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If $\mathrm{z}=\frac{\sqrt{3}}{2}+\frac{\mathrm{i}}{2}, \mathrm{i}=\sqrt{-1}$, then $\left(\mathrm{z}^{201}-\mathrm{i}\right)^{8}$ is equal to [JEE MAIN 23 Jan 2026 S II]
(A) -1
(B) 0
(C) 1
(D) 256

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Solution

$\mathrm{z}=\frac{\sqrt{3}+\mathrm{i}}{2}=\mathrm{e}^{\mathrm{i} \pi / 6} \Rightarrow \mathrm{z}^{201}=\left(\mathrm{e}^{\mathrm{i} \pi / 6}\right)^{201}=\mathrm{e}^{\mathrm{i} \frac{201 \pi}{6}}=\mathrm{e}^{\mathrm{i} \frac{67}{2} \pi}$

\[ \Rightarrow \mathrm{z}^{201}=\mathrm{e}^{\mathrm{i}\left(32 \pi+\frac{3}{2} \pi\right)}=\mathrm{e}^{\mathrm{i} \frac{3 \pi}{2}}=-\mathrm{i} \text { Now }\left(\mathrm{z}^{201}-\mathrm{i}\right)^{8}=(-\mathrm{i}-\mathrm{i})^{8}=(-2 \mathrm{i})^{8}=(-1)^{8} 2^{8} \mathrm{i}^{8}=2^{8}=256 \]

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Answer

D

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