Complex NumbersHard
Question
If $z = \frac{\sqrt{3}}{2} + \frac{i}{2},i = \sqrt{- 1}$, then $\left( z^{201} - i \right)^{8}$ is equal to
Options
A.-1
B.0
C.1
D.256
Solution
$z = cos\frac{\pi}{6} + isin\frac{\pi}{6}$
$$\begin{matrix} & z^{201} = cos\left( 201\frac{\pi}{6} \right) + isin\left( 201\frac{\pi}{6} \right) = - i \\ & \left( z^{201} - i \right)^{8} = ( - 2i)^{8} = 256 \end{matrix}$$
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