The value of left(frac{-1+i sqrt{3}}{1-i}righ | vedclass

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(C) Let $z = \frac{-1+i \sqrt{3}}{1-i}$.First,express the numerator in polar form: $-1+i \sqrt{3} = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{

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$-2^{15} i$

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(C) Let $z = \frac{-1+i \sqrt{3}}{1-i}$.First,express the numerator in polar form: $-1+i \sqrt{3} = 2 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 2e^{i 2\pi/3}$.Next,express the denominator: $1-i = \sqrt{2} \left( \cos \frac{-\pi}{4} + i \sin \frac{-\pi}{4} \right) = \sqrt{2}e^{-i \pi/4}$.Then,$z = \frac{2e^{i 2\pi/3}}{\sqrt{2}e^{-i \pi/4}} = \sqrt{2} e^{i (2\pi/3 + \pi/4)} = \sqrt{2} e^{i 11\pi/12}$.Now,$z^{30} = (\sqrt{2})^{30} e^{i (11\pi/12) \cdot 30} = 2^{15} e^{i 55\pi/2}$.Since $e^{i 55\pi/2} = e^{i (26\pi + 3\pi/2)} = e^{i 3\pi/2} = -i$.Therefore,$z^{30} = 2^{15} \cdot (-i) = -2^{15} i$.

The value of $\left(\frac{-1+i \sqrt{3}}{1-i}\right)^{30}$ is

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