(-1 + isqrt{3})^{20} is equal to | vedclass

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(D) Let $z = -1 + i\sqrt{3}$. The modulus $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. We can write $z = 2 \left( -\frac{1}{2} + i\frac{\sqr

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None of these

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(D) Let $z = -1 + i\sqrt{3}$. The modulus $r = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2$. We can write $z = 2 \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)$. Using De Moivre's Theorem,$z^{20} = 2^{20} \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right)^{20} = 2^{20} \left( \cos \frac{40\pi}{3} + i \sin \frac{40\pi}{3} \right)$. Since $\frac{40\pi}{3} = 13\pi + \frac{\pi}{3} = 12\pi + \pi + \frac{\pi}{3} = 6(2\pi) + \frac{4\pi}{3}$,we have $\cos \frac{40\pi}{3} = \cos \frac{4\pi}{3} = -\frac{1}{2}$ and $\sin \frac{40\pi}{3} = \sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$. Thus,$z^{20} = 2^{20} \left( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) = 2^{19}(-1 - i\sqrt{3})$. None of the given options match this result.

$(-1 + i\sqrt{3})^{20}$ is equal to

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