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TS EAMCET 5-Aug-2021 Shift 2 Question Paper

Section: Mathematics

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Question : 7 of 160

Marks: +1, -0

If

(\sqrt{3}+i)^8 - (\sqrt{3}-i)^8

\alpha + i\beta

\alpha - \frac{\sqrt{3}}{2}\beta

=

$256$
$384\sqrt{3}$
$384$
$256\sqrt{3}$

Solution: 👈: Video Solution

(\sqrt{3}+i)^8 - (\sqrt{3}-i)^8 = \alpha + i\beta

\frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\left(\frac{\pi}{6}\right) + i \sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}

\frac{\sqrt{3}}{2} - \frac{i}{2} = \cos\left(2\pi - \frac{\pi}{6}\right) + i \sin\left(2\pi - \frac{\pi}{6}\right)

= e^{i(2\pi - \frac{\pi}{6})}

From Eq. (i),

(2 e^{i\pi/6})^8 - (2 e^{i11\pi/6})^8 = \alpha + i\beta

\Rightarrow 2^{8}\left(e^{i\frac{8\pi}{6}} - e^{i\frac{88\pi}{6}}\right) = \alpha + i\beta

\Rightarrow 2^{8}\left[e^{i\left(\pi + \frac{\pi}{3}\right)} - e^{i\left(15\pi - \frac{\pi}{3}\right)}\right] = \alpha + i\beta

\Rightarrow 2^{8}\left[\left(-\frac{1}{2} - \frac{\sqrt{3}i}{2}\right) - \left(-\frac{1}{2} + \frac{\sqrt{3}i}{2}\right)\right] = \alpha + i\beta

\Rightarrow 2^{8}(-\sqrt{3} i) = \alpha + i\beta \Rightarrow \alpha = 0, \beta = -\sqrt{3} \cdot 2^{8}

\alpha - \frac{\sqrt{3}}{2} \beta \Rightarrow 0 - \frac{\sqrt{3}}{2}(-\sqrt{3} \cdot 2^{8}) = 3 \cdot 2^{7} = 384

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