Question

If \( z = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2} \), \( i = \sqrt{-1} \), then \[ \left(z^{201} - i\right)^8 \] is equal to

• A. 0
• B. 256
• C. $-1$
• D. 1

Hint:

Powers of $i$: $i^1=i, i^2=-1, i^3=-i, i^4=1$.

Answer 1

The Correct Option is B

$z = \cos 30^\circ + i \sin 30^\circ = e^{i\pi/6}$.
$z^{201} = e^{i 201\pi/6} = e^{i 67\pi/2}$.
$67\pi/2 = 32\pi + 3\pi/2$.
$z^{201} = \cos(3\pi/2) + i \sin(3\pi/2) = -i$.
$(z^{201} - i)^8 = (-i - i)^8 = (-2i)^8$.
$= (-2)^8 i^8 = 256 \times 1 = 256$.