If z = frac{1 + isqrt{3}}{sqrt{3} + i},then ( | vedclass

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

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$III$ quadrant

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(C) Given $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}$.Multiply numerator and denominator by the conjugate of the denominator: $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i} 	imes \frac{\sqrt{3} - i}{\sqrt{3} - i}$.$z = \frac{\sqrt{3} - i + 3i - i^2\sqrt{3}}{3 + 1} = \frac{\sqrt{3} + 2i + \sqrt{3}}{4} = \frac{2\sqrt{3} + 2i}{4} = \frac{\sqrt{3} + i}{2}$.$z = \cos\left(\frac{\pi}{6}ight) + i\sin\left(\frac{\pi}{6}ight)$.Then $\bar{z} = \cos\left(\frac{\pi}{6}ight) - i\sin\left(\frac{\pi}{6}ight)$.Using De Moivre's Theorem,$(\bar{z})^{100} = \left[\cos\left(\frac{\pi}{6}ight) - i\sin\left(\frac{\pi}{6}ight)ight]^{100} = \cos\left(\frac{100\pi}{6}ight) - i\sin\left(\frac{100\pi}{6}ight)$.$(\bar{z})^{100} = \cos\left(\frac{50\pi}{3}ight) - i\sin\left(\frac{50\pi}{3}ight)$.Since $\frac{50\pi}{3} = 16\pi + \frac{2\pi}{3}$,we have $(\bar{z})^{100} = \cos\left(\frac{2\pi}{3}ight) - i\sin\left(\frac{2\pi}{3}ight)$.This is equal to $-\frac{1}{2} - i\frac{\sqrt{3}}{2}$.Since both the real and imaginary parts are negative,$(\bar{z})^{100}$ lies in the $III$ quadrant.

If $z = \frac{1 + i\sqrt{3}}{\sqrt{3} + i}$,then $(\bar{z})^{100}$ lies in

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