The sum of the least positive arguments of th | vedclass

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(D) Let $z = 1 - i \sqrt{3}$.First,express $z$ in polar form: $|z| = \sqrt{1^2 + (-\sqrt{3})^2} = 2$.The argument $	heta$ satisfies $	an 	heta = \f

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$\frac{11 \pi}{3}$

Vedclass · Answer

(D) Let $z = 1 - i \sqrt{3}$.First,express $z$ in polar form: $|z| = \sqrt{1^2 + (-\sqrt{3})^2} = 2$.The argument $	heta$ satisfies $	an 	heta = \frac{-\sqrt{3}}{1} = -\sqrt{3}$. Since $z$ is in the fourth quadrant,$	heta = -\frac{\pi}{3}$ or $2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$.Thus,$z = 2(\cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3})$.The cube roots of $z$ are given by $z^{1/3} = 2^{1/3} \left[ \cos \left( \frac{5\pi/3 + 2k\pi}{3} ight) + i \sin \left( \frac{5\pi/3 + 2k\pi}{3} ight) ight]$ for $k = 0, 1, 2$.The arguments are $	heta_k = \frac{5\pi + 6k\pi}{9}$.For $k=0$,$	heta_0 = \frac{5\pi}{9}$.For $k=1$,$	heta_1 = \frac{11\pi}{9}$.For $k=2$,$	heta_2 = \frac{17\pi}{9}$.All these are positive and less than $2\pi$.The sum is $\frac{5\pi}{9} + \frac{11\pi}{9} + \frac{17\pi}{9} = \frac{33\pi}{9} = \frac{11\pi}{3}$.

The sum of the least positive arguments of the distinct cube roots of the complex number $(1-i \sqrt{3})$ is

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