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

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(C) Let $z = (\sqrt{3}-i)^{2/5} = [2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))]^{2/5}$.Using De Moivre's Theorem,the $5$ values are given by $z_k

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$2(1-\sqrt{3}i)$

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(C) Let $z = (\sqrt{3}-i)^{2/5} = [2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6}))]^{2/5}$.Using De Moivre's Theorem,the $5$ values are given by $z_k = 2^{2/5} [\cos(\frac{2k\pi - \pi/6}{5/2}) + i \sin(\frac{2k\pi - \pi/6}{5/2})]$ for $k = 0, 1, 2, 3, 4$.This simplifies to $z_k = 2^{2/5} [\cos(\frac{4k\pi}{5} - \frac{\pi}{15}) + i \sin(\frac{4k\pi}{5} - \frac{\pi}{15})]$.The product of the $n$ roots of a complex number $z = a^{p/q}$ is given by $(-1)^{q-1} (a^{p/q})^q$ if we consider the roots of the equation $z^q = a^p$.Here,$z^5 = (\sqrt{3}-i)^2 = 3 - 1 - 2\sqrt{3}i = 2 - 2\sqrt{3}i$.The product of the roots of $z^5 - (2 - 2\sqrt{3}i) = 0$ is $(-1)^{5-1} 	imes (-(2 - 2\sqrt{3}i)) = 1 	imes (-2 + 2\sqrt{3}i) = -2 + 2\sqrt{3}i$.Wait,the product of roots $z_k$ of $z^n = c$ is $(-1)^{n-1} c$. Here $n=5$ and $c = (\sqrt{3}-i)^2 = 2 - 2\sqrt{3}i$.Product $= (-1)^{5-1} (2 - 2\sqrt{3}i) = 1 	imes (2 - 2\sqrt{3}i) = 2(1 - \sqrt{3}i)$.

The product of all the values of $(\sqrt{3}-i)^{2/5}$ is

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