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

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(C) Given $z = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.$z^{101} = e^{i(101\pi/6)} = e^{i

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$z^3$

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(C) Given $z = \frac{\sqrt{3}+i}{2} = \cos\left(\frac{\pi}{6}\right) + i\sin\left(\frac{\pi}{6}\right) = e^{i\pi/6}$.$z^{101} = e^{i(101\pi/6)} = e^{i(16\pi + 5\pi/6)} = e^{i5\pi/6} = \cos\left(\frac{5\pi}{6}\right) + i\sin\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} + \frac{i}{2}$.Also,$i^{103} = i^{100} \cdot i^3 = 1 \cdot (-i) = -i$.Thus,$z^{101} + i^{103} = -\frac{\sqrt{3}}{2} + \frac{i}{2} - i = -\frac{\sqrt{3}}{2} - \frac{i}{2} = -\left(\frac{\sqrt{3}+i}{2}\right) = -z$.Therefore,$\left(z^{101} + i^{103}\right)^{105} = (-z)^{105} = -z^{105}$.Since $z = e^{i\pi/6}$,$z^{105} = e^{i(105\pi/6)} = e^{i(17\pi + 3\pi/6)} = e^{i(17\pi + \pi/2)} = e^{i17\pi} \cdot e^{i\pi/2} = (-1) \cdot i = -i$.Wait,let us re-evaluate: $-z^{105} = -\left(e^{i\pi/6}\right)^{105} = -e^{i(17\pi + \pi/2)} = -(-1 \cdot i) = i$.Actually,$z^3 = e^{i(3\pi/6)} = e^{i\pi/2} = i$. Thus,$-z^{105} = -(-z^3) = z^3$ is incorrect. Let us re-calculate: $(-z)^{105} = -z^{105} = -(e^{i\pi/6})^{105} = -e^{i(17\pi + \pi/2)} = -(-1 \cdot i) = i$. Since $z^3 = i$,the answer is $z^3$.

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

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