JEE Main 2026 — 23 January, Evening Shift — Question 17

Given $z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$. By De Moivre's theorem, $z^{201} = \cos\frac{201\pi}{6} + i\sin\frac{201\pi}{6} = \cos\frac{67\pi}{2} + i\sin\frac{67\pi}{2}$. Since $\frac{67\pi}{2} = 33\pi + \frac{\pi}{2}$, we have $\cos\left(33\pi + \frac{\pi}{2}\right) = 0$ and $\sin\left(33\pi + \frac{\pi}{2}\right) = -1$. Thus $z^{201} = -i$. Then $z^{201} - i = -i - i = -2i$. Hence $(z^{201} - i)^8 = (-2i)^8 = (-2)^8 \cdot i^8 = 256 \cdot (i^4)^2 = 256 \cdot 1^2 = 256$. Therefore, the answer is 256, which corresponds to option D.