If x^6 = (sqrt{3} - i)^5,then the product of | vedclass

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(D) The given equation is $x^6 = (\sqrt{3} - i)^5$. Let $z = \sqrt{3} - i$. We can write $z$ in polar form: $z = 2(\cos(-\frac{\pi}{6}) + i \sin(-\fra

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$\frac{2^6}{\sqrt{3} - i}$

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(D) The given equation is $x^6 = (\sqrt{3} - i)^5$. Let $z = \sqrt{3} - i$. We can write $z$ in polar form: $z = 2(\cos(-\frac{\pi}{6}) + i \sin(-\frac{\pi}{6})) = 2e^{-i\pi/6}$. Then $z^5 = (2e^{-i\pi/6})^5 = 2^5 e^{-i5\pi/6}$. The equation is $x^6 = 2^5 e^{-i5\pi/6}$. For an equation of the form $x^n = A$,the product of the roots is given by $(-1)^{n-1} A$. Here $n = 6$ and $A = 2^5 e^{-i5\pi/6}$. Product of roots $= (-1)^{6-1} (2^5 e^{-i5\pi/6}) = -2^5 e^{-i5\pi/6} = 2^5 e^{i\pi} e^{-i5\pi/6} = 2^5 e^{i\pi/6}$. $2^5 e^{i\pi/6} = 2^5 (\cos(\frac{\pi}{6}) + i \sin(\frac{\pi}{6})) = 2^5 (\frac{\sqrt{3}}{2} + i \frac{1}{2}) = 2^4 (\sqrt{3} + i) = 16(\sqrt{3} + i)$. Checking the options,we note that $\frac{2^6}{\sqrt{3} + i} = \frac{64(\sqrt{3} - i)}{3 + 1} = 16(\sqrt{3} - i)$. Wait,let us re-evaluate the product: $(-1)^{n-1} A = -A$. $-2^5 e^{-i5\pi/6} = 2^5 e^{i\pi} e^{-i5\pi/6} = 2^5 e^{i\pi/6} = 2^5 (\frac{\sqrt{3}}{2} + i \frac{1}{2}) = 16(\sqrt{3} + i)$. Since $16(\sqrt{3} + i) = \frac{64}{\sqrt{3} - i} = \frac{2^6}{\sqrt{3} - i}$,the correct option is $D$.

If $x^6 = (\sqrt{3} - i)^5$,then the product of all of its roots is

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