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If $$z = \frac{\sqrt{3}}{2}+\frac{i}{2},i=\sqrt{-1},\text{ then }(z^{201}-i)^{8}\text{ is equal to }$$
$$z = \frac{\sqrt{3}}{2} + \frac{i}{2} = \cos 30° + i\sin 30° = e^{i\pi/6}$$
$$z^{201} = e^{i \cdot 201\pi/6} = e^{i \cdot 33.5\pi} = e^{i(33\pi + \pi/2)} = e^{i\pi/2} \cdot (e^{i\pi})^{33} = i \cdot (-1)^{33} = -i$$
$$(z^{201} - i)^8 = (-i - i)^8 = (-2i)^8 = 2^8 \cdot i^8 = 256 \cdot 1 = 256$$
The answer is Option 3: 256.
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