The product of all the values of (sqrt{3}-i)^ | vedclass

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(D) Let $z = \sqrt{3}-i$. In polar form,$z = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) = 2e^{-i\frac{\pi}{6}}$.We want to find the product of al

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$-8i$

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(D) Let $z = \sqrt{3}-i$. In polar form,$z = 2(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) = 2e^{-i\frac{\pi}{6}}$.We want to find the product of all values of $w = z^{3/7} = (2e^{-i\frac{\pi}{6} + 2k\pi i})^{3/7}$ for $k = 0, 1, 2, 3, 4, 5, 6$.This simplifies to $w_k = 2^{3/7} e^{i(-\frac{\pi}{14} + \frac{6k\pi}{7})}$.The product of these $7$ values is $P = \prod_{k=0}^{6} 2^{3/7} e^{i(-\frac{\pi}{14} + \frac{6k\pi}{7})}$.$P = (2^{3/7})^7 \cdot e^{i \sum_{k=0}^{6} (-\frac{\pi}{14} + \frac{6k\pi}{7})} = 8 \cdot e^{i (-\frac{7\pi}{14} + \frac{6\pi}{7} \cdot \frac{6 \cdot 7}{2})} = 8 \cdot e^{i (-\frac{\pi}{2} + 18\pi)} = 8 \cdot e^{-i\frac{\pi}{2}} = 8(-i) = -8i$.

The product of all the values of $(\sqrt{3}-i)^{\frac{3}{7}}$ is

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