The amplitude of the complex number frac{(sqr | vedclass

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Question

(D) Let $z = \frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(-1+i)(-1-i)}$.First,simplify the numerator: $(\sqrt{3}+i)(1-\sqrt{3} i) = \sqrt{3} - 3i + i - \sqrt{3}

Vedclass · Accepted Answer

$-\frac{\pi}{6}$

Vedclass · Answer

(D) Let $z = \frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(-1+i)(-1-i)}$.First,simplify the numerator: $(\sqrt{3}+i)(1-\sqrt{3} i) = \sqrt{3} - 3i + i - \sqrt{3} i^2 = \sqrt{3} - 2i + \sqrt{3} = 2\sqrt{3} - 2i$.Next,simplify the denominator: $(-1+i)(-1-i) = (-1)^2 - (i)^2 = 1 - (-1) = 2$.Thus,$z = \frac{2\sqrt{3} - 2i}{2} = \sqrt{3} - i$.The amplitude $	heta$ is given by $	an^{-1}(\frac{y}{x}) = 	an^{-1}(\frac{-1}{\sqrt{3}})$.Since the complex number lies in the fourth quadrant $(x > 0, y < 0)$,the amplitude is $-\frac{\pi}{6}$.

The amplitude of the complex number $\frac{(\sqrt{3}+i)(1-\sqrt{3} i)}{(-1+i)(-1-i)}$ is

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