Thus,$z^8 - \bar{z}^8 = (-128 - 128\sqrt{3}i) - (-128 + 128\sqrt{3}i) = -256\sqrt{3}i$. Comparing with $\alpha + i\beta$,we get $\alpha = 0$ and $\beta = -256\sqrt{3}$. Finally,$\alpha - \frac{\sqrt{3}}{2}\beta = 0 - \frac{\sqrt{3}}{2}(-256\sqrt{3}) = \frac{3}{2} 	imes 256 = 384$.

If $(\sqrt{3}+i)^8-(\sqrt{3}-i)^8=\alpha+i \beta$,then $\alpha-\frac{\sqrt{3}}{2} \beta=$

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