If z = frac{sqrt{3} + i}{2},then left(z^{101} | vedclass

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(C) Given,$z = \frac{\sqrt{3} + i}{2} = \cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)$.Using De Moivre's Theorem,$z^{101} = \cos

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$i$

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(C) Given,$z = \frac{\sqrt{3} + i}{2} = \cos \left(\frac{\pi}{6}ight) + i \sin \left(\frac{\pi}{6}ight)$.Using De Moivre's Theorem,$z^{101} = \cos \left(\frac{101\pi}{6}ight) + i \sin \left(\frac{101\pi}{6}ight)$.Since $\frac{101\pi}{6} = 17\pi - \frac{\pi}{6}$,we have $z^{101} = \cos \left(17\pi - \frac{\pi}{6}ight) + i \sin \left(17\pi - \frac{\pi}{6}ight) = -\cos \left(\frac{\pi}{6}ight) + i \sin \left(\frac{\pi}{6}ight) = \frac{-\sqrt{3} + i}{2}$.Also,$i^{103} = i^{100} 	imes i^3 = 1 	imes (-i) = -i$.Thus,$z^{101} + i^{103} = \frac{-\sqrt{3} + i}{2} - i = \frac{-\sqrt{3} - i}{2} = -\left(\frac{\sqrt{3} + i}{2}ight) = -z$.Therefore,$\left(z^{101} + i^{103}ight)^{105} = (-z)^{105} = -z^{105}$.$z^{105} = \left(\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}ight)^{105} = \cos \frac{105\pi}{6} + i \sin \frac{105\pi}{6} = \cos \frac{35\pi}{2} + i \sin \frac{35\pi}{2}$.Since $\frac{35\pi}{2} = 18\pi - \frac{\pi}{2}$,we have $z^{105} = \cos \left(-\frac{\pi}{2}ight) + i \sin \left(-\frac{\pi}{2}ight) = 0 - i = -i$.Finally,$-z^{105} = -(-i) = i$.

If $z = \frac{\sqrt{3} + i}{2}$,then $\left(z^{101} + i^{103}\right)^{105} = $

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