Complex Numbers Part 2

Concept:A complex number in polar form can be expressed as $r e^{i\theta}$ using its modulus and argument. De Moivre's theorem then simplifies powers and trigonometric calculations.Explanation:First, find the modulus and argument of $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$.Modulus: $|z| = \sqrt{\left(\frac{\sqrt{3}}{2}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1$.Argument: $\theta = \tan^{-1}\left(\frac{1/2}{\sqrt{3}/2}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$ (since $z$ lies in the first quadrant).Thus, $z = e^{i\pi/6}$.Now, $z^{201} = \left(e^{i\pi/6}\right)^{201} = e^{i\frac{201\pi}{6}} = e^{i\frac{67\pi}{2}}$.Using Euler's formula: $e^{i\frac{67\pi}{2}} = \cos\frac{67\pi}{2} + i\sin\frac{67\pi}{2}$.Since $\frac{67\pi}{2} = 33\pi + \frac{\pi}{2}$, we have $\cos\frac{67\pi}{2} = 0$ and $\sin\frac{67\pi}{2} = -1$.Therefore, $z^{201} = 0 + i(-1) = -i$.Then, $z^{201} - i = -i - i = -2i$.So, $(z^{201} - i)^8 = (-2i)^8 = (-2)^8 \cdot i^8 = 256 \cdot 1 = 256$.Here $i^8 = (i^4)^2 = 1^2 = 1$.Answer:$256$, which corresponds to option D.