The argument of frac{1+i sqrt{3}}{sqrt{3}+i}, | vedclass

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

Vedclass · Accepted Answer

$\frac{\pi}{6}$

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(C) Let $z = \frac{1+i \sqrt{3}}{\sqrt{3}+i}$.Multiply the numerator and denominator by the conjugate of the denominator $(\sqrt{3}-i)$:$z = \frac{(1+i \sqrt{3})(\sqrt{3}-i)}{(\sqrt{3}+i)(\sqrt{3}-i)}$$z = \frac{\sqrt{3} - i + 3i - i^2 \sqrt{3}}{3 - i^2}$Since $i^2 = -1$,we have:$z = \frac{\sqrt{3} + 2i + \sqrt{3}}{3 + 1} = \frac{2\sqrt{3} + 2i}{4} = \frac{\sqrt{3}}{2} + \frac{1}{2}i$The argument of $z = a + bi$ is given by $	heta = 	an^{-1}\left(\frac{b}{a}ight)$.$	heta = 	an^{-1}\left(\frac{1/2}{\sqrt{3}/2}ight) = 	an^{-1}\left(\frac{1}{\sqrt{3}}ight) = \frac{\pi}{6}$.

The argument of $\frac{1+i \sqrt{3}}{\sqrt{3}+i}$,where $i=\sqrt{-1}$,is

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