The amplitude of frac{1 + sqrt{3}i}{sqrt{3} - | vedclass

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(D) Let $z = \frac{1 + \sqrt{3}i}{\sqrt{3} - i}$.Multiply the numerator and denominator by the conjugate of the denominator,$\sqrt{3} + i$:$z = \frac{

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$\pi / 2$

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(D) Let $z = \frac{1 + \sqrt{3}i}{\sqrt{3} - i}$.Multiply the numerator and denominator by the conjugate of the denominator,$\sqrt{3} + i$:$z = \frac{1 + \sqrt{3}i}{\sqrt{3} - i} 	imes \frac{\sqrt{3} + i}{\sqrt{3} + i}$$z = \frac{\sqrt{3} + i + 3i + \sqrt{3}i^2}{3 - i^2}$Since $i^2 = -1$,we have:$z = \frac{\sqrt{3} + 4i - \sqrt{3}}{3 + 1} = \frac{4i}{4} = i$Now,$z = 0 + 1i$. The amplitude (argument) of $z = i$ is $	heta = 	an^{-1}(\frac{1}{0}) = \pi / 2$.

The amplitude of $\frac{1 + \sqrt{3}i}{\sqrt{3} - i}$ is

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