If z_1 = -sqrt{3} + i and z_2 = -sqrt{3} - i, | vedclass

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(C) Given $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$. We know that the argument of a quotient is given by $	ext{arg}(\frac{z_1}{z_2}) = 	ext{ar

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$-\frac{\pi}{3}$

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(C) Given $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$. We know that the argument of a quotient is given by $	ext{arg}(\frac{z_1}{z_2}) = 	ext{arg}(z_1) - 	ext{arg}(z_2)$. For $z_1 = -\sqrt{3} + i$,the point lies in the second quadrant. $	ext{arg}(z_1) = \pi - 	an^{-1}(\frac{1}{\sqrt{3}}) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. For $z_2 = -\sqrt{3} - i$,the point lies in the third quadrant. $	ext{arg}(z_2) = -(\pi - 	an^{-1}(\frac{1}{\sqrt{3}})) = -(\pi - \frac{\pi}{6}) = -\frac{5\pi}{6}$. Therefore,$	ext{arg}(\frac{z_1}{z_2}) = \frac{5\pi}{6} - (-\frac{5\pi}{6}) = \frac{10\pi}{6} = \frac{5\pi}{3}$. Since the principal amplitude must lie in the interval $(-\pi, \pi]$,we subtract $2\pi$: $\frac{5\pi}{3} - 2\pi = -\frac{\pi}{3}$. Thus,the principal amplitude is $-\frac{\pi}{3}$.

If $z_1 = -\sqrt{3} + i$ and $z_2 = -\sqrt{3} - i$,then the principal amplitude of the complex number $\frac{z_1}{z_2}$ is

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